To show that there are two people in the tournament with the same starting hand and position in a Texas Hold'em poker tournament.

Question

A Texas Hold'em poker tournament has $1557$ players, in $9$ distinct positions at $173$ tables.

Each table has a standard $52$ card deck, and each player is dealt $2$ cards. Hands are considered "the same” if they have the same rank of cards and the same number of different suits. For example, the nine of hearts and three of hearts are the same as the nine of clubs and three of clubs, but neither are the same as the nine of spades and three of diamonds.

$(a)$ Show that there are two people in the tournament with the same starting hand and position.

$(b)$ Show that, for every position, there are two people with the same starting hand.

$(c)$ Show that there are two people who are dealt identical cards (not just "the same”, but with ranks and suits matching).

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What I have done so far?

$(b)$ From Texas hold 'em starting hands I learnt that there are $169$ non-equivalent starting hands in hold 'em. Since there are $173$ tables by the pigeonhole principle, for every position, there are two people with the same starting hand.

$(c)$ Total there are $\binom{52}{2} = 1326$ distinct possible combinations of two hole cards from a standard 52-card deck in hold 'em. Since there are $1557$ players, again using the pigeonhole principle, we can conclude that there are two people who are dealt identical cards.

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Can someone help me in doing the part $(a)$? Also, if the arguments for parts $(b)$ and $(c)$ are correct?

Thanks.

Answer 1

There are $(13)$ distinct ways of having a pair.

There are $\displaystyle \binom{13}{2} \times 2$ distinct ways of not having a pair, where the 2nd factor reflects that the non-pair may or may not be suited.

So, the analysis for part (b) is valid.

The Math in part (c) is accurate and the analysis in part (c) is therefore also valid.