I've never been particularly good at counting, and I was pleasantly surprised to learn today that many typical probability-counting problems can be solved using the equation:
$P(A\cap B) = P(A)P(B|A)$
and it's generalisations, rather than by more traditional counting methods. However, my first attempt to apply this equation to a more difficult counting problem has sadly failed.
I'm not going to go through all the reasoning, but by 'traditional' counting methods, the probability of getting a full house is: $\frac{{13\choose 1} {4\choose 3}{12 \choose 1}{4 \choose 2}}{{52\choose 5}}\approx 0.001440576$
By iteratively applying the new formula, we get: $P(A\cap B\cap C\cap D\cap E) = P(A)P(B|A)P(C|A\cap B)P(D|A\cap B\cap C)P(E|A\cap B\cap C\cap D)$
Now if we say these events are:
A - pick a card, any card
B - pick a card with same value as A
C - pick a card with same value as A
D - pick a card different value as A
E - Pick a card with same value as D
Now if all of these events occur ($A\cap B\cap C\cap D\cap E$), you will have your full house. The relevent probabilities are:
$P(A) = 1$ - certain you will pick a card
$P(B|A) = 3/51$ - given A, you have just 51 cards and only 3 of the same value in the deck remaining
$P(C|A\cap B) = 2/50$ - Same reasoning as above
$P(D|A\cap B\cap C) = 48/49$ - Can pick any of the 49 cards except the 1 card remaining with the same value as A
$P(E|A\cap B\cap C\cap D) = 3/48$ - Pick any of the 3 cards which have same value as D.
However, multiplying all of these together frustratingly gives:
$1\times \frac{3}{51} \times \frac{2}{50} \times \frac{48}{49} \times \frac{3}{48} = 0.0001440576$
which is a factor of 10 out :S Can anyone point me in the right direction???
Thanks