Probability of a Full House for five-card hand.

So I know this can be solved easily by counting the total ways to make a full house and dividing that by the total possible hands, but I want to know why another way I thought of to solve it is wrong.

My calculation is: $$1 \times \frac{3}{51} \times \frac{2}{50} \times \frac{48}{49} \times \frac{3}{48} \times 5!$$

To break this down, the first card can be any. The second card must be the same number as the first ($\frac{3}{51}$) and the third card must also be the same number ($\frac{2}{50}$). The fourth card can be any from the deck with the exception of whatever card makes a 4-of-a-kind ($\frac{48}{49}$). And the fifth card must be the same number as the fourth card ($\frac{3}{48}$).

Since order should not matter for a hand of cards, I multiply this probability by $5!$.

I can't figure out where I went wrong, but evidently this does not give me the correct answer. Can anyone help me find my error?

You are correct that your initial logic under-counts, but the correction of $5!$ is too high. In fact the correct factor to adjust by is $10.$

Before incorrectly multiplying by $5!$, you correctly compute the probability of getting the pattern (xxx)(yy) (hopefully it's clear what I mean by that). Of course this is not the only pattern a full house can come in. It can also be like (yy)(xxx), or (y)(xxx)(y), etc. The way to get the answer would be to compute the probability for all these patterns that constitute a full house, and since they're the mutually exclusive ways to get a full house, just add them up to get the probability of a full house.

Of course the probability is the same for all the patterns (though you may want to talk yourself through a weird one like (x)(y)(x)(y)(x) to make sure it is obvious why it must be the same as for your choice.) So it's just a matter of counting all the patterns. Well, there are five spots and you need to choose three to be x so that's ${5\choose 3} = 10.$

In multiplying by $5!$ you forgot that you had already taken into account the permutations of the cards that only interchange the x's and y's amongst themselves. After all, $(1)(3/51)(2/50)$ is the correct probability of drawing three of a kind in a three card hand... you wouldn't multiply that by $3!.$

• Thank you! This makes perfect sense! – ToGzGaming Jan 7 '18 at 20:50

Because the second card doesn't need to be another card in the triplet. It can be any other card in the deck, which starts the construction of the pair. Likewise, at every step, it doesn't need to be part of the pair or needs to be part of the triplet, it needs be part of {the pair or the triplet}.

• But I account for this when I multiply by 5!, do I not? – ToGzGaming Jan 7 '18 at 3:12

Your initial fractions are correct, but you adjusted too much. If you have the answer, you would notice being off by a factor of 12. For this answer, I am going to call the two same cards as a pair, and the three same cards as the group of three.

The total number of ways to rearrange a specific full house is actually $5 \choose 2$, not $5!$. This is because you are rearranging two identical items, with three other identical items, not five different items. The result would be $5 \choose 2$, since you are choosing two spots for the pair, out of five total spots.

You can think about it this way too, $5!$ over-counts because swapping the positions of the two paired items changes nothing about your hand. E.g., if your hand is $JJ555$, swapping the two $J$'s is redundant. This over-counts by a factor of two. If you do the same with the group of three, you will end up with $\frac{5!}{2*6}$, which is the same result as above.