how can I calculate the probability to get triples or better when throwing n 6-sided dice?

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I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.

I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.

Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...

I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).

Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.

probability combinatorics probability-distributions

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asked Jul 31 at 13:16

user1747281

112

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  What does 'triples or better' mean precisely?
  – Stefan Mesken
  Jul 31 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many dice participate in one roll?
  – Vasya
  Jul 31 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StefanMesken Triples or better is 3 or more of a kind
  – user1747281
  Jul 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
  – user1747281
  Jul 31 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
  – MJD
  Jul 31 at 14:16
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

favorite

I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.

I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.

Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...

I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).

Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.

probability combinatorics probability-distributions

share|cite|improve this question

asked Jul 31 at 13:16

user1747281

112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does 'triples or better' mean precisely?
  – Stefan Mesken
  Jul 31 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many dice participate in one roll?
  – Vasya
  Jul 31 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StefanMesken Triples or better is 3 or more of a kind
  – user1747281
  Jul 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
  – user1747281
  Jul 31 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
  – MJD
  Jul 31 at 14:16
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.

I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.

Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...

I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).

Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.

probability combinatorics probability-distributions

share|cite|improve this question

asked Jul 31 at 13:16

user1747281

112

I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.

I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.

Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...

I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).

Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.

probability combinatorics probability-distributions

share|cite|improve this question

asked Jul 31 at 13:16

user1747281

112

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 13:16

user1747281

112

asked Jul 31 at 13:16

user1747281

112

asked Jul 31 at 13:16

user1747281

112

112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does 'triples or better' mean precisely?
  – Stefan Mesken
  Jul 31 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many dice participate in one roll?
  – Vasya
  Jul 31 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StefanMesken Triples or better is 3 or more of a kind
  – user1747281
  Jul 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
  – user1747281
  Jul 31 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
  – MJD
  Jul 31 at 14:16
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does 'triples or better' mean precisely?
  – Stefan Mesken
  Jul 31 at 13:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  How many dice participate in one roll?
  – Vasya
  Jul 31 at 13:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StefanMesken Triples or better is 3 or more of a kind
  – user1747281
  Jul 31 at 13:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
  – user1747281
  Jul 31 at 13:44
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
  – MJD
  Jul 31 at 14:16
  

  
  
  
  

What does 'triples or better' mean precisely?
– Stefan Mesken
Jul 31 at 13:26

What does 'triples or better' mean precisely?
– Stefan Mesken
Jul 31 at 13:26

How many dice participate in one roll?
– Vasya
Jul 31 at 13:28

How many dice participate in one roll?
– Vasya
Jul 31 at 13:28

@StefanMesken Triples or better is 3 or more of a kind
– user1747281
Jul 31 at 13:43

@StefanMesken Triples or better is 3 or more of a kind
– user1747281
Jul 31 at 13:43

@Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
– user1747281
Jul 31 at 13:44

@Vasya the number of dice is variable. I want to make an excel sheet with the probability for each number of dice But it's always going to be between 3 and 12
– user1747281
Jul 31 at 13:44

1

1

There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
– MJD
Jul 31 at 14:16

There is a.form at the bottom of this page that will do the calculation automatically, and the attached article explains how the calculation is done: blog.plover.com/math/yahtzee.html
– MJD
Jul 31 at 14:16

 | 
show 2 more comments

4 Answers
4

active

oldest

votes

up vote
2
down vote



accepted

In what follows I will refer to “pattern” in the dice rolls. For example the pattern AAABC means that three of the dice show the same number and the other two dice are different from the first three and also different from each other. The roll $1 2 2 4 2$ has this pattern, but $1 2 2 1 2$ does not; that has the pattern AAABB, and similarly $2 2 2 2 1$ is not patternAAABC but AAAAB.

There are seven patterns possible with five dice:

AAAAA
AAAAB
AAABB
AAABC

AABBC
AABCD
ABCDE

If you want three of a kind “or better” on five dice, you are interested in the sum of the first four patterns and you want to disregard the other three. What follows is an explanation of how to compute the probability for each pattern.

We will represent the patterns numerically like this: AAABC will be $(2, 0, 1)$ because there are two letters that appear once, no letters that appear twice, and one letter that appears three times. AAABB will be $(0, 1, 1)$ because there are no letters that appear once, one that appears twice, and one that appears three times. AABCD is $(3, 1)$ (we omit the trailing zero) and ABCDE is $(5)$. We'll refer to these numbers as $n_1, n_2, ldots$. For AAABC, we have $n_1 = 2, n_2 = 0, n_3 = 1$. For five dice, the representations of the seven patterns are:

$$beginarraylrrrrr
& n_1 & n_2 & n_3 & n_4 & n_5 \
AAAAA & 0 & 0 & 0 & 0 & 1 \
AAAAB & 1 & 0 & 0 & 1 \
AAABB & 0 & 1 & 1 \
AAABC & 2 & 0 & 1 \
AABBC & 1 & 2 \
AABCD & 3 & 1 \
ABCDE & 5
endarray
$$

Now suppose we're rolling $N$ dice each with $d$ sides. If there are $N$ dice, we should always have $sum icdot n_i = N$. We'll also take $k=sum n_i$; this is just the number of different letters in the pattern.

Then it transpires that the number of ways of rolling any pattern is:

$$
colormaroondchoose kcolordarkbluek!
colordarkgreenN!over colorpurpleprod i!^n_in_i!
$$

To get the probability, just divide by $d^N$.

I'll work through one example to demonstrate the formula. How many ways are there to roll the pattern AAABC, which is three of a kind, but not counting full house, for of a kind, or five of a kind. For AAABC we have $n_1=2, n_2=0, n_3 = 1$, so $k=n_1+n_2+n_3 = 3$, and the formula gives:

$$colormaroon6choose 3colordarkblue3!
colordarkgreen5!over colorpurple(1!^2cdot2!)(2!^0cdot0!)(3!^1cdot1!) =
colormaroon20cdot colordarkblue6cdot
colordarkgreen120over colorpurple2cdot1cdot6 = mathbf1200
$$

There are 1200 ways to roll the pattern AAABC, so the probability is $frac12006^5 approx 15.43%$. Similar calculations for the other three patterns of interest give:

$$beginarraylrrl
A A A B C & 1200 & 15.43 & % \
A A A B B & 300 & 3.86 \
A A A A B & 150 & 1.93 \
A A A A A & 6 & 0.08 \ hline
textTotal & 1656 & 21.30 & %
endarray
$$

So you can expect to get three of a kind or better around one time in five.

(By far the most common pattern is the single pair AABCD, which occurs almost half the time.)

I explained the formula in more detail in a blog post. This page tabulates the probabilities for every pattern of up to 12 dice. You program that generated these tables is available online: the URL

 https://perl.plover.com/misc/enumeration/tabulate-dice.cgi?N=7&S=11

generates a table for seven dice each with 11 sides. You can adjust the 7 and 11 to suit yourself. Related materials, including program source code, are available also.

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edited Aug 1 at 5:33

community wiki

3 revs
MJD

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  Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
  – BruceET
  Aug 1 at 5:06
  
  

  
  
  
  


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  Is there a nice way to calculate $n_i$s ?
  – Jaroslaw Matlak
  Aug 1 at 8:42
  

  
  
  
  


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  @jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
  – MJD
  Aug 1 at 12:34
  

  
  
  
  


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  I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
  – Jaroslaw Matlak
  Aug 1 at 12:45
  

  
  
  
  


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  Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
  – MJD
  Aug 1 at 14:05
  
  

  
  
  
  

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show 6 more comments

up vote
0
down vote

If I understand your problem correctly, you're looking for a generalization of the binomial distribution to $n > 2$ where $n$ is the number of possible classes. The term for this is a multinomial distribution.

We can view a $n = 2$ side analog of a dice as a coin. The probability of a single outcome is a Bernoulli distribution (e.g. $Pr(H)$). The probability of multiple outcomes is a Binomial distribution (e.g. $Pr(HT)$).

For $n > 2$, the probability of a single outcome is a Categorical distribution (e.g. $Pr(1)$). The probability of multiple outcomes is a Multinomial distribution (e.g. $Pr(1126)$).

Remember that a multinomial distribution will only tell you the distribution of, say 5 of the rolls being $1$. So if you're looking for pairs or triplets regardless of the class -- whether its 1 or 2, you don't care -- you have to multiply your probability (if its a fair dice) or construct a sum in the case the dice is not fair.

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answered Aug 1 at 2:36

Sentient

439516

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up vote
0
down vote

Hint

You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = binom61\
f_2 = binom622!+binom61\
f_3=binom633!+binom61binom51frac3!2!\
f_4=binom644!+binom61binom52frac4!2!+binom62frac4!2!2!\
f_5=binom655!+binom61binom53frac5!2!+binom62binom41frac5!2!2!\
f_6=binom666!+binom61binom54frac6!2!+binom62binom42frac6!2!2!+binom63frac6!2!2!2!\
f_7=binom61frac7!2!+binom62binom43frac7!2!2!+binom63binom31frac7!2!2!2!\
f_8=binom62frac8!2!2!+binom63binom32frac8!2!2!2!+binom64frac8!2!2!2!2!\
f_9=binom63frac9!2!2!2!+binom64binom21frac9!2!2!2!2!\
f_10 = binom64frac10!2!2!2!2!+binom65frac10!2!2!2!2!2!\
f_11=binom65frac11!2!2!2!2!2!\
f_12=binom66frac12!2!2!2!2!2!2!$$
For $k>13$ we have $f_k=0$

The number of all possible throws $omega_n$ is:
$$omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=fracf_nomega_n$$
and probability of getting triple:
$$q_n=1-p_n$$

share|cite|improve this answer

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

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  Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
  – user1747281
  Jul 31 at 16:41
  

  
  
  
  


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  Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
  – Jaroslaw Matlak
  Jul 31 at 21:36
  

  
  
  
  


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  What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
  – BruceET
  Aug 1 at 5:24
  
  

  
  
  
  


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  @BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
  – Jaroslaw Matlak
  Aug 1 at 8:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

The OP is correct in saying that this is a generalization of the Birthday Problem and that it is harder because we are interested in cases where the number of people with the same birthday is greater than two. One approach is by way of exponential generating functions.

Let's take a concrete example. Suppose we want to roll $10$ six-sided dice and find the probability that at least one number comes up $5$ or more times. It seems easier to consider the complementary event, i.e. all numbers come up $4$ or fewer times. There are $6^10$ possible sequences of die rolls, all of which we assume are equally likely. We would like to count the number of sequences in which no number comes up more than $4$ times. We generalize the problem a bit and think about a sequence of $n$ die rolls, and let the number of sequences in which no number appears more than $4$ times be $a_n$.

The exponential generating function of $a_n$ is defined to be
$$f(x) = sum_n=0^infty frac1n! a_n x^n$$
Since each die has $6$ sides and each number appears no more than $4$ times,
$$f(x) = left( sum_i=0^4 frac1i! x^i right)^6$$
The number we want in our problem, $a_10$, is $10!$ times the coefficient of $x^10$ when $f(x)$ is expanded. I suppose a pencil and paper solution is possible, but I took the easy way out and used a computer algebra system to find that the coefficient of $x^10$ is $2177 / 144$. Therefore the number of sequences of $10$ die rolls in which no number appears more than $4$ times is $$a_10 = 10! times frac2177 144 = 54,860,400$$

So the probability that no number appears more than $4$ times in $10$ rolls is
$$p = fraca_106^10 = 0.907291$$
and the probability that at least one number appears $5$ times or more is
$$1-p = 0.0927093$$

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answered Aug 1 at 18:00

awkward

5,05611021

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

In what follows I will refer to “pattern” in the dice rolls. For example the pattern AAABC means that three of the dice show the same number and the other two dice are different from the first three and also different from each other. The roll $1 2 2 4 2$ has this pattern, but $1 2 2 1 2$ does not; that has the pattern AAABB, and similarly $2 2 2 2 1$ is not patternAAABC but AAAAB.

There are seven patterns possible with five dice:

AAAAA
AAAAB
AAABB
AAABC

AABBC
AABCD
ABCDE

If you want three of a kind “or better” on five dice, you are interested in the sum of the first four patterns and you want to disregard the other three. What follows is an explanation of how to compute the probability for each pattern.

We will represent the patterns numerically like this: AAABC will be $(2, 0, 1)$ because there are two letters that appear once, no letters that appear twice, and one letter that appears three times. AAABB will be $(0, 1, 1)$ because there are no letters that appear once, one that appears twice, and one that appears three times. AABCD is $(3, 1)$ (we omit the trailing zero) and ABCDE is $(5)$. We'll refer to these numbers as $n_1, n_2, ldots$. For AAABC, we have $n_1 = 2, n_2 = 0, n_3 = 1$. For five dice, the representations of the seven patterns are:

$$beginarraylrrrrr
& n_1 & n_2 & n_3 & n_4 & n_5 \
AAAAA & 0 & 0 & 0 & 0 & 1 \
AAAAB & 1 & 0 & 0 & 1 \
AAABB & 0 & 1 & 1 \
AAABC & 2 & 0 & 1 \
AABBC & 1 & 2 \
AABCD & 3 & 1 \
ABCDE & 5
endarray
$$

Now suppose we're rolling $N$ dice each with $d$ sides. If there are $N$ dice, we should always have $sum icdot n_i = N$. We'll also take $k=sum n_i$; this is just the number of different letters in the pattern.

Then it transpires that the number of ways of rolling any pattern is:

$$
colormaroondchoose kcolordarkbluek!
colordarkgreenN!over colorpurpleprod i!^n_in_i!
$$

To get the probability, just divide by $d^N$.

I'll work through one example to demonstrate the formula. How many ways are there to roll the pattern AAABC, which is three of a kind, but not counting full house, for of a kind, or five of a kind. For AAABC we have $n_1=2, n_2=0, n_3 = 1$, so $k=n_1+n_2+n_3 = 3$, and the formula gives:

$$colormaroon6choose 3colordarkblue3!
colordarkgreen5!over colorpurple(1!^2cdot2!)(2!^0cdot0!)(3!^1cdot1!) =
colormaroon20cdot colordarkblue6cdot
colordarkgreen120over colorpurple2cdot1cdot6 = mathbf1200
$$

There are 1200 ways to roll the pattern AAABC, so the probability is $frac12006^5 approx 15.43%$. Similar calculations for the other three patterns of interest give:

$$beginarraylrrl
A A A B C & 1200 & 15.43 & % \
A A A B B & 300 & 3.86 \
A A A A B & 150 & 1.93 \
A A A A A & 6 & 0.08 \ hline
textTotal & 1656 & 21.30 & %
endarray
$$

So you can expect to get three of a kind or better around one time in five.

(By far the most common pattern is the single pair AABCD, which occurs almost half the time.)

I explained the formula in more detail in a blog post. This page tabulates the probabilities for every pattern of up to 12 dice. You program that generated these tables is available online: the URL

 https://perl.plover.com/misc/enumeration/tabulate-dice.cgi?N=7&S=11

generates a table for seven dice each with 11 sides. You can adjust the 7 and 11 to suit yourself. Related materials, including program source code, are available also.

share|cite|improve this answer

edited Aug 1 at 5:33

community wiki

3 revs
MJD

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
  – BruceET
  Aug 1 at 5:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there a nice way to calculate $n_i$s ?
  – Jaroslaw Matlak
  Aug 1 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
  – MJD
  Aug 1 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
  – Jaroslaw Matlak
  Aug 1 at 12:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
  – MJD
  Aug 1 at 14:05
  
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote



accepted

In what follows I will refer to “pattern” in the dice rolls. For example the pattern AAABC means that three of the dice show the same number and the other two dice are different from the first three and also different from each other. The roll $1 2 2 4 2$ has this pattern, but $1 2 2 1 2$ does not; that has the pattern AAABB, and similarly $2 2 2 2 1$ is not patternAAABC but AAAAB.

There are seven patterns possible with five dice:

AAAAA
AAAAB
AAABB
AAABC

AABBC
AABCD
ABCDE

If you want three of a kind “or better” on five dice, you are interested in the sum of the first four patterns and you want to disregard the other three. What follows is an explanation of how to compute the probability for each pattern.

We will represent the patterns numerically like this: AAABC will be $(2, 0, 1)$ because there are two letters that appear once, no letters that appear twice, and one letter that appears three times. AAABB will be $(0, 1, 1)$ because there are no letters that appear once, one that appears twice, and one that appears three times. AABCD is $(3, 1)$ (we omit the trailing zero) and ABCDE is $(5)$. We'll refer to these numbers as $n_1, n_2, ldots$. For AAABC, we have $n_1 = 2, n_2 = 0, n_3 = 1$. For five dice, the representations of the seven patterns are:

$$beginarraylrrrrr
& n_1 & n_2 & n_3 & n_4 & n_5 \
AAAAA & 0 & 0 & 0 & 0 & 1 \
AAAAB & 1 & 0 & 0 & 1 \
AAABB & 0 & 1 & 1 \
AAABC & 2 & 0 & 1 \
AABBC & 1 & 2 \
AABCD & 3 & 1 \
ABCDE & 5
endarray
$$

Now suppose we're rolling $N$ dice each with $d$ sides. If there are $N$ dice, we should always have $sum icdot n_i = N$. We'll also take $k=sum n_i$; this is just the number of different letters in the pattern.

Then it transpires that the number of ways of rolling any pattern is:

$$
colormaroondchoose kcolordarkbluek!
colordarkgreenN!over colorpurpleprod i!^n_in_i!
$$

To get the probability, just divide by $d^N$.

I'll work through one example to demonstrate the formula. How many ways are there to roll the pattern AAABC, which is three of a kind, but not counting full house, for of a kind, or five of a kind. For AAABC we have $n_1=2, n_2=0, n_3 = 1$, so $k=n_1+n_2+n_3 = 3$, and the formula gives:

$$colormaroon6choose 3colordarkblue3!
colordarkgreen5!over colorpurple(1!^2cdot2!)(2!^0cdot0!)(3!^1cdot1!) =
colormaroon20cdot colordarkblue6cdot
colordarkgreen120over colorpurple2cdot1cdot6 = mathbf1200
$$

There are 1200 ways to roll the pattern AAABC, so the probability is $frac12006^5 approx 15.43%$. Similar calculations for the other three patterns of interest give:

$$beginarraylrrl
A A A B C & 1200 & 15.43 & % \
A A A B B & 300 & 3.86 \
A A A A B & 150 & 1.93 \
A A A A A & 6 & 0.08 \ hline
textTotal & 1656 & 21.30 & %
endarray
$$

So you can expect to get three of a kind or better around one time in five.

(By far the most common pattern is the single pair AABCD, which occurs almost half the time.)

I explained the formula in more detail in a blog post. This page tabulates the probabilities for every pattern of up to 12 dice. You program that generated these tables is available online: the URL

 https://perl.plover.com/misc/enumeration/tabulate-dice.cgi?N=7&S=11

generates a table for seven dice each with 11 sides. You can adjust the 7 and 11 to suit yourself. Related materials, including program source code, are available also.

share|cite|improve this answer

edited Aug 1 at 5:33

community wiki

3 revs
MJD

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
  – BruceET
  Aug 1 at 5:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there a nice way to calculate $n_i$s ?
  – Jaroslaw Matlak
  Aug 1 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
  – MJD
  Aug 1 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
  – Jaroslaw Matlak
  Aug 1 at 12:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
  – MJD
  Aug 1 at 14:05
  
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

In what follows I will refer to “pattern” in the dice rolls. For example the pattern AAABC means that three of the dice show the same number and the other two dice are different from the first three and also different from each other. The roll $1 2 2 4 2$ has this pattern, but $1 2 2 1 2$ does not; that has the pattern AAABB, and similarly $2 2 2 2 1$ is not patternAAABC but AAAAB.

There are seven patterns possible with five dice:

AAAAA
AAAAB
AAABB
AAABC

AABBC
AABCD
ABCDE

If you want three of a kind “or better” on five dice, you are interested in the sum of the first four patterns and you want to disregard the other three. What follows is an explanation of how to compute the probability for each pattern.

We will represent the patterns numerically like this: AAABC will be $(2, 0, 1)$ because there are two letters that appear once, no letters that appear twice, and one letter that appears three times. AAABB will be $(0, 1, 1)$ because there are no letters that appear once, one that appears twice, and one that appears three times. AABCD is $(3, 1)$ (we omit the trailing zero) and ABCDE is $(5)$. We'll refer to these numbers as $n_1, n_2, ldots$. For AAABC, we have $n_1 = 2, n_2 = 0, n_3 = 1$. For five dice, the representations of the seven patterns are:

$$beginarraylrrrrr
& n_1 & n_2 & n_3 & n_4 & n_5 \
AAAAA & 0 & 0 & 0 & 0 & 1 \
AAAAB & 1 & 0 & 0 & 1 \
AAABB & 0 & 1 & 1 \
AAABC & 2 & 0 & 1 \
AABBC & 1 & 2 \
AABCD & 3 & 1 \
ABCDE & 5
endarray
$$

Now suppose we're rolling $N$ dice each with $d$ sides. If there are $N$ dice, we should always have $sum icdot n_i = N$. We'll also take $k=sum n_i$; this is just the number of different letters in the pattern.

Then it transpires that the number of ways of rolling any pattern is:

$$
colormaroondchoose kcolordarkbluek!
colordarkgreenN!over colorpurpleprod i!^n_in_i!
$$

To get the probability, just divide by $d^N$.

I'll work through one example to demonstrate the formula. How many ways are there to roll the pattern AAABC, which is three of a kind, but not counting full house, for of a kind, or five of a kind. For AAABC we have $n_1=2, n_2=0, n_3 = 1$, so $k=n_1+n_2+n_3 = 3$, and the formula gives:

$$colormaroon6choose 3colordarkblue3!
colordarkgreen5!over colorpurple(1!^2cdot2!)(2!^0cdot0!)(3!^1cdot1!) =
colormaroon20cdot colordarkblue6cdot
colordarkgreen120over colorpurple2cdot1cdot6 = mathbf1200
$$

There are 1200 ways to roll the pattern AAABC, so the probability is $frac12006^5 approx 15.43%$. Similar calculations for the other three patterns of interest give:

$$beginarraylrrl
A A A B C & 1200 & 15.43 & % \
A A A B B & 300 & 3.86 \
A A A A B & 150 & 1.93 \
A A A A A & 6 & 0.08 \ hline
textTotal & 1656 & 21.30 & %
endarray
$$

So you can expect to get three of a kind or better around one time in five.

(By far the most common pattern is the single pair AABCD, which occurs almost half the time.)

I explained the formula in more detail in a blog post. This page tabulates the probabilities for every pattern of up to 12 dice. You program that generated these tables is available online: the URL

 https://perl.plover.com/misc/enumeration/tabulate-dice.cgi?N=7&S=11

generates a table for seven dice each with 11 sides. You can adjust the 7 and 11 to suit yourself. Related materials, including program source code, are available also.

share|cite|improve this answer

edited Aug 1 at 5:33

community wiki

3 revs
MJD

In what follows I will refer to “pattern” in the dice rolls. For example the pattern AAABC means that three of the dice show the same number and the other two dice are different from the first three and also different from each other. The roll $1 2 2 4 2$ has this pattern, but $1 2 2 1 2$ does not; that has the pattern AAABB, and similarly $2 2 2 2 1$ is not patternAAABC but AAAAB.

There are seven patterns possible with five dice:

AAAAA
AAAAB
AAABB
AAABC

AABBC
AABCD
ABCDE

If you want three of a kind “or better” on five dice, you are interested in the sum of the first four patterns and you want to disregard the other three. What follows is an explanation of how to compute the probability for each pattern.

We will represent the patterns numerically like this: AAABC will be $(2, 0, 1)$ because there are two letters that appear once, no letters that appear twice, and one letter that appears three times. AAABB will be $(0, 1, 1)$ because there are no letters that appear once, one that appears twice, and one that appears three times. AABCD is $(3, 1)$ (we omit the trailing zero) and ABCDE is $(5)$. We'll refer to these numbers as $n_1, n_2, ldots$. For AAABC, we have $n_1 = 2, n_2 = 0, n_3 = 1$. For five dice, the representations of the seven patterns are:

$$beginarraylrrrrr
& n_1 & n_2 & n_3 & n_4 & n_5 \
AAAAA & 0 & 0 & 0 & 0 & 1 \
AAAAB & 1 & 0 & 0 & 1 \
AAABB & 0 & 1 & 1 \
AAABC & 2 & 0 & 1 \
AABBC & 1 & 2 \
AABCD & 3 & 1 \
ABCDE & 5
endarray
$$

Now suppose we're rolling $N$ dice each with $d$ sides. If there are $N$ dice, we should always have $sum icdot n_i = N$. We'll also take $k=sum n_i$; this is just the number of different letters in the pattern.

Then it transpires that the number of ways of rolling any pattern is:

$$
colormaroondchoose kcolordarkbluek!
colordarkgreenN!over colorpurpleprod i!^n_in_i!
$$

To get the probability, just divide by $d^N$.

I'll work through one example to demonstrate the formula. How many ways are there to roll the pattern AAABC, which is three of a kind, but not counting full house, for of a kind, or five of a kind. For AAABC we have $n_1=2, n_2=0, n_3 = 1$, so $k=n_1+n_2+n_3 = 3$, and the formula gives:

$$colormaroon6choose 3colordarkblue3!
colordarkgreen5!over colorpurple(1!^2cdot2!)(2!^0cdot0!)(3!^1cdot1!) =
colormaroon20cdot colordarkblue6cdot
colordarkgreen120over colorpurple2cdot1cdot6 = mathbf1200
$$

There are 1200 ways to roll the pattern AAABC, so the probability is $frac12006^5 approx 15.43%$. Similar calculations for the other three patterns of interest give:

$$beginarraylrrl
A A A B C & 1200 & 15.43 & % \
A A A B B & 300 & 3.86 \
A A A A B & 150 & 1.93 \
A A A A A & 6 & 0.08 \ hline
textTotal & 1656 & 21.30 & %
endarray
$$

So you can expect to get three of a kind or better around one time in five.

(By far the most common pattern is the single pair AABCD, which occurs almost half the time.)

I explained the formula in more detail in a blog post. This page tabulates the probabilities for every pattern of up to 12 dice. You program that generated these tables is available online: the URL

 https://perl.plover.com/misc/enumeration/tabulate-dice.cgi?N=7&S=11

generates a table for seven dice each with 11 sides. You can adjust the 7 and 11 to suit yourself. Related materials, including program source code, are available also.

share|cite|improve this answer

edited Aug 1 at 5:33

community wiki

3 revs
MJD

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 5:33

edited Aug 1 at 5:33

edited Aug 1 at 5:33

community wiki

3 revs
MJD

community wiki

3 revs
MJD

community wiki

3 revs
MJD

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
  – BruceET
  Aug 1 at 5:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there a nice way to calculate $n_i$s ?
  – Jaroslaw Matlak
  Aug 1 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
  – MJD
  Aug 1 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
  – Jaroslaw Matlak
  Aug 1 at 12:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
  – MJD
  Aug 1 at 14:05
  
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
  – BruceET
  Aug 1 at 5:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is there a nice way to calculate $n_i$s ?
  – Jaroslaw Matlak
  Aug 1 at 8:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
  – MJD
  Aug 1 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
  – Jaroslaw Matlak
  Aug 1 at 12:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
  – MJD
  Aug 1 at 14:05
  
  

  
  
  
  

Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
– BruceET
Aug 1 at 5:06

Confirmed by simulation (+1): If you sort the outcomes from 5 dice, you're asking for the longest run length to be at least three. In R, the function rle (for 'run length encoding') does most of the work: three lines of code: set.seed(731); m=10^5; die = 1:6, x=replicate(m,max(rle(sort(sample(die,5,rep=T)))$lengths)), and mean(x >= 3) return 0.21219, which matches your 0.213 within the margin of simulation error.
– BruceET
Aug 1 at 5:06

Is there a nice way to calculate $n_i$s ?
– Jaroslaw Matlak
Aug 1 at 8:42

Is there a nice way to calculate $n_i$s ?
– Jaroslaw Matlak
Aug 1 at 8:42

@jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
– MJD
Aug 1 at 12:34

@jaroslaw count the number of occurrences of each letter and store the counts in an array. Then count the occurrences of each value in the array and store those counts in a second array. The second array is $n$.
– MJD
Aug 1 at 12:34

I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
– Jaroslaw Matlak
Aug 1 at 12:45

I see, but is there a simple formula? Because for large $N$ (10,11,12) this array might be very big.
– Jaroslaw Matlak
Aug 1 at 12:45

1

1

Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
– MJD
Aug 1 at 14:05

Note that even for large $N$, you do not have to calculate all the patterns and store them. It is not hard to calculate them one at a time, each one from the previous, so that only one is in memory at any time. The Higher-Order Perl link above explains how to do this. Also, for $n=12, d=6$ there are 58 patterns, not 37.
– MJD
Aug 1 at 14:05

 | 
show 6 more comments

up vote
0
down vote

If I understand your problem correctly, you're looking for a generalization of the binomial distribution to $n > 2$ where $n$ is the number of possible classes. The term for this is a multinomial distribution.

We can view a $n = 2$ side analog of a dice as a coin. The probability of a single outcome is a Bernoulli distribution (e.g. $Pr(H)$). The probability of multiple outcomes is a Binomial distribution (e.g. $Pr(HT)$).

For $n > 2$, the probability of a single outcome is a Categorical distribution (e.g. $Pr(1)$). The probability of multiple outcomes is a Multinomial distribution (e.g. $Pr(1126)$).

Remember that a multinomial distribution will only tell you the distribution of, say 5 of the rolls being $1$. So if you're looking for pairs or triplets regardless of the class -- whether its 1 or 2, you don't care -- you have to multiply your probability (if its a fair dice) or construct a sum in the case the dice is not fair.

share|cite|improve this answer

answered Aug 1 at 2:36

Sentient

439516

add a comment | 

up vote
0
down vote

If I understand your problem correctly, you're looking for a generalization of the binomial distribution to $n > 2$ where $n$ is the number of possible classes. The term for this is a multinomial distribution.

We can view a $n = 2$ side analog of a dice as a coin. The probability of a single outcome is a Bernoulli distribution (e.g. $Pr(H)$). The probability of multiple outcomes is a Binomial distribution (e.g. $Pr(HT)$).

For $n > 2$, the probability of a single outcome is a Categorical distribution (e.g. $Pr(1)$). The probability of multiple outcomes is a Multinomial distribution (e.g. $Pr(1126)$).

Remember that a multinomial distribution will only tell you the distribution of, say 5 of the rolls being $1$. So if you're looking for pairs or triplets regardless of the class -- whether its 1 or 2, you don't care -- you have to multiply your probability (if its a fair dice) or construct a sum in the case the dice is not fair.

share|cite|improve this answer

answered Aug 1 at 2:36

Sentient

439516

add a comment | 

up vote
0
down vote

up vote
0
down vote

If I understand your problem correctly, you're looking for a generalization of the binomial distribution to $n > 2$ where $n$ is the number of possible classes. The term for this is a multinomial distribution.

We can view a $n = 2$ side analog of a dice as a coin. The probability of a single outcome is a Bernoulli distribution (e.g. $Pr(H)$). The probability of multiple outcomes is a Binomial distribution (e.g. $Pr(HT)$).

For $n > 2$, the probability of a single outcome is a Categorical distribution (e.g. $Pr(1)$). The probability of multiple outcomes is a Multinomial distribution (e.g. $Pr(1126)$).

Remember that a multinomial distribution will only tell you the distribution of, say 5 of the rolls being $1$. So if you're looking for pairs or triplets regardless of the class -- whether its 1 or 2, you don't care -- you have to multiply your probability (if its a fair dice) or construct a sum in the case the dice is not fair.

share|cite|improve this answer

answered Aug 1 at 2:36

Sentient

439516

If I understand your problem correctly, you're looking for a generalization of the binomial distribution to $n > 2$ where $n$ is the number of possible classes. The term for this is a multinomial distribution.

We can view a $n = 2$ side analog of a dice as a coin. The probability of a single outcome is a Bernoulli distribution (e.g. $Pr(H)$). The probability of multiple outcomes is a Binomial distribution (e.g. $Pr(HT)$).

For $n > 2$, the probability of a single outcome is a Categorical distribution (e.g. $Pr(1)$). The probability of multiple outcomes is a Multinomial distribution (e.g. $Pr(1126)$).

Remember that a multinomial distribution will only tell you the distribution of, say 5 of the rolls being $1$. So if you're looking for pairs or triplets regardless of the class -- whether its 1 or 2, you don't care -- you have to multiply your probability (if its a fair dice) or construct a sum in the case the dice is not fair.

share|cite|improve this answer

answered Aug 1 at 2:36

Sentient

439516

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answered Aug 1 at 2:36

Sentient

439516

answered Aug 1 at 2:36

Sentient

439516

answered Aug 1 at 2:36

Sentient

439516

439516

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add a comment | 

up vote
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down vote

Hint

You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = binom61\
f_2 = binom622!+binom61\
f_3=binom633!+binom61binom51frac3!2!\
f_4=binom644!+binom61binom52frac4!2!+binom62frac4!2!2!\
f_5=binom655!+binom61binom53frac5!2!+binom62binom41frac5!2!2!\
f_6=binom666!+binom61binom54frac6!2!+binom62binom42frac6!2!2!+binom63frac6!2!2!2!\
f_7=binom61frac7!2!+binom62binom43frac7!2!2!+binom63binom31frac7!2!2!2!\
f_8=binom62frac8!2!2!+binom63binom32frac8!2!2!2!+binom64frac8!2!2!2!2!\
f_9=binom63frac9!2!2!2!+binom64binom21frac9!2!2!2!2!\
f_10 = binom64frac10!2!2!2!2!+binom65frac10!2!2!2!2!2!\
f_11=binom65frac11!2!2!2!2!2!\
f_12=binom66frac12!2!2!2!2!2!2!$$
For $k>13$ we have $f_k=0$

The number of all possible throws $omega_n$ is:
$$omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=fracf_nomega_n$$
and probability of getting triple:
$$q_n=1-p_n$$

share|cite|improve this answer

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
  – user1747281
  Jul 31 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
  – Jaroslaw Matlak
  Jul 31 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
  – BruceET
  Aug 1 at 5:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
  – Jaroslaw Matlak
  Aug 1 at 8:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

Hint

You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = binom61\
f_2 = binom622!+binom61\
f_3=binom633!+binom61binom51frac3!2!\
f_4=binom644!+binom61binom52frac4!2!+binom62frac4!2!2!\
f_5=binom655!+binom61binom53frac5!2!+binom62binom41frac5!2!2!\
f_6=binom666!+binom61binom54frac6!2!+binom62binom42frac6!2!2!+binom63frac6!2!2!2!\
f_7=binom61frac7!2!+binom62binom43frac7!2!2!+binom63binom31frac7!2!2!2!\
f_8=binom62frac8!2!2!+binom63binom32frac8!2!2!2!+binom64frac8!2!2!2!2!\
f_9=binom63frac9!2!2!2!+binom64binom21frac9!2!2!2!2!\
f_10 = binom64frac10!2!2!2!2!+binom65frac10!2!2!2!2!2!\
f_11=binom65frac11!2!2!2!2!2!\
f_12=binom66frac12!2!2!2!2!2!2!$$
For $k>13$ we have $f_k=0$

The number of all possible throws $omega_n$ is:
$$omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=fracf_nomega_n$$
and probability of getting triple:
$$q_n=1-p_n$$

share|cite|improve this answer

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
  – user1747281
  Jul 31 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
  – Jaroslaw Matlak
  Jul 31 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
  – BruceET
  Aug 1 at 5:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
  – Jaroslaw Matlak
  Aug 1 at 8:37
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint

You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = binom61\
f_2 = binom622!+binom61\
f_3=binom633!+binom61binom51frac3!2!\
f_4=binom644!+binom61binom52frac4!2!+binom62frac4!2!2!\
f_5=binom655!+binom61binom53frac5!2!+binom62binom41frac5!2!2!\
f_6=binom666!+binom61binom54frac6!2!+binom62binom42frac6!2!2!+binom63frac6!2!2!2!\
f_7=binom61frac7!2!+binom62binom43frac7!2!2!+binom63binom31frac7!2!2!2!\
f_8=binom62frac8!2!2!+binom63binom32frac8!2!2!2!+binom64frac8!2!2!2!2!\
f_9=binom63frac9!2!2!2!+binom64binom21frac9!2!2!2!2!\
f_10 = binom64frac10!2!2!2!2!+binom65frac10!2!2!2!2!2!\
f_11=binom65frac11!2!2!2!2!2!\
f_12=binom66frac12!2!2!2!2!2!2!$$
For $k>13$ we have $f_k=0$

The number of all possible throws $omega_n$ is:
$$omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=fracf_nomega_n$$
and probability of getting triple:
$$q_n=1-p_n$$

share|cite|improve this answer

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

Hint

You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = binom61\
f_2 = binom622!+binom61\
f_3=binom633!+binom61binom51frac3!2!\
f_4=binom644!+binom61binom52frac4!2!+binom62frac4!2!2!\
f_5=binom655!+binom61binom53frac5!2!+binom62binom41frac5!2!2!\
f_6=binom666!+binom61binom54frac6!2!+binom62binom42frac6!2!2!+binom63frac6!2!2!2!\
f_7=binom61frac7!2!+binom62binom43frac7!2!2!+binom63binom31frac7!2!2!2!\
f_8=binom62frac8!2!2!+binom63binom32frac8!2!2!2!+binom64frac8!2!2!2!2!\
f_9=binom63frac9!2!2!2!+binom64binom21frac9!2!2!2!2!\
f_10 = binom64frac10!2!2!2!2!+binom65frac10!2!2!2!2!2!\
f_11=binom65frac11!2!2!2!2!2!\
f_12=binom66frac12!2!2!2!2!2!2!$$
For $k>13$ we have $f_k=0$

The number of all possible throws $omega_n$ is:
$$omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=fracf_nomega_n$$
and probability of getting triple:
$$q_n=1-p_n$$

share|cite|improve this answer

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 8:33

edited Aug 1 at 8:33

edited Aug 1 at 8:33

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

answered Jul 31 at 14:39

Jaroslaw Matlak

3,870830

3,870830

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
  – user1747281
  Jul 31 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
  – Jaroslaw Matlak
  Jul 31 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
  – BruceET
  Aug 1 at 5:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
  – Jaroslaw Matlak
  Aug 1 at 8:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
  – user1747281
  Jul 31 at 16:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
  – Jaroslaw Matlak
  Jul 31 at 21:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
  – BruceET
  Aug 1 at 5:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
  – Jaroslaw Matlak
  Aug 1 at 8:37
  

  
  
  
  

Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
– user1747281
Jul 31 at 16:41

Thanks a lot for the suggestion. It does allow to solve the "triples" problem but not the "or better" part. Still pretty interesting take on the problem.
– user1747281
Jul 31 at 16:41

Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
– Jaroslaw Matlak
Jul 31 at 21:36

Actually this solution consider "or better" as a triple. $p_n$ denotes the number of events, where there at most pairs.
– Jaroslaw Matlak
Jul 31 at 21:36

What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
– BruceET
Aug 1 at 5:24

What am I getting wrong? If $n=5,$ then $f_5 = 6+60+60=126$ and $omega_n = 252,$ so $q_n = p_n = .5.$ Which is already larger than @MJD's answer. (Seems exactly a triple has probability $approx 0.19.)$
– BruceET
Aug 1 at 5:24

1

1

@BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
– Jaroslaw Matlak
Aug 1 at 8:37

@BruceET You're right - my answer was not complete (I comsidered each situation as one, while there are more possibilities of throwing AAABB for given A and B than AAAAA. I've edited my answer.
– Jaroslaw Matlak
Aug 1 at 8:37

add a comment | 

up vote
0
down vote

The OP is correct in saying that this is a generalization of the Birthday Problem and that it is harder because we are interested in cases where the number of people with the same birthday is greater than two. One approach is by way of exponential generating functions.

Let's take a concrete example. Suppose we want to roll $10$ six-sided dice and find the probability that at least one number comes up $5$ or more times. It seems easier to consider the complementary event, i.e. all numbers come up $4$ or fewer times. There are $6^10$ possible sequences of die rolls, all of which we assume are equally likely. We would like to count the number of sequences in which no number comes up more than $4$ times. We generalize the problem a bit and think about a sequence of $n$ die rolls, and let the number of sequences in which no number appears more than $4$ times be $a_n$.

The exponential generating function of $a_n$ is defined to be
$$f(x) = sum_n=0^infty frac1n! a_n x^n$$
Since each die has $6$ sides and each number appears no more than $4$ times,
$$f(x) = left( sum_i=0^4 frac1i! x^i right)^6$$
The number we want in our problem, $a_10$, is $10!$ times the coefficient of $x^10$ when $f(x)$ is expanded. I suppose a pencil and paper solution is possible, but I took the easy way out and used a computer algebra system to find that the coefficient of $x^10$ is $2177 / 144$. Therefore the number of sequences of $10$ die rolls in which no number appears more than $4$ times is $$a_10 = 10! times frac2177 144 = 54,860,400$$

So the probability that no number appears more than $4$ times in $10$ rolls is
$$p = fraca_106^10 = 0.907291$$
and the probability that at least one number appears $5$ times or more is
$$1-p = 0.0927093$$

share|cite|improve this answer

answered Aug 1 at 18:00

awkward

5,05611021

add a comment | 

up vote
0
down vote

The OP is correct in saying that this is a generalization of the Birthday Problem and that it is harder because we are interested in cases where the number of people with the same birthday is greater than two. One approach is by way of exponential generating functions.

Let's take a concrete example. Suppose we want to roll $10$ six-sided dice and find the probability that at least one number comes up $5$ or more times. It seems easier to consider the complementary event, i.e. all numbers come up $4$ or fewer times. There are $6^10$ possible sequences of die rolls, all of which we assume are equally likely. We would like to count the number of sequences in which no number comes up more than $4$ times. We generalize the problem a bit and think about a sequence of $n$ die rolls, and let the number of sequences in which no number appears more than $4$ times be $a_n$.

The exponential generating function of $a_n$ is defined to be
$$f(x) = sum_n=0^infty frac1n! a_n x^n$$
Since each die has $6$ sides and each number appears no more than $4$ times,
$$f(x) = left( sum_i=0^4 frac1i! x^i right)^6$$
The number we want in our problem, $a_10$, is $10!$ times the coefficient of $x^10$ when $f(x)$ is expanded. I suppose a pencil and paper solution is possible, but I took the easy way out and used a computer algebra system to find that the coefficient of $x^10$ is $2177 / 144$. Therefore the number of sequences of $10$ die rolls in which no number appears more than $4$ times is $$a_10 = 10! times frac2177 144 = 54,860,400$$

So the probability that no number appears more than $4$ times in $10$ rolls is
$$p = fraca_106^10 = 0.907291$$
and the probability that at least one number appears $5$ times or more is
$$1-p = 0.0927093$$

share|cite|improve this answer

answered Aug 1 at 18:00

awkward

5,05611021

add a comment | 

up vote
0
down vote

up vote
0
down vote

The OP is correct in saying that this is a generalization of the Birthday Problem and that it is harder because we are interested in cases where the number of people with the same birthday is greater than two. One approach is by way of exponential generating functions.

Let's take a concrete example. Suppose we want to roll $10$ six-sided dice and find the probability that at least one number comes up $5$ or more times. It seems easier to consider the complementary event, i.e. all numbers come up $4$ or fewer times. There are $6^10$ possible sequences of die rolls, all of which we assume are equally likely. We would like to count the number of sequences in which no number comes up more than $4$ times. We generalize the problem a bit and think about a sequence of $n$ die rolls, and let the number of sequences in which no number appears more than $4$ times be $a_n$.

The exponential generating function of $a_n$ is defined to be
$$f(x) = sum_n=0^infty frac1n! a_n x^n$$
Since each die has $6$ sides and each number appears no more than $4$ times,
$$f(x) = left( sum_i=0^4 frac1i! x^i right)^6$$
The number we want in our problem, $a_10$, is $10!$ times the coefficient of $x^10$ when $f(x)$ is expanded. I suppose a pencil and paper solution is possible, but I took the easy way out and used a computer algebra system to find that the coefficient of $x^10$ is $2177 / 144$. Therefore the number of sequences of $10$ die rolls in which no number appears more than $4$ times is $$a_10 = 10! times frac2177 144 = 54,860,400$$

So the probability that no number appears more than $4$ times in $10$ rolls is
$$p = fraca_106^10 = 0.907291$$
and the probability that at least one number appears $5$ times or more is
$$1-p = 0.0927093$$

share|cite|improve this answer

answered Aug 1 at 18:00

awkward

5,05611021

The OP is correct in saying that this is a generalization of the Birthday Problem and that it is harder because we are interested in cases where the number of people with the same birthday is greater than two. One approach is by way of exponential generating functions.

Let's take a concrete example. Suppose we want to roll $10$ six-sided dice and find the probability that at least one number comes up $5$ or more times. It seems easier to consider the complementary event, i.e. all numbers come up $4$ or fewer times. There are $6^10$ possible sequences of die rolls, all of which we assume are equally likely. We would like to count the number of sequences in which no number comes up more than $4$ times. We generalize the problem a bit and think about a sequence of $n$ die rolls, and let the number of sequences in which no number appears more than $4$ times be $a_n$.

The exponential generating function of $a_n$ is defined to be
$$f(x) = sum_n=0^infty frac1n! a_n x^n$$
Since each die has $6$ sides and each number appears no more than $4$ times,
$$f(x) = left( sum_i=0^4 frac1i! x^i right)^6$$
The number we want in our problem, $a_10$, is $10!$ times the coefficient of $x^10$ when $f(x)$ is expanded. I suppose a pencil and paper solution is possible, but I took the easy way out and used a computer algebra system to find that the coefficient of $x^10$ is $2177 / 144$. Therefore the number of sequences of $10$ die rolls in which no number appears more than $4$ times is $$a_10 = 10! times frac2177 144 = 54,860,400$$

So the probability that no number appears more than $4$ times in $10$ rolls is
$$p = fraca_106^10 = 0.907291$$
and the probability that at least one number appears $5$ times or more is
$$1-p = 0.0927093$$

share|cite|improve this answer

answered Aug 1 at 18:00

awkward

5,05611021

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 18:00

awkward

5,05611021

answered Aug 1 at 18:00

awkward

5,05611021

answered Aug 1 at 18:00

awkward

5,05611021

5,05611021

add a comment | 

add a comment | 

 

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