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How many compositions does 40 have (with up to 5 addends), with each positive integer addend between 2 and 12?
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I was on Youtube and found a show called Monopoly Millionaires' Club. I thought it would be interesting to try to calculate the probability of winning the million dollars.
The contestant starts on the Go square, and if they reach the Go square again in less than 5 rolls of the pair of dice, they become a millionaire. Here are the givens:
- The rolls must sum to 40 to get back to Go (I counted on an image of a board)
- You are allowed at most 5 rolls (although we know it's impossible to get to 40 with 3 rolls, so the partition must be at least 4 in length and at most 5).
- Each addend in the partition must be at least 2 and at most 12.
I think if we could answer that question, we could get an approximate answer to the probability, but I am not sure how to do it; it's been some time I've done combinatorics, and I don't even know if I would have been able to do it to start with due to the restriction on the addends.
Could someone possibly please show me how I could do this? Possibly without using the exhahustive method, but by using generating functions? I'd also be interested in subtracting the possibilities that would land the contestant in jail, but it seems that we would have to phrase the question in terms of composition since order matters (if you get 12+12+6 you land in jail, but you don't if you get 6+12+8+12+2 since summing to 30 will get you to jail).
(For the record, I already graduated with my math degree and am no longer studying math. This is not homework.)
Thanks!
combinatorics generating-functions integers integer-partitions
asked Jul 27 at 9:03
August
1,077617
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up vote
2
down vote
favorite
I was on Youtube and found a show called Monopoly Millionaires' Club. I thought it would be interesting to try to calculate the probability of winning the million dollars.
The contestant starts on the Go square, and if they reach the Go square again in less than 5 rolls of the pair of dice, they become a millionaire. Here are the givens:
- The rolls must sum to 40 to get back to Go (I counted on an image of a board)
- You are allowed at most 5 rolls (although we know it's impossible to get to 40 with 3 rolls, so the partition must be at least 4 in length and at most 5).
- Each addend in the partition must be at least 2 and at most 12.
I think if we could answer that question, we could get an approximate answer to the probability, but I am not sure how to do it; it's been some time I've done combinatorics, and I don't even know if I would have been able to do it to start with due to the restriction on the addends.
Could someone possibly please show me how I could do this? Possibly without using the exhahustive method, but by using generating functions? I'd also be interested in subtracting the possibilities that would land the contestant in jail, but it seems that we would have to phrase the question in terms of composition since order matters (if you get 12+12+6 you land in jail, but you don't if you get 6+12+8+12+2 since summing to 30 will get you to jail).
(For the record, I already graduated with my math degree and am no longer studying math. This is not homework.)
Thanks!
combinatorics generating-functions integers integer-partitions
asked Jul 27 at 9:03
August
1,077617
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was on Youtube and found a show called Monopoly Millionaires' Club. I thought it would be interesting to try to calculate the probability of winning the million dollars.
The contestant starts on the Go square, and if they reach the Go square again in less than 5 rolls of the pair of dice, they become a millionaire. Here are the givens:
- The rolls must sum to 40 to get back to Go (I counted on an image of a board)
- You are allowed at most 5 rolls (although we know it's impossible to get to 40 with 3 rolls, so the partition must be at least 4 in length and at most 5).
- Each addend in the partition must be at least 2 and at most 12.
I think if we could answer that question, we could get an approximate answer to the probability, but I am not sure how to do it; it's been some time I've done combinatorics, and I don't even know if I would have been able to do it to start with due to the restriction on the addends.
Could someone possibly please show me how I could do this? Possibly without using the exhahustive method, but by using generating functions? I'd also be interested in subtracting the possibilities that would land the contestant in jail, but it seems that we would have to phrase the question in terms of composition since order matters (if you get 12+12+6 you land in jail, but you don't if you get 6+12+8+12+2 since summing to 30 will get you to jail).
(For the record, I already graduated with my math degree and am no longer studying math. This is not homework.)
Thanks!
combinatorics generating-functions integers integer-partitions
asked Jul 27 at 9:03
August
1,077617
I was on Youtube and found a show called Monopoly Millionaires' Club. I thought it would be interesting to try to calculate the probability of winning the million dollars.
The contestant starts on the Go square, and if they reach the Go square again in less than 5 rolls of the pair of dice, they become a millionaire. Here are the givens:
- The rolls must sum to 40 to get back to Go (I counted on an image of a board)
- You are allowed at most 5 rolls (although we know it's impossible to get to 40 with 3 rolls, so the partition must be at least 4 in length and at most 5).
- Each addend in the partition must be at least 2 and at most 12.
I think if we could answer that question, we could get an approximate answer to the probability, but I am not sure how to do it; it's been some time I've done combinatorics, and I don't even know if I would have been able to do it to start with due to the restriction on the addends.
Could someone possibly please show me how I could do this? Possibly without using the exhahustive method, but by using generating functions? I'd also be interested in subtracting the possibilities that would land the contestant in jail, but it seems that we would have to phrase the question in terms of composition since order matters (if you get 12+12+6 you land in jail, but you don't if you get 6+12+8+12+2 since summing to 30 will get you to jail).
(For the record, I already graduated with my math degree and am no longer studying math. This is not homework.)
Thanks!
combinatorics generating-functions integers integer-partitions
asked Jul 27 at 9:03
August
1,077617
edited Jul 27 at 9:32
asked Jul 27 at 9:03
August
1,077617
asked Jul 27 at 9:03
August
1,077617
asked Jul 27 at 9:03
August
1,077617
1,077617
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2 Answers
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Note that the possible addends 2, 3, ... , 11, 12 should be weighted with the corresponding probabilities to get them, else the model is not accurate.
If i correctly understand the question, a short version of it is as follows.
- $8$ dices are thrown together independently. If the sum of the faces is $40$ we win and stop here.
- If there was no win above, $2$ further dices are thrown together independently. It the sum of all ten dices is $40$ we win. Else we loose.
For the $8$ dices, we can consider the following polynomial:
$$
P_8(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^8
%=frac x^86^8Big( 1+x+x^2+x^3+x^4+x^5 Big)^8
.
$$
Then the corresponding probability is the coefficient of $x^40$ in the expansion. Using sage, we can ask for the polynomial $P$ (but it would be a mess to insert it here) and the needed coefficient. It is
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^8
sage: P.expand().coefficient(x^40)
683/186624
sage: _.n()
0.00365976508916324
This is a relatively big probability to become a millionaire. And if we also consider the probability to hit the $40$ with $10$ dices, we get with the similar polynomial
$$
P_10(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^10
$$
the result
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^10
sage: P.expand().coefficient(x^40)
2930455/60466176
sage: _.n()
0.0484643679137242
This is even a better probability to become a millionaire!
To handle also the jail complications, it is a good idea to introduce the following notations:
$$
beginaligned
P_N(x)
&=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^N ,\
P_N,k
&=text coefficient of $x^k$ in $P_N(x)$.
endaligned
$$
Then we have the following probabilities:
- The probability to get $40$ by throwing $8$ dices is $P_8,40=683/186624$.
- The probability to get $40$ by throwing $10$ dices is $P_10,40=2930455/60466176$.
- The probability to get $40$ by throwing either $8$ dices and getting it, or else after throwing $2$ further dices is
$P_8,40+P_10,40=3151747/60466176approx 0.0521241330028874dots$. - The probability to get the jail in $3,4,5$ rounds of throwing two dices is $P_6,30+P_8,30+P_10,30=8042599/60466176approx 0.133009883079095dots$ . (Most traders will agree to play for the million, some mathematicians will also do, conditioned by the constraint to have their own library in the cell, even if they have to pay the million.)
- The probability to get $40$ by throwing $8$ dices and not landing in jail is $P_8,40-P_6,30cdot P_2,10=59/20736
approx 0.00284529320987654dots$. - The probability to get $40$ by throwing $10$ dices and not landing in jail is $P_10,40-P_6,30cdot P_4,10-P_6,30cdot P_4,10
=2517211/60466176
approx 0.0416300676927213dots$.
answered Jul 27 at 14:46
dan_fulea
4,1171211
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
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up vote
3
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It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
beginalign*
[x^k](1+x)^n=binomnk
endalign*
Hint: Here is a starter which shows how to obtain the coefficient of $x^40$ in a generating function representing five rolls with two dice.
beginalign*
colorblue[x^40]&colorblueleft(x^2+x^3+cdots+x^12right)^5\
&=[x^40]x^10left(1+x^2+cdots x^10right)^5\
&=[x^30]left(frac1-x^111-xright)^5tag1\
&=[x^30]left(1-5x^11+10x^22right)sum_j=0^inftybinom-5j(-x)^jtag2\
&=left([x^30]-5[x^19]+10[x^8]right)sum_j=0^inftybinomj+44x^jtag3\
&=binom344-5binom234+10binom124tag4\
&,,colorblue=7,051
endalign*
Comment:
In (1) we apply the rule $[x^p-q]A(x)=[x^p]x^qA(x)$ and we use the finite geometric series formula.
In (2) we expand the numerator up to terms with powers less or equal $30$ since other terms do not contribute. We also use the binomial series expansion.
In (3) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (4) we select the coefficients of $x^j$ accordingly.
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Note that the possible addends 2, 3, ... , 11, 12 should be weighted with the corresponding probabilities to get them, else the model is not accurate.
If i correctly understand the question, a short version of it is as follows.
- $8$ dices are thrown together independently. If the sum of the faces is $40$ we win and stop here.
- If there was no win above, $2$ further dices are thrown together independently. It the sum of all ten dices is $40$ we win. Else we loose.
For the $8$ dices, we can consider the following polynomial:
$$
P_8(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^8
%=frac x^86^8Big( 1+x+x^2+x^3+x^4+x^5 Big)^8
.
$$
Then the corresponding probability is the coefficient of $x^40$ in the expansion. Using sage, we can ask for the polynomial $P$ (but it would be a mess to insert it here) and the needed coefficient. It is
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^8
sage: P.expand().coefficient(x^40)
683/186624
sage: _.n()
0.00365976508916324
This is a relatively big probability to become a millionaire. And if we also consider the probability to hit the $40$ with $10$ dices, we get with the similar polynomial
$$
P_10(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^10
$$
the result
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^10
sage: P.expand().coefficient(x^40)
2930455/60466176
sage: _.n()
0.0484643679137242
This is even a better probability to become a millionaire!
To handle also the jail complications, it is a good idea to introduce the following notations:
$$
beginaligned
P_N(x)
&=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^N ,\
P_N,k
&=text coefficient of $x^k$ in $P_N(x)$.
endaligned
$$
Then we have the following probabilities:
- The probability to get $40$ by throwing $8$ dices is $P_8,40=683/186624$.
- The probability to get $40$ by throwing $10$ dices is $P_10,40=2930455/60466176$.
- The probability to get $40$ by throwing either $8$ dices and getting it, or else after throwing $2$ further dices is
$P_8,40+P_10,40=3151747/60466176approx 0.0521241330028874dots$. - The probability to get the jail in $3,4,5$ rounds of throwing two dices is $P_6,30+P_8,30+P_10,30=8042599/60466176approx 0.133009883079095dots$ . (Most traders will agree to play for the million, some mathematicians will also do, conditioned by the constraint to have their own library in the cell, even if they have to pay the million.)
- The probability to get $40$ by throwing $8$ dices and not landing in jail is $P_8,40-P_6,30cdot P_2,10=59/20736
approx 0.00284529320987654dots$. - The probability to get $40$ by throwing $10$ dices and not landing in jail is $P_10,40-P_6,30cdot P_4,10-P_6,30cdot P_4,10
=2517211/60466176
approx 0.0416300676927213dots$.
answered Jul 27 at 14:46
dan_fulea
4,1171211
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
add a comment |Â
up vote
1
down vote
accepted
Note that the possible addends 2, 3, ... , 11, 12 should be weighted with the corresponding probabilities to get them, else the model is not accurate.
If i correctly understand the question, a short version of it is as follows.
- $8$ dices are thrown together independently. If the sum of the faces is $40$ we win and stop here.
- If there was no win above, $2$ further dices are thrown together independently. It the sum of all ten dices is $40$ we win. Else we loose.
For the $8$ dices, we can consider the following polynomial:
$$
P_8(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^8
%=frac x^86^8Big( 1+x+x^2+x^3+x^4+x^5 Big)^8
.
$$
Then the corresponding probability is the coefficient of $x^40$ in the expansion. Using sage, we can ask for the polynomial $P$ (but it would be a mess to insert it here) and the needed coefficient. It is
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^8
sage: P.expand().coefficient(x^40)
683/186624
sage: _.n()
0.00365976508916324
This is a relatively big probability to become a millionaire. And if we also consider the probability to hit the $40$ with $10$ dices, we get with the similar polynomial
$$
P_10(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^10
$$
the result
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^10
sage: P.expand().coefficient(x^40)
2930455/60466176
sage: _.n()
0.0484643679137242
This is even a better probability to become a millionaire!
To handle also the jail complications, it is a good idea to introduce the following notations:
$$
beginaligned
P_N(x)
&=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^N ,\
P_N,k
&=text coefficient of $x^k$ in $P_N(x)$.
endaligned
$$
Then we have the following probabilities:
- The probability to get $40$ by throwing $8$ dices is $P_8,40=683/186624$.
- The probability to get $40$ by throwing $10$ dices is $P_10,40=2930455/60466176$.
- The probability to get $40$ by throwing either $8$ dices and getting it, or else after throwing $2$ further dices is
$P_8,40+P_10,40=3151747/60466176approx 0.0521241330028874dots$. - The probability to get the jail in $3,4,5$ rounds of throwing two dices is $P_6,30+P_8,30+P_10,30=8042599/60466176approx 0.133009883079095dots$ . (Most traders will agree to play for the million, some mathematicians will also do, conditioned by the constraint to have their own library in the cell, even if they have to pay the million.)
- The probability to get $40$ by throwing $8$ dices and not landing in jail is $P_8,40-P_6,30cdot P_2,10=59/20736
approx 0.00284529320987654dots$. - The probability to get $40$ by throwing $10$ dices and not landing in jail is $P_10,40-P_6,30cdot P_4,10-P_6,30cdot P_4,10
=2517211/60466176
approx 0.0416300676927213dots$.
answered Jul 27 at 14:46
dan_fulea
4,1171211
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Note that the possible addends 2, 3, ... , 11, 12 should be weighted with the corresponding probabilities to get them, else the model is not accurate.
If i correctly understand the question, a short version of it is as follows.
- $8$ dices are thrown together independently. If the sum of the faces is $40$ we win and stop here.
- If there was no win above, $2$ further dices are thrown together independently. It the sum of all ten dices is $40$ we win. Else we loose.
For the $8$ dices, we can consider the following polynomial:
$$
P_8(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^8
%=frac x^86^8Big( 1+x+x^2+x^3+x^4+x^5 Big)^8
.
$$
Then the corresponding probability is the coefficient of $x^40$ in the expansion. Using sage, we can ask for the polynomial $P$ (but it would be a mess to insert it here) and the needed coefficient. It is
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^8
sage: P.expand().coefficient(x^40)
683/186624
sage: _.n()
0.00365976508916324
This is a relatively big probability to become a millionaire. And if we also consider the probability to hit the $40$ with $10$ dices, we get with the similar polynomial
$$
P_10(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^10
$$
the result
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^10
sage: P.expand().coefficient(x^40)
2930455/60466176
sage: _.n()
0.0484643679137242
This is even a better probability to become a millionaire!
To handle also the jail complications, it is a good idea to introduce the following notations:
$$
beginaligned
P_N(x)
&=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^N ,\
P_N,k
&=text coefficient of $x^k$ in $P_N(x)$.
endaligned
$$
Then we have the following probabilities:
- The probability to get $40$ by throwing $8$ dices is $P_8,40=683/186624$.
- The probability to get $40$ by throwing $10$ dices is $P_10,40=2930455/60466176$.
- The probability to get $40$ by throwing either $8$ dices and getting it, or else after throwing $2$ further dices is
$P_8,40+P_10,40=3151747/60466176approx 0.0521241330028874dots$. - The probability to get the jail in $3,4,5$ rounds of throwing two dices is $P_6,30+P_8,30+P_10,30=8042599/60466176approx 0.133009883079095dots$ . (Most traders will agree to play for the million, some mathematicians will also do, conditioned by the constraint to have their own library in the cell, even if they have to pay the million.)
- The probability to get $40$ by throwing $8$ dices and not landing in jail is $P_8,40-P_6,30cdot P_2,10=59/20736
approx 0.00284529320987654dots$. - The probability to get $40$ by throwing $10$ dices and not landing in jail is $P_10,40-P_6,30cdot P_4,10-P_6,30cdot P_4,10
=2517211/60466176
approx 0.0416300676927213dots$.
answered Jul 27 at 14:46
dan_fulea
4,1171211
Note that the possible addends 2, 3, ... , 11, 12 should be weighted with the corresponding probabilities to get them, else the model is not accurate.
If i correctly understand the question, a short version of it is as follows.
- $8$ dices are thrown together independently. If the sum of the faces is $40$ we win and stop here.
- If there was no win above, $2$ further dices are thrown together independently. It the sum of all ten dices is $40$ we win. Else we loose.
For the $8$ dices, we can consider the following polynomial:
$$
P_8(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^8
%=frac x^86^8Big( 1+x+x^2+x^3+x^4+x^5 Big)^8
.
$$
Then the corresponding probability is the coefficient of $x^40$ in the expansion. Using sage, we can ask for the polynomial $P$ (but it would be a mess to insert it here) and the needed coefficient. It is
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^8
sage: P.expand().coefficient(x^40)
683/186624
sage: _.n()
0.00365976508916324
This is a relatively big probability to become a millionaire. And if we also consider the probability to hit the $40$ with $10$ dices, we get with the similar polynomial
$$
P_10(x)=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^10
$$
the result
sage: P = ( (x+x^2+x^3+x^4+x^5+x^6)/6 )^10
sage: P.expand().coefficient(x^40)
2930455/60466176
sage: _.n()
0.0484643679137242
This is even a better probability to become a millionaire!
To handle also the jail complications, it is a good idea to introduce the following notations:
$$
beginaligned
P_N(x)
&=left( frac 16(x+x^2+x^3+x^4+x^5+x^6) right)^N ,\
P_N,k
&=text coefficient of $x^k$ in $P_N(x)$.
endaligned
$$
Then we have the following probabilities:
- The probability to get $40$ by throwing $8$ dices is $P_8,40=683/186624$.
- The probability to get $40$ by throwing $10$ dices is $P_10,40=2930455/60466176$.
- The probability to get $40$ by throwing either $8$ dices and getting it, or else after throwing $2$ further dices is
$P_8,40+P_10,40=3151747/60466176approx 0.0521241330028874dots$. - The probability to get the jail in $3,4,5$ rounds of throwing two dices is $P_6,30+P_8,30+P_10,30=8042599/60466176approx 0.133009883079095dots$ . (Most traders will agree to play for the million, some mathematicians will also do, conditioned by the constraint to have their own library in the cell, even if they have to pay the million.)
- The probability to get $40$ by throwing $8$ dices and not landing in jail is $P_8,40-P_6,30cdot P_2,10=59/20736
approx 0.00284529320987654dots$. - The probability to get $40$ by throwing $10$ dices and not landing in jail is $P_10,40-P_6,30cdot P_4,10-P_6,30cdot P_4,10
=2517211/60466176
approx 0.0416300676927213dots$.
answered Jul 27 at 14:46
dan_fulea
4,1171211
answered Jul 27 at 14:46
dan_fulea
4,1171211
answered Jul 27 at 14:46
dan_fulea
4,1171211
answered Jul 27 at 14:46
dan_fulea
4,1171211
4,1171211
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
add a comment |Â
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
Wow! This is such an awesome answer! So it looks like the probability of winning the million and not landing in jail is 4%, and winning anything at all (ie. not landing in jail within the 5 rolls but passing go - you still get some prizes, I believe) is 1 - 0.13 = 87%. Those are great chances. So if we were to weight the addends using the generating functions as done by the first contributor, it would be something like: $[x^40]((1/36)x^2+(2/36)x^3+(3/36)x^4+...+(1/6)x^12)^5$? Thanks again! This was crystal clear!
– August
Jul 29 at 5:39
add a comment |Â
up vote
3
down vote
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
beginalign*
[x^k](1+x)^n=binomnk
endalign*
Hint: Here is a starter which shows how to obtain the coefficient of $x^40$ in a generating function representing five rolls with two dice.
beginalign*
colorblue[x^40]&colorblueleft(x^2+x^3+cdots+x^12right)^5\
&=[x^40]x^10left(1+x^2+cdots x^10right)^5\
&=[x^30]left(frac1-x^111-xright)^5tag1\
&=[x^30]left(1-5x^11+10x^22right)sum_j=0^inftybinom-5j(-x)^jtag2\
&=left([x^30]-5[x^19]+10[x^8]right)sum_j=0^inftybinomj+44x^jtag3\
&=binom344-5binom234+10binom124tag4\
&,,colorblue=7,051
endalign*
Comment:
In (1) we apply the rule $[x^p-q]A(x)=[x^p]x^qA(x)$ and we use the finite geometric series formula.
In (2) we expand the numerator up to terms with powers less or equal $30$ since other terms do not contribute. We also use the binomial series expansion.
In (3) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (4) we select the coefficients of $x^j$ accordingly.
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
add a comment |Â
up vote
3
down vote
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
beginalign*
[x^k](1+x)^n=binomnk
endalign*
Hint: Here is a starter which shows how to obtain the coefficient of $x^40$ in a generating function representing five rolls with two dice.
beginalign*
colorblue[x^40]&colorblueleft(x^2+x^3+cdots+x^12right)^5\
&=[x^40]x^10left(1+x^2+cdots x^10right)^5\
&=[x^30]left(frac1-x^111-xright)^5tag1\
&=[x^30]left(1-5x^11+10x^22right)sum_j=0^inftybinom-5j(-x)^jtag2\
&=left([x^30]-5[x^19]+10[x^8]right)sum_j=0^inftybinomj+44x^jtag3\
&=binom344-5binom234+10binom124tag4\
&,,colorblue=7,051
endalign*
Comment:
In (1) we apply the rule $[x^p-q]A(x)=[x^p]x^qA(x)$ and we use the finite geometric series formula.
In (2) we expand the numerator up to terms with powers less or equal $30$ since other terms do not contribute. We also use the binomial series expansion.
In (3) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (4) we select the coefficients of $x^j$ accordingly.
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
beginalign*
[x^k](1+x)^n=binomnk
endalign*
Hint: Here is a starter which shows how to obtain the coefficient of $x^40$ in a generating function representing five rolls with two dice.
beginalign*
colorblue[x^40]&colorblueleft(x^2+x^3+cdots+x^12right)^5\
&=[x^40]x^10left(1+x^2+cdots x^10right)^5\
&=[x^30]left(frac1-x^111-xright)^5tag1\
&=[x^30]left(1-5x^11+10x^22right)sum_j=0^inftybinom-5j(-x)^jtag2\
&=left([x^30]-5[x^19]+10[x^8]right)sum_j=0^inftybinomj+44x^jtag3\
&=binom344-5binom234+10binom124tag4\
&,,colorblue=7,051
endalign*
Comment:
In (1) we apply the rule $[x^p-q]A(x)=[x^p]x^qA(x)$ and we use the finite geometric series formula.
In (2) we expand the numerator up to terms with powers less or equal $30$ since other terms do not contribute. We also use the binomial series expansion.
In (3) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (4) we select the coefficients of $x^j$ accordingly.
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
beginalign*
[x^k](1+x)^n=binomnk
endalign*
Hint: Here is a starter which shows how to obtain the coefficient of $x^40$ in a generating function representing five rolls with two dice.
beginalign*
colorblue[x^40]&colorblueleft(x^2+x^3+cdots+x^12right)^5\
&=[x^40]x^10left(1+x^2+cdots x^10right)^5\
&=[x^30]left(frac1-x^111-xright)^5tag1\
&=[x^30]left(1-5x^11+10x^22right)sum_j=0^inftybinom-5j(-x)^jtag2\
&=left([x^30]-5[x^19]+10[x^8]right)sum_j=0^inftybinomj+44x^jtag3\
&=binom344-5binom234+10binom124tag4\
&,,colorblue=7,051
endalign*
Comment:
In (1) we apply the rule $[x^p-q]A(x)=[x^p]x^qA(x)$ and we use the finite geometric series formula.
In (2) we expand the numerator up to terms with powers less or equal $30$ since other terms do not contribute. We also use the binomial series expansion.
In (3) we use the linearity of the coefficient of operator and apply the same rule as in (1).
In (4) we select the coefficients of $x^j$ accordingly.
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
edited Jul 27 at 13:02
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
answered Jul 27 at 12:36
Markus Scheuer
55.9k450135
55.9k450135
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
add a comment |Â
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
Thank you for your answer, Markus! It looks like I'll have to review my combinatorics to be able to do this; some of it is coming back to me, some of it is not. Thanks again!
– August
Jul 29 at 5:43
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
@August: You're welcome.
– Markus Scheuer
Jul 29 at 6:36
add a comment |Â
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