Probability question

D

Daniel_Feldman

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Sep 30, 2005
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252
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.
 
Daniel_Feldman said:
No, never.
Google isn't going to make me understand these concepts if I have it wrong.
Click to expand...
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"?
 
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)
Click to expand...

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

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

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

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


(b) One Pair
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\(\displaystyle \text{There are 13 choices for the value of the Pair.}\)

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

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

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

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