# How does opponent's flush affect odds of your full house?

A cute probability intuition test:

Let $f$ be the probability of being dealt a full house in a five-card poker hand, from a randomly shuffled standard deck. ($f \approx 0.001468$).

Now look at the case of two players being dealt hands, and player one shows that she has a flush. Now what is the probability $f_2$ that the five cards dealt to player 2 is a full house?

It seems clear that we will have $f_2 < f$ because if one player has a full house it must be a tiny bit more likely that the other has pairs and other clumps of the same rank, and a flush has none of those. But to test your intuition, how small is that effect?

Is $f_2$ more than 99% of $f$? More than 95%? More than 90%? or less than 90%?

EDIT I had written $h$ for "house" in the first sentence. Then I used $f$ for "full" later.

• What does $f$ refer to? Is it the same as $h$? – Barry Cipra Mar 24 '16 at 1:00

## 1 Answer

Nice question. I'll give my spontaneous intuitive answer, a quick estimate and then a spoiler-proof exact calculation.

My spontaneous intuitive answer was, like you wrote: "a tiny bit" – something like $99\%$ or $98\%$.

Then I did a quick estimate: The hard part of the full house is the three of a kind. Instead of $13$ ranks of $4$, there are now only $8$ ranks of $4$ and $5$ ranks of $3$, and it's only $\frac14$ as likely to get three of a kind in the ranks of $3$, so the factor should be something like $(8+\frac54)\div13\approx71\%$.

Here's the exact calculation:

There are $5\cdot4\cdot\binom33\cdot\binom32=60$ ways to form both the triple and the pair from a depleted rank. There are $5\cdot8\cdot\binom33\cdot\binom42=240$ ways to form only the triple from a depleted rank.
There are $5\cdot8\cdot\binom43\cdot\binom32=480$ ways to form only the pair from a depleted rank.
There are $8\cdot7\cdot\binom43\cdot\binom42=1344$ ways to form neither the triple nor the pair from a depleted rank. Thus the probability is $$\frac{60+240+480+1344}{\binom{47}5}=\frac{708}{511313}\approx0.001385\;,$$ so the factor is about $94\%$. That raises the question why my quick estimate was so wrong. :-)

• It sounds like in your quick estimate, you are thinking about the likelihood of "getting a full house if there are only 8 ranks of 4" vs. "getting a full house when all cards are available." Rather, shouldn't we be comparing it to "getting a full house when five random cards are missing"? In this sense, while having only 8 ranks of 4 is severe, it is only a little more severe than, for example, your opponent having 1 pair. – Michael Harrison Mar 24 '16 at 1:58
• @MichaelHarrison: Quite right, well spotted :-) – joriki Mar 24 '16 at 2:01
• @MichaelHarrison: So I was dropping a factor $\frac{\binom{47}5}{\binom{52}5}\approx59\%$, which is in fact more significant than the estimated change in the triple probabilities -- talk about failing a probability intuition test :-) So we need to take the pairs into account -- an updated "quick" estimate: The pairs are only half as likely in the depleted ranks, so they contribute a factor of about $(8+\frac52)\div13\approx81\%$. Combining the three factors yields a factor of about $97\%$ -- not quite as bad :-) – joriki Mar 24 '16 at 2:08
• Nice answer; this is what I was getting at -- why our estimation was so hinky. BTW, my gut poker instinct said this would be 98-99% which is why I was a bit surprised at the answer (yours is correct). – Mark Fischler Mar 27 '16 at 13:28
• But @MichaelHarrison, the likelihood of a full house when all cards are available is exactly the same as the likelihood of a full house with five random cards missing. Just as the likelihood of the first five cards being a full house is the same as the likelihood of the sixth through tenth cards being a full house. – TonyK Mar 27 '16 at 13:48