Mixing
What is the smallest number of cards one must take from a deck?
Clash Royale CLAN TAG#URR8PPP
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I can assume by elimination that 5 or 6 is the correct answer and I tried using combination and the hyper geometric distribution considering a one face as being the M category and the rest of the cards as being the N. However, I am still not able to get 6. I tried to substitute 6 in the hypergeometric formula and I got 0.74. How should I proceed then?
probability
asked Jul 14 at 17:06
Roy Rizk
887
add a comment |Â
up vote
0
down vote
favorite
I can assume by elimination that 5 or 6 is the correct answer and I tried using combination and the hyper geometric distribution considering a one face as being the M category and the rest of the cards as being the N. However, I am still not able to get 6. I tried to substitute 6 in the hypergeometric formula and I got 0.74. How should I proceed then?
probability
asked Jul 14 at 17:06
Roy Rizk
887
I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I can assume by elimination that 5 or 6 is the correct answer and I tried using combination and the hyper geometric distribution considering a one face as being the M category and the rest of the cards as being the N. However, I am still not able to get 6. I tried to substitute 6 in the hypergeometric formula and I got 0.74. How should I proceed then?
probability
asked Jul 14 at 17:06
Roy Rizk
887
I can assume by elimination that 5 or 6 is the correct answer and I tried using combination and the hyper geometric distribution considering a one face as being the M category and the rest of the cards as being the N. However, I am still not able to get 6. I tried to substitute 6 in the hypergeometric formula and I got 0.74. How should I proceed then?
probability
asked Jul 14 at 17:06
Roy Rizk
887
asked Jul 14 at 17:06
Roy Rizk
887
asked Jul 14 at 17:06
Roy Rizk
887
asked Jul 14 at 17:06
Roy Rizk
887
887
I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16
add a comment |Â
I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16
I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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Look at the complementary probability. The probability that $k$ cards are of different ranks is $$fracbinom13k4^kbinom52k,$$ because there are $binom13k$ ways to choose the ranks, and four cards of each rank. When $k=5,$ we get $0.50708$ and when $k=6,$ we get $0.345247.$
EDIT
To answer the OP's question about the factor of $4^k:$ This is just a direct application of the multiplication principle. Say that $k=2.$ I have $binom132=78$ ways in which I can choose the ranks of the two cards. Say I choose Aces and Kings. Then I can choose Ace of Spades, Ace of Hearts, Ace of Diamonds, or Ace of Clubs. Similarly, there are $4$ Kings I can choose. So there are $4^2=16$ choices in all; for each of the $4$ Aces, there are $4$ possible Kings.
answered Jul 14 at 17:19
saulspatz
10.7k21323
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
add a comment |Â
up vote
0
down vote
Let us take $n$ cards. What needs to happen for them to all be different?
After two cards, the chances are $48/51$. The second card just need to have a different face to the first one and there are 48 cards with a different face.
After three, they are $48/51 times 44/50$. As above but the third card needs to have a different face to the first two.
After four, the chances are $48/51 times 44/50 times 40/49$.
Now we need this to fall below 0.4. Let's see what we have.
After 2
$12/13=0.94$
After 3
$48/51*44/50=0.82$
After 4, we multiply the above by $40/49$ to get $0.68$
After 5 $0.51$
So...after $6$
$0.345$ kerching. If we take $6$ cards, the chances of having two the same is $1-0.345$ which is $0.655$
answered Jul 14 at 17:22
Simon Terrington
1466
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Look at the complementary probability. The probability that $k$ cards are of different ranks is $$fracbinom13k4^kbinom52k,$$ because there are $binom13k$ ways to choose the ranks, and four cards of each rank. When $k=5,$ we get $0.50708$ and when $k=6,$ we get $0.345247.$
EDIT
To answer the OP's question about the factor of $4^k:$ This is just a direct application of the multiplication principle. Say that $k=2.$ I have $binom132=78$ ways in which I can choose the ranks of the two cards. Say I choose Aces and Kings. Then I can choose Ace of Spades, Ace of Hearts, Ace of Diamonds, or Ace of Clubs. Similarly, there are $4$ Kings I can choose. So there are $4^2=16$ choices in all; for each of the $4$ Aces, there are $4$ possible Kings.
answered Jul 14 at 17:19
saulspatz
10.7k21323
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
add a comment |Â
up vote
1
down vote
accepted
Look at the complementary probability. The probability that $k$ cards are of different ranks is $$fracbinom13k4^kbinom52k,$$ because there are $binom13k$ ways to choose the ranks, and four cards of each rank. When $k=5,$ we get $0.50708$ and when $k=6,$ we get $0.345247.$
EDIT
To answer the OP's question about the factor of $4^k:$ This is just a direct application of the multiplication principle. Say that $k=2.$ I have $binom132=78$ ways in which I can choose the ranks of the two cards. Say I choose Aces and Kings. Then I can choose Ace of Spades, Ace of Hearts, Ace of Diamonds, or Ace of Clubs. Similarly, there are $4$ Kings I can choose. So there are $4^2=16$ choices in all; for each of the $4$ Aces, there are $4$ possible Kings.
answered Jul 14 at 17:19
saulspatz
10.7k21323
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Look at the complementary probability. The probability that $k$ cards are of different ranks is $$fracbinom13k4^kbinom52k,$$ because there are $binom13k$ ways to choose the ranks, and four cards of each rank. When $k=5,$ we get $0.50708$ and when $k=6,$ we get $0.345247.$
EDIT
To answer the OP's question about the factor of $4^k:$ This is just a direct application of the multiplication principle. Say that $k=2.$ I have $binom132=78$ ways in which I can choose the ranks of the two cards. Say I choose Aces and Kings. Then I can choose Ace of Spades, Ace of Hearts, Ace of Diamonds, or Ace of Clubs. Similarly, there are $4$ Kings I can choose. So there are $4^2=16$ choices in all; for each of the $4$ Aces, there are $4$ possible Kings.
answered Jul 14 at 17:19
saulspatz
10.7k21323
Look at the complementary probability. The probability that $k$ cards are of different ranks is $$fracbinom13k4^kbinom52k,$$ because there are $binom13k$ ways to choose the ranks, and four cards of each rank. When $k=5,$ we get $0.50708$ and when $k=6,$ we get $0.345247.$
EDIT
To answer the OP's question about the factor of $4^k:$ This is just a direct application of the multiplication principle. Say that $k=2.$ I have $binom132=78$ ways in which I can choose the ranks of the two cards. Say I choose Aces and Kings. Then I can choose Ace of Spades, Ace of Hearts, Ace of Diamonds, or Ace of Clubs. Similarly, there are $4$ Kings I can choose. So there are $4^2=16$ choices in all; for each of the $4$ Aces, there are $4$ possible Kings.
answered Jul 14 at 17:19
saulspatz
10.7k21323
edited Jul 14 at 17:34
answered Jul 14 at 17:19
saulspatz
10.7k21323
answered Jul 14 at 17:19
saulspatz
10.7k21323
answered Jul 14 at 17:19
saulspatz
10.7k21323
10.7k21323
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
add a comment |Â
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
Why the multiplication with $4^k$?
– Vizag
Jul 14 at 17:21
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
@Vizag Because in each of the $k$ ranks there are four cards I can choose -- one of each suit.
– saulspatz
Jul 14 at 17:23
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
I did not really get the rule of the 4 times k
– Roy Rizk
Jul 14 at 17:29
add a comment |Â
up vote
0
down vote
Let us take $n$ cards. What needs to happen for them to all be different?
After two cards, the chances are $48/51$. The second card just need to have a different face to the first one and there are 48 cards with a different face.
After three, they are $48/51 times 44/50$. As above but the third card needs to have a different face to the first two.
After four, the chances are $48/51 times 44/50 times 40/49$.
Now we need this to fall below 0.4. Let's see what we have.
After 2
$12/13=0.94$
After 3
$48/51*44/50=0.82$
After 4, we multiply the above by $40/49$ to get $0.68$
After 5 $0.51$
So...after $6$
$0.345$ kerching. If we take $6$ cards, the chances of having two the same is $1-0.345$ which is $0.655$
answered Jul 14 at 17:22
Simon Terrington
1466
add a comment |Â
up vote
0
down vote
Let us take $n$ cards. What needs to happen for them to all be different?
After two cards, the chances are $48/51$. The second card just need to have a different face to the first one and there are 48 cards with a different face.
After three, they are $48/51 times 44/50$. As above but the third card needs to have a different face to the first two.
After four, the chances are $48/51 times 44/50 times 40/49$.
Now we need this to fall below 0.4. Let's see what we have.
After 2
$12/13=0.94$
After 3
$48/51*44/50=0.82$
After 4, we multiply the above by $40/49$ to get $0.68$
After 5 $0.51$
So...after $6$
$0.345$ kerching. If we take $6$ cards, the chances of having two the same is $1-0.345$ which is $0.655$
answered Jul 14 at 17:22
Simon Terrington
1466
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us take $n$ cards. What needs to happen for them to all be different?
After two cards, the chances are $48/51$. The second card just need to have a different face to the first one and there are 48 cards with a different face.
After three, they are $48/51 times 44/50$. As above but the third card needs to have a different face to the first two.
After four, the chances are $48/51 times 44/50 times 40/49$.
Now we need this to fall below 0.4. Let's see what we have.
After 2
$12/13=0.94$
After 3
$48/51*44/50=0.82$
After 4, we multiply the above by $40/49$ to get $0.68$
After 5 $0.51$
So...after $6$
$0.345$ kerching. If we take $6$ cards, the chances of having two the same is $1-0.345$ which is $0.655$
answered Jul 14 at 17:22
Simon Terrington
1466
Let us take $n$ cards. What needs to happen for them to all be different?
After two cards, the chances are $48/51$. The second card just need to have a different face to the first one and there are 48 cards with a different face.
After three, they are $48/51 times 44/50$. As above but the third card needs to have a different face to the first two.
After four, the chances are $48/51 times 44/50 times 40/49$.
Now we need this to fall below 0.4. Let's see what we have.
After 2
$12/13=0.94$
After 3
$48/51*44/50=0.82$
After 4, we multiply the above by $40/49$ to get $0.68$
After 5 $0.51$
So...after $6$
$0.345$ kerching. If we take $6$ cards, the chances of having two the same is $1-0.345$ which is $0.655$
answered Jul 14 at 17:22
Simon Terrington
1466
edited Jul 14 at 17:50
answered Jul 14 at 17:22
Simon Terrington
1466
answered Jul 14 at 17:22
Simon Terrington
1466
answered Jul 14 at 17:22
Simon Terrington
1466
1466
add a comment |Â
add a comment |Â
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I think you want to look at the complement: the probability that all cards are different ranks is less than $.4$
– saulspatz
Jul 14 at 17:11
How do I compute the complement ?
– Roy Rizk
Jul 14 at 17:16