Probability

[llitle]

Calculate the probability of being dealt the following poker hand. (Recall that a poker player is dealt 5 cards at random from a standard deck of 52.) Express your answer as a decimal rounded to four decimal places. HINT [See Example 3.]
Two pair: Two cards with one denomination, two with another, and one with a third. Example: 3, 3, Q, Q, J

 

[Deleted member 4993]

Calculate the probability of being dealt the following poker hand. (Recall that a poker player is dealt 5 cards at random from a standard deck of 52.) Express your answer as a decimal rounded to four decimal places. HINT [See Example 3.]
Two pair: Two cards with one denomination, two with another, and one with a third. Example: 3, 3, Q, Q, J

First card can be any card ? P = 1

second card must one of those 3 ? P = 3/51

Third card can be anything else ? P = 48/50

Fourth card must be one of 3 ? P = 3/49

Now what .....

 

[soroban]

Hello, llitle!

Calculate the probability of being dealt the following poker hand.
Express your answer as a decimal rounded to four decimal places.

Two pair: Two cards with one denomination, two with another, and one with a third.
Example: 3, 3, Q, Q, J

\(\displaystyle \text{There are: }\,{52\choose5} \:=\:2,\!598,\!960\text{ possible poker hands.}\)


\(\displaystyle \text{We want a hand with the pattern: }\:\{X,X,Y,Y,Z\}\)

\(\displaystyle \text{There are: }\:{13\choose2} \,=\,78\text{ choices for the values of }X\text{ and }Y.\)
. . . \(\displaystyle \text{There are: }\:{4\choose2} \,=\,6\text{ ways to get the two }X\text{'s.}\)
. . . . . . \(\displaystyle \text{There are: }\:{4\choose2} \,=\,6\text{ ways to get the two }Y\text{'s.}\)
. . . . . . . . . \(\displaystyle \text{There are }{44\text{ choices for }Z.\)

\(\displaystyle \text{Hence, there are: }\:78\cdot6\cdot6\cdot44 \:=\:123,\!552\text{ hands with Two Pairs.}\)


\(\displaystyle \text{Therefore: }\(\text{Two pair}) \:=\:\frac{123,\!552}{2,\!598,\!960} \;\approx\;0.0475\)

 

[Deleted member 4993]

Soroban,

Does your analysis exclude "four of a kind"?

 

[soroban]

Hello, Subhotosh!

Does your analysis exclude "four of a kind"?

\(\displaystyle \text{Yes, I deliberately chose }two\;di\!f\!\!f\!erent\;values\text{ for the two pairs by using }\,{13\choose2}\)