# poker probability

• Oct 27th 2012, 01:18 PM
Phoebert
poker probability
In order to find the probability of dealing a full house in a 5 card poker hand, you use the folowing:
(13-choose-1)x(4-choose-3)x(12-choose-1)x(4-choose-2) / (52-choose-5)

But in order to find the probability of dealing 2 pairs, you use the this:
(13-choose-2)x(4-choose-2)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)

What I don't understand is why doesn't the 2 pairs probability method work the same way as the full house method (choosing 1 from 13 then 1 from 12 rather than choosing 2 from the 13 at the start) and work as follows:
(13-choose-1)x(4-choose-2)x(12-choose-1)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)
• Oct 27th 2012, 02:13 PM
Plato
Re: poker probability
Quote:

Originally Posted by Phoebert
What I don't understand is why doesn't the 2 pairs probability method work the same way as the full house method (choosing 1 from 13 then 1 from 12 rather than choosing 2 from the 13 at the start) and work as follows:
(13-choose-1)x(4-choose-2)x(12-choose-1)x(4-choose-2)x(11-choose-1)(4-choose-1) / (52-choose-5)

Have a look at this webpage.
• Oct 27th 2012, 02:59 PM
Soroban
Re: poker probability
Hello, Phoebert!

I assume you understand the purpose of each of the factors.

There is a subtle difference in the two methods you mentioned.

Quote:

In order to find the probability of dealing a Full House in a 5-card poker hand,

you use the following: . $\dfrac{{13\choose1}{4\choose3}{12\choose1}{4 \choose2}} {{52\choose5}}$

Choose one of the 13 values for the Triple: ${13\choose1}$ ways.
Choose 3 of the 4 cards of that value: ${4\choose3}$ ways.
Choose one of the other 12 values for the Pair: ${12\choose1}$ ways.
Choose 2 of the 4 cards of that value: ${4\choose2}$ ways.

There are: . ${13\choose1}{4\choose3}{12\choose1}{4\choose2}$ ways to get a Full House.
. . and divide by the number of 5-card hands: ${52\choose5}$
. . to get the probability.

Quote:

But in order to find the probability of dealing Two Pairs,

you use the this: . $\dfrac{{13\choose2}{4\choose2}{4\choose2}{11 \choose1}{4\choose1}}{{52\choose5}}$

Choose 2 of the 13 values for the Two Pairs: ${13\choose2}$ ways.
Choose 2 of the 4 cards of one value: ${4\choose2}$ ways.
Choose 2 of the 4 cards of the other value: ${4\choose2}$ ways.
Choose 1 of the other 11 values: ${11\choose1}$ way.
Choose 1 of the 4 cards of that value: ${4\choose1}$ way.

There are: . ${13\choose2}{4\choose2}{4\choose2}{11\choose1} {4\choose1}$ ways to get two Two Pairs.
. . divide by the number of 5-cards hands: ${52\choose5}$
. . to get the probability.

Quote:

What I don't understand is why doesn't the 2 pairs probability method work the same way as the Full House method
(choosing 1 from 13 then 1 from 12, rather than choosing 2 from the 13 at the start) and work as follows:

. . $\frac{{13\choose1}{4\choose2}{12\choose1}{4\choose 2}{11\choose1}{4\choose1}}{{52\choose5}}$

Your method produces a number twice as large as necessary.
Here's the reason why.

The original method chooses 2 values from the 13 available values.
There are: ${13\choose2} \:=\:78$ possible choices for values.

We can list them if we like:

. . $\begin{array}{c}A2\;A3\;A4\;A5\;\cdots\;AQ\;AK \\ 23\;24\;25\;\cdots\;2Q\;2K \\ 34\;35\;\cdots\;3Q\;3K \\ \vdots \\ JQ\;JK \\ QK \end{array}$

Your method has: ${13\choose1}{12\choose1} \:=\:156$ possible choices for values.

. . $\begin{array}{c}A2\,A3\,A4\,A5\,\cdots\,AQ\,AK \\ 2A\;23\;24\;25\;\cdots\;2Q\;2K \\ 3A\;32\;34\;35\;\cdots\;3Q\;3K \\ \vdots \\ K\!A\,K\!2\,K\!3\,K\!4\,\cdots\,K\!J\,K\!Q \end{array}$
Your list considers $A2$ to be different from $2A$.