Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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

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

With

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

and

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

and

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

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

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

With

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

and

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

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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.