Combinatorics question. Where am I wrong in my reasoning?

A 5-card poker hand is said to be a full house if it consists of 3 cards of the same denomination and 2 other cards of the same denomination (of course, different from the first denomination). Thus, one kind of full house is three of a kind plus a pair. What is the probability that one is dealt a full house?

I solved this as follows: Sample space is 52 choose 5

For the first card out of 5 we have 13x4 possible choices (any card); For second 3 choices (3 different suits of same card); For third 2 choice (2 different suits of same card); For fourth we have (13 x 4 - 4); For fifth we have 3 choices;

thus: ( 13 x 4 x 3 x 2 x (13 x 4 - 4) x 3 ) / (52 choose 5)

The answer is approx : 0.017

While the solution in book : Where was I wrong in my reasoning?

Thanks

• For second card you have actually 51 choices because what ever card it is you can still get a full house. – Harto Saarinen Oct 11 '14 at 16:41

What is wrong in your reasoning is that in your description (first we draw a card, then we draw a card of the same denomination, etc) the order is fixed. That is, you are not considering that the 5 cards of a full house can be drawn in any order.

There are basically two ways to solve the problem:

1. Combinations. There are $\binom{52}5$ deals. How many of these deals are a full house? First: choose the denomination of the three: 13 possibilities. Then, choose the denomination of the pair: 12 possibilities. Now choose the cards of the three: $\binom{4}3$ possibilities. Now, choose the two other cards: $\binom{4}2$ possibilities.
Then, the probability is $$\frac{13\cdot12\cdot\binom{4}3\binom{4}2}{\binom{52}5}$$

2. Variations. There are $52\cdot51\cdot50\cdot49\cdot48$ deals. Now, we are considering that two deals are different if the cards were drawn in a different order, so there are much more deals. Namely, $5!=120$ times more. Now you can count your way. The probability is $$120\cdot \frac{52\cdot3\cdot2\cdot48\cdot3}{52\cdot51\cdot50\cdot49\cdot48}$$

You can check that they are, in fact, the same number.

Another way is the following:

• The probability for the first card is $1$, because no matter which card is, the full house can be achieved.
• The probaility of that the second card is of the same kind as first is $3/51$, and the pro. of that the third card is also the same kind is $2/50$
• The pro. of that the four card is not of the same kind as the former cards is $48/49$, and the probability of that the last card is the same kind as fourth is $3/48$

Then, the probability of drawing a full house, drawing the three and after the pair is $$p=\frac{52}{52}\frac{3}{51}\frac{2}{50}\frac{48}{49}\frac{3}{48}$$

But since the cards of the full house can be drawn at any order, we must multiply by $120$.