I am very bad at probability problems in math. There is a homework question such that:
During a game of poker, you are dealt a five-card hand at random. With the convention that aces may count high or low, show that;
P(one pair) = 0.46
I dont know this is easy or not, but I am working on this question for a day. Can you give me some intution or show me how to solve this?
Any help would be appreciated.
On
To calculate the probability of having a pair: $P(\textrm{pair}) = 1 - P(\textrm{no pair}) = 1 - \frac{48}{51} \frac{44}{50} \frac{40}{49} \frac{36}{48} = \frac{2053}{4165} \approx 0.4929$.
To calculate the probability of having exactly one pair (and no better hand): There are $13$ possibilities for the rank of the pair, and $6$ possibilities for the suits of the cards in the pair. Furthermore, there are $12 \choose 3$ possibilities for the ranks of the other cards, which can be of any suit. So the probability is: $$P(\textrm{exactly one pair}) = \frac{13 \cdot 6 \cdot {12 \choose 3} \cdot 4^3}{52 \choose 5} = \frac{352}{833} \approx 0.4226$$.
Here is all solutions (one pair, flush etc.) :
http://en.wikipedia.org/wiki/Poker_probability