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I've got two questions:

What’s the probability that someone else has a flush, given that you have a flush?

Notes: - there are four people at the table - we don't know anything about the kind of flush. all five cards could be out in front

If you have a flush, what’s the probability that someone has a better flush?

Both of these seem of quite a bit more complexity than any of the other such questions form I have done.

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If this is about Texas Hold'em then you should say so in the question. –  Henry Jan 25 '12 at 8:17
2  
It depends, how you phrase the question. If we are talking about Texas Hold'em which I assume we are. Even then what do you mean flush? Do you mean you have two spades and there are three spades on the board. Do you mean you have 1 spade and there is 4 spades on the board? This makes a big difference. Especially to your second question about probability someone has a better flush. –  simplicity Jan 25 '12 at 9:21
    
@simplicity, it sounds like the question is saying that all you know about your hand is that it is a flush. So it looks like you'd have to compute separate probabilities for all the cases of board/hand combinations. It seems like it'll be a lot of terms, but wont be too complicated. –  rapidash Jan 25 '12 at 12:18
    
This seems like the kind of question that is better suited to monte carlo approximation than it is to direct computation. –  Chris Taylor Jul 30 '12 at 13:56

1 Answer 1

I'll only answer the first question. The probability that someone has a better flush can be calculated with similar methods, but the calculation is considerably more complicated.

I presume that you don't mean an actual game of Texas hold 'em, in which there would have been betting and folding at earlier stages, which would render the assumption of uniform distribution of hands highly unrealistic; so this is about a fictitious game in which five cards are dealt to the table and two cards are dealt to each hand, with no intermediate betting.

There are three possibilities: There might be $3$, $4$ or $5$ cards of the same suit on the table.

The probability for $5$ cards of the same suit on the table is

$$4\frac{\binom{13}5}{\binom{52}5}=\frac{33}{16660}\approx0.002\;.$$

In this case everyone has a flush.

The probability for exactly $4$ cards of the same suit on the table is

$$4\frac{\binom{13}4\binom{39}1}{\binom{52}5}=\frac{143}{3332}\approx0.043\;.$$

In this case, your probability of having one card of the suit is

$$\frac{\binom91\binom{38}1}{\binom{47}2}=\frac{342}{1081}\approx0.316\;,$$

and

$$\frac{\binom92}{\binom{47}2}=\frac{36}{1081}\approx0.033$$

for two cards of the suit. If you have one card, the probability of someone else having at least one is

$$ 1-\frac{\binom{37}6}{\binom{45}6}=\frac{6299}{8815}\approx0.715\;, $$

whereas if you have two cards, it is

$$ 1-\frac{\binom{38}6}{\binom{45}6}=\frac{23309}{35260}\approx0.661\;. $$

The probability for exactly $3$ cards of the same suit on the table is

$$4\frac{\binom{13}3\binom{39}2}{\binom{52}5}=\frac{2717}{8330}\approx0.326\;.$$

In this case your probability of having two cards of the suit is

$$\frac{\binom{10}2}{\binom{47}2}=\frac{45}{1081}\approx0.042\;.$$

The probability of someone else then also having two cards of the suit is a bit more complicated to calculate. We can get it by applying the inclusion–exclusion principle:

$$3\frac{\binom82}{\binom{45}2}-3\frac{\binom84}{\binom{45}4}+\frac{\binom86}{\binom{45}6}=\frac{8091}{96965}\approx0.083\;.$$

To put this all together, we have to divide the probability of you and someone else having a flush by the probability of you having a flush:

$$\frac{\frac{33}{16660}+\frac{143}{3332}\left(\frac{342}{1081}\cdot\frac{6299}{8815}+\frac{36}{1081}\cdot\frac{23309}{35260}\right)+\frac{2717}{8330}\cdot\frac{45}{1081}\cdot\frac{8091}{96965}}{\frac{33}{16660}+\frac{143}{3332}\left(\frac{342}{1081}+\frac{36}{1081}\right)+\frac{2717}{8330}\cdot\frac{45}{1081}}=\frac{145640554}{323494633}\approx0.450$$

(computation). Thus the chance of someone having a flush if you have one is about $45\%$. This is confirmed by computer simulations.

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