# Given a pair of $2$'s from a deck with $52$ cards, count the ways that among $9$ other cards certainly restricted there is a higher pair

Suppose we have a deck with $52$ cards, as usual in four suits with ranks $2,3,\ldots, 10, J, Q, K, A$. Suppose we take two $2$'s aside, so that we have $50$ cards left. Then we take two times two random cards more aside, so that we have stacks of cards with two cards each. Call them stack 1 and stack 2. So in total we have put $6$ cards aside, the initial two $2$'s, and two separate stacks consisting of two cards each.

Then we draw five cards more, lets call them community cards.

Now how many ways are there that any of the two stacks together with the community cards contain two cards of the same rank higher than $2$, where as a further restriction at least one of the cards must be from the stack?

We count ways in such a way that the order among the stacks, and the community cards does not matter. This question came up in connection with the game of poker (the Texas Hold'em variant), to phrase this problem in poker language: Suppose we have a pocket pair with $2$'s, and we play against two opponents. How many ways are there that any opponent hits a pair higher than mine.

To give an example. If we have the following distribution

Community cards: 7 J K K 2 Stack 1: J 3 Stack 2: A 4

Then Stack 1 can form the higher pair J J. If for example Stack 1 contains 5 3 instead of J 3, then there is no higher pair among the stack. Note that the pair K K in the community cards does not count, as one requirement is that at least one card from the stacks must be used. Furthermore we cannot form a pair from the two stack, i.e. just Stack 1 + Community cards, or Stack 2 + Community cards.

I have computed it for the case of one stack (or one opponent), but I run into trouble when trying to count the possibilites for two opponets (all my calculations must be wrong as when I determine the probability it gives me wrong results).

We are just looking at who has the highest pair. i.e. your opponent pairs up his aces, but you pick up another 2. Pair of Aces wins this game.

Lets look at a 2 player game first.

Scenario 1.

Your opponents has a higher pocket pair.

$\frac {48\cdot3}{50 \cdot 49})\approx 5.6\%$

Scenario 1.1 (remote but not impossible).

Your opponent also has a pair of 2s, and you push $\frac {2}{50\cdot 49} \approx 0.08\%$

Scenario 2.

Your opponent does not hold a pair but one of his hole cards is a 2.

$\frac {2\cdot2\cdot 48}{50\cdot 49}$

And he fails pairs the non-2 hole card

$\frac {45 \choose 5}{48\choose 5} \approx 81\%$

Pairs the non-2 hole card

$1-\frac {45 \choose 5}{48\choose 5} \approx 28.6%$

Scenario 3

Your opponent does not pair and neither card is a 2. \frac {48\cdot 44}{50\cdot 49}

he pairs either hole card

$1-\frac {42 \choose 5}{48\choose 5} \approx 50.3\%$

You have a $51\%$ chance of losing heads up.

Make this a 3 player game.

You have a $51.5\%$ chance of losing to either opponent, (and a chance of losing to both)

$\approx 0.485^2 = 23.5\%$ chance of winning or tying.

• Thank you, but could you please give more details about your derivation for three players, as this was the part I stuck. PS: Regarding your first sentence you are right! Commented Jan 12, 2017 at 0:46
• Independent expectation. Player 1 has a 51% chance of beating you regardless of the cards in Player 2's hand. And Player 2 has a 51% chance of beating you. Either 1 beats you and 2 doesn't (0.51)(0.49) or 2 beats you and 1 doesn't + (0.49)(0.51) or both beat you (0.51)(0.51) and that sum equals 1-(0.49)^2 Commented Jan 12, 2017 at 0:51
• Your sum does not equals $1-0.49^2$... and thanks for providing the probabilities, but I am also interesting in how you count the events, as written in the question I am asking for the number of ways the opponent wins. Commented Jan 12, 2017 at 1:10
• I am also unsure if the events are independent, as I guess that is used in your comment. For if Player 1 hits something, he takes outs of Player 2 away... Commented Jan 12, 2017 at 10:18