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You deal your friend five cards from a standard shuffled deck. He looks at his hand and says either "Oh! I have at least one $X$!" or "I don't have any $X$s," where $X$ is the name of a rank. Your friend never lies. What is the probability that he has a flush in each case?

A little thought reveals that this problem is not well-posed, at least without some understanding of how your friend is choosing the rank to tell you about. Four natural choices are:

  1. He chose $X$ before looking at his hand.
  2. He looked at his hand, chose a random card in it, and let $X$ be that card's rank.
  3. He looked at his hand, then chose $X$ at random from among the ranks in it.
  4. He looked at his hand, then chose $X$ at random from among the ranks not in it.

In which of these cases do you gain any information about whether he has a flush? How much?

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up vote 1 down vote accepted

As Ross has pointed out, the parity of the statement is known beforehand in cases $2$ to $4$, and the value of $X$ is irrelevant, so there's no information gain in these cases.

In case $1$, the probability of having a certain rank in a uniformly random poker hand is $1-(1-1/13)^5=122461/371293$. The probability of having a certain rank given a flush is $5/13$. The information you gain if he has the rank is measured by the Kullback–Leibler divergence

$$ P(\text{flush}\mid X)\log\frac{P(\text{flush}\mid X)}{P(\text{flush})}+P(\overline{\text{flush}}\mid X)\log\frac{P(\overline{\text{flush}}\mid X)}{P(\overline{\text{flush}})}\;. $$


$$ P(\text{flush})=\frac{4\binom{13}5}{\binom{52}5}=\frac{33}{16660}\approx0.00198 $$


$$ \frac{P(\text{flush}\mid X)}{P(\text{flush})}=\frac{P(X\mid\text{flush})}{P(X)}=\frac{5\cdot371293}{13\cdot122461}=\frac{142805}{122461}\approx1.166 $$


$$ \frac{P(\overline{\text{flush}}\mid X)}{P(\overline{\text{flush}})}=\frac{1-P(\text{flush}\mid X)}{1-P(\text{flush})}=\frac{16660-33\cdot142805/122461}{16660-33}=\frac{2035487695}{2036159047}\approx0.99967\;, $$

the divergence is

$$ \Delta_X=\frac{33}{16660}\cdot\frac{142805}{122461}\log\frac{142805}{122461}+\left(1-\frac{33}{16660}\cdot\frac{142805}{122461}\right)\log\frac{2035487695}{2036159047}\approx0.000025988 $$

(computation). If he doesn't have the rank, the divergence is

$$ P(\text{flush}\mid\overline X)\log\frac{P(\text{flush}\mid\overline X)}{P(\text{flush})}+P(\overline{\text{flush}}\mid\overline X)\log\frac{P(\overline{\text{flush}}\mid\overline X)}{P(\overline{\text{flush}})}\;. $$


$$ \frac{P(\text{flush}\mid\overline X)}{P(\text{flush})}=\frac{P(\overline X\mid\text{flush})}{P(\overline X)}=\frac{8\cdot371293}{13\cdot248832}=\frac{28561}{31104}\approx0.918 $$


$$ \frac{P(\overline{\text{flush}}\mid\overline X)}{P(\overline{\text{flush}})}=\frac{1-P(\text{flush}\mid\overline X)}{1-P(\text{flush})}=\frac{16660-33\cdot28561/31104}{16660-33}=\frac{172416709}{172388736}\approx1.00016\;, $$

the divergence is

$$ \Delta_{\overline X}=\frac{33}{16660}\cdot\frac{28561}{31104}\log\frac{28561}{31104}+\left(1-\frac{33}{16660}\cdot\frac{28561}{31104}\right)\log\frac{172416709}{172388736}\approx0.0000068215 $$

(computation). The expected information gain is therefore

$$ P(X)\Delta_X+P(\overline X)\Delta_{\overline X}\approx0.00001314\;, $$

or roughly $0.000019$ bits.

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In none of these cases do you gain information about a flush. For 1, he can always make the appropriate statement. For the others, you know the sense of the statement (have or don't have) before he looks at his hand.

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I agree about cases $2$ to $4$, but I don't understand your argument for case $1$. I thought I was in the process of calculating the information gain :-) – joriki Nov 2 '12 at 4:10
@joriki: Suppose $X$ is ace, selected without looking. Then he looks and has a flush with the expected probability. Then he says whether he has an ace. Where is the asymmetry between ranks that could supply any information? – Ross Millikan Nov 2 '12 at 4:15
That's why it's a conundrum. There isn't any asymmetry between ranks, but it still tells you something to know that he has no aces. I'm not even sure I agree with you about cases 2 to 4, but you're certainly not right about case 1. – mjqxxxx Nov 2 '12 at 4:32
@mjqxxxx: I see. The argument would be that having an ace is more probable if he has a flush as we know he has five different ranks. – Ross Millikan Nov 2 '12 at 4:37
Right, or vice versa: having no aces means he's more likely to have a pair (of something else) and hence no flush. – mjqxxxx Nov 2 '12 at 4:39

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