1

Suppose three people are playing poker. Player1 has the cards 3(diamonds) and 10(spades).

The three cards on the table are 10(clubs), King(spades) and 3(spades).

How many hands can Player2 or Player3 have that so that both of their hands beat Player1s hand?

I have calculated that, for example Player2, alone can have 33 different pokerhands which beats Player1s hand. However, I can't figure out how this will change if the third player also must have a hand which beats Player1s hand.

Rules of poker: https://www.contrib.andrew.cmu.edu/~gc00/reviews/pokerrules

  • 1
    A flush is impossible given the cards on the table. A straight requires a jack and a king. Four of a kind and full house are impossible. A triple requires two aces, two queens or two tens. Beating player $1$ with two pairs requires an ace and either a queen or a ten. There are three aces, two queens and two tens left in the deck. Thus there are $4\cdot4+\binom32+\binom22+\binom22+3\cdot4=33$ hands that beat player $1$. How did you arrive at $27$? – joriki Sep 23 '18 at 21:23
  • Thanks @joriki I completely forgot that pair of aces and pair of tens also beats Player1. I was only counting with pair of aces and pair of queens. Good that your were attentive! –  Sep 23 '18 at 21:28
  • 2
    This is not how this site is to be used. You are not allowed to deface your question and you are not allowed to delete it after you gotten your answer! – Winther Sep 24 '18 at 08:37

1 Answers1

0

A flush is impossible given the cards on the table. A straight requires a jack and a king. Four of a kind and full house are impossible. A triple requires two aces, two queens or two tens. Beating player $1$ with two pairs requires an ace and either a queen or a ten. There are three aces, two queens and two tens left in the deck. Thus there are $4\cdot4+\binom32+\binom22+\binom22+3\cdot4=33$ hands that beat player $1$. As a first approximation, players $2$ and $3$ can have any ordered pair of these, of which there are $33^2=1089$.

Now we need to subtract the pairs that they can't have simultaneously because they overlap, either in two cards or in one card.

The overlap in two cards is easy; that means that they actually have the same hand, so for that we just have to subtract $33$.

The overlap in one card is a bit more complicated. There are $4$ winning hands with each jack, $4$ with each king, $6$ with each ace, $4$ with each queen and $4$ with each ten. (We can check this by checking that, as each hand contains $2$ cards, counting the hands by cards yields twice the number of hands: $4\cdot4+4\cdot4+3\cdot6+2\cdot4+2\cdot4=2\cdot33$.). So, since to form a pair of hands with one overlapping card we can choose the overlapping card and then an ordered pair of distinct cards that form a winning hand with it, there are $4\cdot4\cdot3+4\cdot4\cdot3+3\cdot6\cdot5+2\cdot4\cdot3+2\cdot4\cdot3=234$ ways for the hands to overlap in one card.

Thus, the total number of combinations of hands that players $2$ and $3$ can have such that they both beat player $1$ is $33^2-33-234=822$.

The probabilities for these events to occur are

$$ \frac{33}{\binom{47}2}=\frac{33}{1081}\approx3\% $$

for a single player to beat player $1$ and

$$ \frac{822}{\binom{47}2\binom{45}2}=\frac{137}{179365}\approx0.08\% $$

for two players to both beat player $1$.

joriki
  • 238,052
  • Very well done! Thank you @joriki –  Sep 24 '18 at 08:10
  • Could you be so kind to explain this part again:

    "So, since to form a pair of hands with one overlapping card we can choose the overlapping card and then an ordered pair of distinct cards that form a winning hand with it, there are 4⋅4⋅3+4⋅4⋅3+3⋅6⋅5+2⋅4⋅3+2⋅4⋅3"

    For example were do we get the "5" from in 365. I get that there are 6 winning hands for each of the 3 aces, but why multiply by 5? @joriki

     –  Sep 24 '18 at 20:55