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Drawing cards in poker - why is my answer wrong?

  1. Jan 18, 2014 #1
    1. The problem statement, all variables and given/known data
    When drawing a poker hand (5 cards), what is the probability to get a single pair?

    3. The attempt at a solution

    This is my (wrong) line of thought: The first draw can be any card, so you have 52 options. For each of these cards you have 3 options to get a pair, so 52*3 in total ways to draw a pair. Now, for each of these 52*3 ways to draw a pair, you got 48*47*46 ways to draw the remaining cards. This makes a total of 52*3*48*47*46 ways to draw, and thus the probability for a pair is P=(52*3*48*47*46)/(52*51*50*49*48)=1/17.

    1/17 is obviously wrong, but why is my logic wrong? Does it have something to do with the order the cards are drawn in?
  2. jcsd
  3. Jan 18, 2014 #2

    D H

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    That's the probability of drawing a pair if you are dealt just two cards. That's not a poker hand, which comprises five cards.
  4. Jan 18, 2014 #3
    I know a poker hand is 5 cards, but why is the formula from OP wrong? P=(52*3*48*47*46)/(52*51*50*49*48) ?
  5. Jan 18, 2014 #4
    I just realized two mistakes:

    1) I didn't take into account that the newly drawn cards must not be pairs.

    2) I sense I need to look at the drawing process as a combination, not permutation, to get the correct probability. So I need to divide the nominator and denominator by the amount of ways they can be sorted. Why is this the case? Why should it matter?

    If I adjust my formula for 1), I get P=(52*3*48*44*40)/(52*51*50*49*48) = 0.0422569, which is 1/10 of the correct answer (0.422569)...

    Last edited: Jan 18, 2014
  6. Jan 18, 2014 #5
    You are close. Surely you mean 48 44 40 in your last post.

    I would have written it out combination style as:
    Numerator= 13c1(pick the type of pair)* 4c2 (pick 2 cards from 4)
    *12c3(pick the 3 other non pair types)*(4c1)^3 (pick one of 4 cards for each of these)
    Denominator = 52c5

    If you write that out it comes to pretty much what you had, except that it takes into account your second point about reordering. This matters because the hand is the same no matter in what order it is dealt.
  7. Jan 18, 2014 #6
    thanks for reply! Btw I might have edited my post while you were writing.

    I agree, but shouldn't the reording issue "cancel itself" when you divide two permutations by each other?
  8. Jan 18, 2014 #7

    D H

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    That's the probability of your first two cards containing a pair and the next three cards containing neither a different pair nor one of the two remaining cards in the deck that match that initial pair. Those paired cards can appear anywhere in that hand, not just the first two cards. You need to multiply that result by (5 choose 2)=10 and voila! you have the correct answer.
  9. Jan 18, 2014 #8
    The DH answer gives the correct " permutation style" method. I believe the "combination style method is easier to think about, but as you say they must come to the same thing.
  10. Jan 18, 2014 #9
    yeah you're right David. In fact, I found the correct combation-style solution on the web, learned it (I hope.. :P) and wrote it down as my answer. At first though I tried the permutation style method as that is what my professor usually used to solve example problems on the blackboard, but I didn't get it to work. And so I came here

    DH: thanks for the reply.

    At any rate, I think I understand now. Thanks :)
    Last edited: Jan 18, 2014
  11. Jan 18, 2014 #10
    OK seems I didn't understand after-all.

    When drawing two pairs, why do you have to take into account the ordering of the pairs themselves? Shouldn't it be enough to just order the cards? I think my major issue with this type of probabilities is that I don't understand how ordering works. To top it off, it seems I have to order the pairs only when the two groups are of the same "type"? I mean, in Full House you have one pair and one three-of-a-kind, making two groups, but you don't have to order these?

    To illustrate:

    Two Pairs
    Correct P(two pairs) = ##\frac{{13 \choose 2} \cdot {4 \choose 2}^2 \cdot 4 ({11 \choose 1})}{({52 \choose 5})}##. What I thought was correct: P(two pairs) = ##\frac{13 \cdot 12 \cdot {4 \choose 2}^2 \cdot 4 {11 \choose 1}}{{52 \choose 5}}##. Notice my (wrong) formula got 13*12 instead of (13 choose 2).

    Full House
    Correct P(house)= ##\frac{13 {4 \choose 2} \cdot 12 {4 \choose 3}}{{52 \choose 5}}##
    Notice how I don't use (13 choose 2) here like I do in the two pairs formula.

    Why is it like this?
    Last edited: Jan 18, 2014
  12. Jan 18, 2014 #11

    D H

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    One way to look at it is counting. Of the 52 choose 5 possible hands, how many are two pair? There are 13 numbers in a deck. The two pair are any 2 of these 13 numbered cards, so 13 choose 2. For each pair, there are 4 cards to choose from; we're taking 2. So (4 choose 2)2. The fifth card is any one of the 11 remaining numbered cards, and any one of the four suits. So (11 choose 1)*(4 choose 1). Altogether, this is ##\binom {13} 2 {\binom 4 2}^2 \binom {11} 1 \binom 4 1## hands that contain exactly two pair. Dividing by ##\binom {52} 5## yields the probability of two pair, 198/4165.

    Another way to look at it is to compute the probability of getting a pair in the first two cards, a different pair in the next two cards, and an unrelated card on the last card. The probability of this particular hand is ##1\cdot \frac 3 {51} \cdot \frac{48}{50} \cdot \frac 3 {49} \cdot \frac {44}{48} = \frac {66}{20825}##. That isn't the only way to be dealt two pair. With this approach one has to look at the number of ways we can be dealt an equivalent hand, and one has to be careful in doing so. That unrelated card can be any one of the five cards, so this multiplies the probability by five. The four paired cards is where one needs to take care. There are six permutations of {A,A,B,B}, but multiplying by six would be over counting. {B,B,A,A} is the same as {A,A,B,B}. Fixing the first card, there are only three permutations of {A,B,B}. Thus we need to multiply the probability of that specifically ordered hand by 15, yielding 198/4165.
  13. Jan 18, 2014 #12
    Dick, I edited my previous, asking why I don't have to order the two groups (pair and three-of-a-kind) for P(house). Dunno if it changes your point, though. At any rate thanks for the reply, I'll read it now.
  14. Jan 18, 2014 #13

    D H

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    Because in the two pair situation, the two pair are indistinguishable. Using ## 13 \cdot 12 = \binom {13} 1 \cdot \binom {12} 1 ## overcounts the possibilities, as opposed to ##\binom {13} 2## , which gets it correct. You are already accounting for the fact that these combinations can be arranged any which way by multiplying these binomial coefficients.

    In the full house situation, the trips versus pair are distinguishable. Hence here it is correct to use ##\binom {13} 1 \cdot \binom {12} 1 ## .
  15. Jan 18, 2014 #14


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    Here's another way to look at it:

    The probability of getting a pair is 10 times the probability of your pair being in the first two cards. Because there are 10 = 5-choose-2 equally likely positions of the pair.

    So, calculate the probability of getting a pair in the first two cards, then nothing in the next 3 (no more pairs) and multiply this by 10.

    The probability of getting a pair in the first two cards is 1/17.

    The next card can be anything, except the number you have, so 48/50. The next card can be anything that doesn't match the first 3, so 44/49, and the last card can't match any of the first four, so 40/48.

    So the probability of getting a pair is 10*(48*44*40)/(17*50*49*48) = 10*(44*4)/(17*5*49) = 42.2%.
  16. Jan 18, 2014 #15
    thanks for the replies :) And I will stop with the annoying editing now (promise).

    Yes, I know that the two groups (pairs, in this case) are indistinguishable, unlike the full house event. What I can't see though is the formal reasoning behind why you will overcount if you use 13*12 instead of 13 choose 2.

    I mean, the order of the elements isn't an issue because you use 4 choose 2 when you draw the two pairs, so why do you need to use 13 choose 2?
  17. Jan 18, 2014 #16

    D H

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    Rhetorical question: What's the difference between a pair of twos and a pair of fives versus a pair of fives and a pair of twos? The answer is that there is no difference. They are the same two pair. Using (13 choose 1)*(12 choose 1) treats a pair of twos and a pair of fives as distinct from a pair of fives and a pair of twos. Using (13 choose 2) treats them as one and the same.

    If all else fails, count. Ignoring that fifth card and ignoring suits, how many distinct sets of two pairs are there in a deck? Let's count them. There's a pair of twos and a pair of threes, a pair of twos and a pair of fours, a pair of twos and a pair of fives, …, all the way up to a pair of twos and a pair of aces. That's 12 other pairs that can be paired with a pair of twos. Next, look at a pair of threes. They can be paired with a pair of fours, etc. Note that we already counted pair of threes and a pair of twos as a pair of twos and a pair of threes. A pair of threes and a pair of twos don't count as a distinct two pair. Add these up and you get 12+11+10+…+3+2+1=78, or 13 choose 2.
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