In poker, there is a 52 card deck with 4 cards each of each of 13 face values. A full house is a hand of 5 cards with 3 of one face value, and 2 of another. What is the probability that a random poker hand is a full house? You can leave your answer in terms of bionomial co-efficients and similar factors, but please explain each term.

Respuesta :

Answer:

[tex]P(FH) = \frac{\binom{13}{1}*\binom{4}{3}*\binom{12}{1}*\binom{4}{2}}{\binom{52}{5}}[/tex]

P(FH) =0.00144

Step-by-step explanation:

The total number of possible poker hands (n) is the combination of 5 out of 52 cards:

[tex]n=\binom{52}{5}[/tex]

In order to get a full house, first pick one out of 13 cards (13 choose 1), then pick 3 out of the 4 cards of the chosen type to form three of a kind (4 choose 3), now pick another card from the remaining 12 different numbers or faces (12 choose 1), then pick 2 out of the 4 suits to form a pair (4 choose 2)

The probability of getting a fullhouse, in binomial coefficients is:

[tex]P(FH) = \frac{\binom{13}{1}*\binom{4}{3}*\binom{12}{1}*\binom{4}{2}}{\binom{52}{5}}[/tex]

Expanding the coefficients and solving:

[tex]P(FH) = \frac{13*4*12*\frac{4*3}{2*1} }{\frac{52*51*50*49*48}{5*4*3*2}}\\P(FH) =0.00144[/tex]