I have a problem working why these two different probabilities are calculated slightly differently.
5 dice are rolled simultaneously.
Work out:
$P(\text{two pairs})$
$P(\text{full house})$
For two pairs, the working commonly shown is:
AABBC
Two different values in 6 * possible values of last die * ways to position first pair * ways to position second pair / total outcomes:
$${6 \choose 2} \cdot 4 \cdot {5 \choose 2}\cdot {3\choose 2} \cdot \frac{1}{6^5}$$
So I would expect the calculation for the full house to be:
two different values in 6 * ways to position triple * ways to position pair / total outcomes:
$$ {6\choose 2} \cdot {5\choose 3}\cdot {2\choose 2}\cdot \frac{1}{6^5}$$
This incorrect way of working it out gives me exactly half the expected answer.
Does it have something to do with the pairs being the same whereas a pair and triple is different? If it does why does that matter? Can you give an intuitive explanation?
Best Answer
That's exactly correct, since triples are different then pairs, you need to differentiate $AAABB$ from $AABBB$, so you need to choose which one of the values is your triple. So your calculation should be: Choose a number to be the triple (6), Choose a number to be the pair (5), Ordenate them 5C3*2C2, which differs from your calculation exactly by the factor of two inclueded in 6C2.