0
$\begingroup$

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 : enter image description here

Where was I wrong in my reasoning?

Thanks

$\endgroup$
  • 1
    $\begingroup$ For second card you have actually 51 choices because what ever card it is you can still get a full house. $\endgroup$ – Harto Saarinen Oct 11 '14 at 16:41
2
$\begingroup$

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 $120$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.