Probability question

[Daniel_Feldman]

Never mind, I've figured it out.

The probability of getting a two pair needs to have a 2! in the denominator, because getting pair A and then pair B is the same as B then A. That was my error.

 

[Denis]

Ever heard of google?

http://en.wikipedia.org/wiki/Poker_probability

 

[Daniel_Feldman]

No, never.

Google isn't going to make me understand these concepts if I have it wrong.

 

[Denis]

No, never.
Google isn't going to make me understand these concepts if I have it wrong.

Huh? A google search will give you loads of sites that explain concepts.

To make you "understand" requires classroom environment...not available here.

Did you read "Read before posting"?

 

[soroban]

Hello, Daniel!

I'll re-order the questions . . . in order of increasing difficulty.

$\displaystyle \text{Yes, the number of possible hands is: }\:{52\choose5} \:=\:2,\!598,\!960$

This is the denominator for all the questions.

Assuming that all poker hands are equally likely (i.e., 52C5), then find the probability of being dealt:

(a) Flush (5 cards, all of the same suit)

Your reasoning is correct!

$\displaystyle \text{There are }4\text{ choices of suits, then }{13\choose5} \text{ ways to get a flush in that suit.}$

$\displaystyle \text{Therefore, the numerator of }P(\text{Flush})\text{ is: }\:4\cdot{13\choose5}$

(e) Four of a Kind

There are 13 choices for the value of the four-of-a-kind.
. . And 48 choices for the fifth card.

$\displaystyle \text{Therefore, the numerator of }P(\text{4-of-a-Kind})\text{ is: }\:13\cdot48$

(d) Three of a Kind

There are 13 choices for the value of the Triple.
$\displaystyle \text{There are: }{4\choose3} = 4\text{ ways to get the Triple.}$

The last two cards must not match the Triple or each other.
. . $\displaystyle \text{There are: }\frac{48\cdot44}{2}\text{ ways.}$

$\displaystyle \text{Therefore, the numerator of }P(\text{3-of-a-Kind})\text{ is: }\:13\cdot4\cdot\frac{48\cdot44}{2}$

(c) Two Pairs

$\displaystyle \text{There are: }\:{13\choose2}\text{ choices of the values of the Two Pairs.}$

$\displaystyle \text{There are: }\:{4\choose2}{4\choose2}\text{ ways to get the Two Pairs.}$

$\displaystyle \text{The fifth card card must not match either of the Pairs: }\:44\text{ choices.}$

$\displaystyle \text{Therefore, the numerator of }P(\text{Two Pairs}) \text{ is: }\:{13\choose2}{4\choose2}{4\choose2}\cdot44$

(b) One Pair

$\displaystyle \text{There are 13 choices for the value of the Pair.}$

$\displaystyle \text{There are }{4\choose2}\text{ ways to get the Pair.}$

$\displaystyle \text{The last three cards must not match the Pair or each other: }\:\frac{48\cdot44\cdot40}{3!}\text{ ways.}$

$\displaystyle \text{Threfore, the numerator of }P(\text{One Pair}) \text{ is: }\:13\cdot{4\choose2}\cdot\frac{48\cdot44\cdot40}{3!}$

 

[Denis]

THERE!! Soroban just added another "site" for google to find... :wink: