0
$\begingroup$

What is the intuition behind, 'What is the probability of obtaining a hand with TWO PAIRS in a standard 5 card game of poker?'


I know the solution is,

$$\frac{\binom{13}{2}\binom{4}{2}\binom{4}{2}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}} = 0.047539$$

I understand the reasoning behind $\binom{11}{1}\binom{4}{1}$. That is, the number of ways a single card (After the two pairs have been selected) can be selected is $11\cdot4$.

However, the $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ part confuses me a little. Why doesn't $\frac{\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$ work? I've read that it counts solutions twice, but I don't understand how? How is that $\binom{13}{2}$ - 78 cards can be selected amongst $\binom{4}{2}\binom{4}{2}$ - 36 suits?

$\endgroup$
0
$\begingroup$

You have to look at it like this:

$\binom{13}2 = \frac{13\cdot12}{2} = 78$ ways of choosing $2$ cards of different ranks.

If you do $\binom{13}1\binom{12}1$, you get double the value of $156$.
You are double counting, say ace - king and king - ace

Similarly, $\binom42 = 6$ gives ways of selecting $2$ suits from $4$
This you can easily count as $S-H,\; S-D,\; S-C,\; H-D,\; H-C\;$ and $D-C$
The way you were writing will obviously overcount

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think I just got a little caught up with the $\binom{13}2$ part I didn't actually think about what $\binom{4}2$ meant. Thank you! $\endgroup$ – user301837 Jan 2 '16 at 9:09
0
$\begingroup$

$\binom {13}2$ is the number of ways to choose the two denominations for the two pairs. $\binom 42$ in each case is the number of ways of choosing a pair of the given denomination.

For your own computation you have $\binom 41$ for the pairs when you need $\binom 42$. Also $\binom {13}1\binom {12}1$ counts the choices $(2,1)$ and $(1,2)$ for the pairs as separate, but they produce the same hand - hence the need to divide by $2$ as in $\binom {13}2$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So by doing $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ simultaneously I effectively compensate for the repetition? i.e. $(2, 1)$ being the same as $(1,2)$ $\endgroup$ – user301837 Jan 2 '16 at 9:06
  • $\begingroup$ @user301837 That's it. $\endgroup$ – Mark Bennet Jan 2 '16 at 9:08
0
$\begingroup$

First pick the ranks of the first pair and second pair. There are 13 choices, then choose which two you want of those of each rank. This gives $\binom{13}{2}\binom{4}{2}\binom{4}{2}$ in total.

$\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}$ doesn't work because choosing one rank and then choose another separately implies that the pairs aren't the same type in the sense that the order is not interchangeable. For example, it is saying that $AAKKQ$ is not the same as $KKAAQ$. Also, you are choosing two per rank, not one, thus it should be $\binom{4}{2}$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Blockquote Why doesn't $\frac{\binom{13}{1}\binom{4}{1}\binom{12}{1}\binom{4}{1}\cdot\binom{11}{1}\binom{4}{1}}{\binom{52}{5}}$ work? I've read that it counts solutions twice, but I don't understand how?

I think that you are understanding the problem "step by step", and I'll give you a "step by step" solution:

1) Choose the "type-number" for the first pair: $\binom{13}{1}$

2) For this pair, choose the "type-symbol" for the first card: $\binom{4}{1}$. For the second card, now you have only 3 type-symbols for choose. Then, there are $\binom{3}{1}$ ways to select the second card. Thus, in principle, there are $\binom{4}{1}\binom{3}{1}$ ways to select the "type-symbol" of the first pair. But there is not difference between select the first or the second card for this pair. Then, really there are $\binom{4}{1}\binom{3}{1}/2$ ways to do this selection. Thus, there are $\frac{\binom{13}{1}\binom{4}{1}\binom{3}{1}}{2}$ ways to choose the first pair.

3) In the same spirit, there are $\frac{\binom{12}{1}\binom{4}{1}\binom{3}{1}}{2}$ ways to choose the second pair. Note that the first factor is $\binom{12}{1}$ and not $\binom{13}{1}$. This is due to the fact that, once you choose the type-number for the first pair, there are only 12 kinds of type-symbol for the second pair.

Then, there are, in principle, $\frac{\binom{13}{1}\binom{4}{1}\binom{3}{1}}{2}*\frac{\binom{12}{1}\binom{4}{1}\binom{3}{1}}{2}$ ways to choose both pairs.

4) But, as in step (2), there isn't difference between select the first and the second pair or viceversa. Then, dividing by 2, the answer is $$\dfrac{\frac{\binom{13}{1}\binom{4}{1}\binom{3}{1}}{2}*\frac{\binom{12}{1}\binom{4}{1}\binom{3}{1}}{2}}{2}$$

5) Finally, for the last card there are $\binom{4}{1}\binom{11}{1}$ ways to select it.

Thus, the result is $$\dfrac{\frac{\binom{13}{1}\binom{4}{1}\binom{3}{1}}{2}*\frac{\binom{12}{1}\binom{4}{1}\binom{3}{1}}{2}*\binom{4}{1}\binom{11}{1}}{2}=123552$$

As you can see, the trick is that, in order to avoid the "order select of the types", you can select the distinc types in only one step using correctly the binomial coefficients

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

The key to correctly applying the combinatorics is to acknowledge how many kinds of MULTIPLICITY - four-suit ("quad"), three-suit ("triple" or "set"), two-suit ("pair" ), or one-suit ("singleton") - exist in a hand AND how many times EACH one occurs.

For example, in a five-card full-house hand, TWO multiplicities exist, once each, a triple and a pair. Thus TWO ranks are selected from the thirteen available using $\binom{13}{2} = 78$. Then, each selected rank (say, K and Q) can be used in EITHER multiplicity to make DISTINCT full-house hands - KKK full of QQ, or vice versa - which is SECOND choice, $\binom{2}{1}=2$.

But, in a six-card full house of the form AAABBB of ranks A and B, both of which are triples (and, of course, the higher of which is strategically played as the rank "full of" the lower) the second choice would not exist, so the "$\binom{2}{1}=2$" multiplication would not be performed.

Likewise in a six-card three-pair scenario $\binom{13}{3} = 286$ would indeed select the three ranks of each pair, but because they are ALL pairs (the same multiplicity) no second choice under "$\binom{3}{1}=3$" or "$\binom{3}{2}=3$" could possibly attach.

But a seven-card THREE-pair hand with ONE singleton would require $\binom{13}{4}\binom{4}{3} = 715*4 = 2860$ ways to fully assign ALL ranks to ALL possible variants.

One final example: ten-card four-of-a-kind of the form AAAABBBCCD: it contains four multiplicities - a quad, a triple, a pair, and a singleton, once each. So $\binom{13}{4}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1} = 17160$ ways to conceive of distributing all selectable ranks through all multiplicities.

If it had been a quad of the form AAAABBCCDD, the result would be only $\binom{13}{4}\binom{4}{1}\binom{3}{3} = 2860$ distinct possible rank-multiplicity assignments. This explains why our (intuitive) experience is that triples and quads with singletons, rather than with multiple pairs, are the most likely.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Please consider formatting your post in order to make it readable. $\endgroup$ – Mårten W Jun 23 '16 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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