In poker, What is the chance of a pair?
0
$begingroup$
While playing poker last night with my housemates I started wondering, "What is the chance of a pair coming up?" (To clarify, at least one pair).
I started solving it like the birthday problem:
- Probability that the first card is not a pair is
1. (Since two cards are required for a pair).
$A$: Probability that the second card is not a pair of the 1st card is $frac{48}{51}$. (Since there are three other cards which could produce a pair)
$B$: Probability that the third card is not a pair of the 1st or the 2nd is $frac{44}{50} * A$. (Since there are now 6 cards which could produce a pair).
$C$: Probability that the fourth card is not a pair of the 1st or the 2nd or the 3rd is $frac{40}{49} * B * A$
$D$: Probability that the fifth card is not a pair of the 1st or the 2nd or the 3rd or the 4th is $frac{36}{48} * C * B * A$
- Therefore the probability that there is a pair given 5 cards is $1 - (frac{36}{48} . frac{40}{49} . frac{44}{50} . frac{48}{51})$
More generally the probability that there are no pairs with n cards is $1 - (prod_{i=1}^{n-1}{frac{(52-i) - 3i}{(52 - i)}})$
This morning I found solutions online like this one but they use combinatorics. Is my solution valid? Have I made any mistakes?
probability probability-theory
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
$endgroup$
add a comment |
0
$begingroup$
While playing poker last night with my housemates I started wondering, "What is the chance of a pair coming up?" (To clarify, at least one pair).
I started solving it like the birthday problem:
- Probability that the first card is not a pair is
1. (Since two cards are required for a pair).
$A$: Probability that the second card is not a pair of the 1st card is $frac{48}{51}$. (Since there are three other cards which could produce a pair)
$B$: Probability that the third card is not a pair of the 1st or the 2nd is $frac{44}{50} * A$. (Since there are now 6 cards which could produce a pair).
$C$: Probability that the fourth card is not a pair of the 1st or the 2nd or the 3rd is $frac{40}{49} * B * A$
$D$: Probability that the fifth card is not a pair of the 1st or the 2nd or the 3rd or the 4th is $frac{36}{48} * C * B * A$
- Therefore the probability that there is a pair given 5 cards is $1 - (frac{36}{48} . frac{40}{49} . frac{44}{50} . frac{48}{51})$
More generally the probability that there are no pairs with n cards is $1 - (prod_{i=1}^{n-1}{frac{(52-i) - 3i}{(52 - i)}})$
This morning I found solutions online like this one but they use combinatorics. Is my solution valid? Have I made any mistakes?
probability probability-theory
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
$endgroup$
2
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
1
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49
add a comment |
0
0
0
$begingroup$
While playing poker last night with my housemates I started wondering, "What is the chance of a pair coming up?" (To clarify, at least one pair).
I started solving it like the birthday problem:
- Probability that the first card is not a pair is
1. (Since two cards are required for a pair).
$A$: Probability that the second card is not a pair of the 1st card is $frac{48}{51}$. (Since there are three other cards which could produce a pair)
$B$: Probability that the third card is not a pair of the 1st or the 2nd is $frac{44}{50} * A$. (Since there are now 6 cards which could produce a pair).
$C$: Probability that the fourth card is not a pair of the 1st or the 2nd or the 3rd is $frac{40}{49} * B * A$
$D$: Probability that the fifth card is not a pair of the 1st or the 2nd or the 3rd or the 4th is $frac{36}{48} * C * B * A$
- Therefore the probability that there is a pair given 5 cards is $1 - (frac{36}{48} . frac{40}{49} . frac{44}{50} . frac{48}{51})$
More generally the probability that there are no pairs with n cards is $1 - (prod_{i=1}^{n-1}{frac{(52-i) - 3i}{(52 - i)}})$
This morning I found solutions online like this one but they use combinatorics. Is my solution valid? Have I made any mistakes?
probability probability-theory
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
$endgroup$
While playing poker last night with my housemates I started wondering, "What is the chance of a pair coming up?" (To clarify, at least one pair).
I started solving it like the birthday problem:
- Probability that the first card is not a pair is
1. (Since two cards are required for a pair).
$A$: Probability that the second card is not a pair of the 1st card is $frac{48}{51}$. (Since there are three other cards which could produce a pair)
$B$: Probability that the third card is not a pair of the 1st or the 2nd is $frac{44}{50} * A$. (Since there are now 6 cards which could produce a pair).
$C$: Probability that the fourth card is not a pair of the 1st or the 2nd or the 3rd is $frac{40}{49} * B * A$
$D$: Probability that the fifth card is not a pair of the 1st or the 2nd or the 3rd or the 4th is $frac{36}{48} * C * B * A$
- Therefore the probability that there is a pair given 5 cards is $1 - (frac{36}{48} . frac{40}{49} . frac{44}{50} . frac{48}{51})$
More generally the probability that there are no pairs with n cards is $1 - (prod_{i=1}^{n-1}{frac{(52-i) - 3i}{(52 - i)}})$
This morning I found solutions online like this one but they use combinatorics. Is my solution valid? Have I made any mistakes?
probability probability-theory
probability probability-theory
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
edited Jan 25 at 23:48
david_adler
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
asked Jan 25 at 16:38
david_adlerdavid_adler
1286
1286
2
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
1
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49
add a comment |
2
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
1
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49
2
2
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
1
1
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49
add a comment |
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2
$begingroup$
Does four-of-a-kind count as a pair?
$endgroup$
– Henry
Jan 25 at 16:43
1
$begingroup$
The video you linked is calculating the probability of exactly 1 pair. What you found is the probability that at least two cards in the hand have the same rank, which could mean there's one pair, two pairs, three of a kind, four of a kind. Also note that your formula only works up to $n=13$ because if $n=14$ there must be a pair by the pidgeonhole principle.
$endgroup$
– kccu
Jan 25 at 16:46
$begingroup$
@Henry yes at least one pair to clarify
$endgroup$
– david_adler
Jan 25 at 23:49