1. ## cards

A gambler has been dealt with 5 cards: 2 aces, 1 king, 1 five and 1 nine. he discards the 5 and 9 and takes 2 more cards. what is the probability that he ends with a full house?

my working:

to get a full house: its either

3 aces and 2 kings or 2 aces and 3 kings
P ( 3 aces and 2 kings ) = 2/47 x 3 /46 x 2
p (2 aces and 3 kings ) = 2/47 x 3 /46

so i think the answer should be the sum of both...but the answer given was 0.0083..

2. Originally Posted by alexandrabel90
A gambler has been dealt with 5 cards: 2 aces, 1 king, 1 five and 1 nine. he discards the 5 and 9 and takes 2 more cards. what is the probability that he ends with a full house?

my working:

to get a full house: its either

3 aces and 2 kings or 2 aces and 3 kings
P ( 3 aces and 2 kings ) = 2/47 x 3 /46 x 2
p (2 aces and 3 kings ) = 2/47 x 3 /46

so i think the answer should be the sum of both...but the answer given was 0.0083..
Hi alexandrabel90,

he will have a full house if he chooses 1 ace from the 2 remaining and 1 king from the 3 remaining..

The probability of this is $\displaystyle \frac{\binom{2}{1}\binom{3}{1}}{\binom{47}{2}}$

he will also have a full house if he choose 2 kings from the remaining 3.

The probability of this is $\displaystyle \frac{\binom{3}{2}}{\binom{47}{2}}$

3. When using the logic you were following, you must break it up into the different possibilities or ways the event can happen.

In choosing an ace and a king,
the ace can be chosen first followed by the king,
or the king can be chosen first followed by the ace.

Remember, 2 aces and a king have already been taken from the pack,
so 2 aces and 3 kings remain to choose from among the available 47.

hence the probability of picking a king and ace is

$\displaystyle \frac{3}{47}\frac{2}{46}+\frac{2}{47}\frac{3}{46}$

The probability of picking 2 more kings instead is

$\displaystyle \frac{3}{47}\frac{2}{46}$

Hence the probability of a full house is $\displaystyle \frac{3(6)}{47(46)}$