# Texas holdem meets the game of hearts

The actual issues of my question are complex but can be reduced to the metaphor of a game of texas holdem with a twist. The holder of the queen of spades always looses.

The actual scenario is that there are two players and all the cards are dealt for one hand only and then the players never see each other ever again. Suddenly one player says "you can have every card in the deck but in exchange if you have the queen of spades you must turn it over, which means that you lose"

The other player declines the deal.

My question is how to show mathematically that such declination (assuming the imperative of rational and self-interested behavior of winning) is mathematically equivalent to holding the queen of spades

• What does having every card in the deck mean? Might not that give me the queen of spades? – jlimahaverford Sep 16 '15 at 21:04
• I am regrettably not as precise as I should be. The point is that the person declining the deal has prior knowledge of the whether he or she is holding the queen of spades. Therefore the act of rationally declining the deal is equivalent (assuming rational behavior and a one time test) should be able to be shown as mathematically equivalent to possessing the queen of spades. (bayesian prior and likelihood analysis is the only way I am aware of showing this but I am unskilled at using this formulation to show proof of equivalency. – Lost In Space Sep 16 '15 at 21:57
• "you can have every card in the deck". What is the significance of having every card in the deck? – jlimahaverford Sep 16 '15 at 22:01
• The significance is that the declining side has absolute guarantee of winning if they do not have the queen of spades and absolute guarantee of losing if they do. – Lost In Space Sep 16 '15 at 22:06
• Then haven't you just answered your own question? – Glen_b Sep 17 '15 at 5:31

## 1 Answer

Let our probability space be ($\Omega, \mathscr{F}, \mathbb{P}$) where $\Omega$ be the sample space containing the possible hands in this game, $\mathscr{F} = 2^{\Omega}$ and $\mathbb{P}$ is however probabilities are assigned.

Let A be the event of having all the cards in the deck in one's hand, and let B be the event of having the queen of spades in one's hand ($A, B \subseteq \Omega; A, B \in \mathscr{F}$).

Note that $A \subseteq B$

By the monotonicity of probability, $P(A) \leq P(B)$. If P(A) = 1, then P(B) = 1.

Another way to look at it is $P(B|A) = P(A \cap B)/P(A) = P(A)/P(A) = 1$