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???


  • $\begingroup$ As Mathematics hints at below, you are not just counting the probability of getting a full house, but the probability of getting a full house where the first three cards you are dealt form a three-of-a-kind, and the last two a pair. Of course, being dealt A-K-A-K-A (in that order) will also give you a full house, but it is not counted above. $\endgroup$ – user642796 Mar 7 '12 at 10:39

An event is a subset of the sample space. In order to make sure that our calculations are correct, we need to explicitly identify the sample space, and the events you have called $A$ to $E$.

The sample space you are using has not been explicitly described, but it looks like all ordered $5$-tuples of cards.

The sample space used in the traditional counting method consists of all possible hands, that is, sets of $5$ cards.

It is perfectly fine to use ordered $5$-tuples. Your $B$ is the event that the second card chosen has the same value as the first card chosen, or to use more traditional cards language, is of the same kind as the first.

Your $C$ is the event that the third card chosen is of the same kind as the first two chosen cards. Your $D$ is the event that the fourth card chosen is of a new kind. And your $E$ is the event that the fifth card chosen is of the same kind as the fourth.

You have correctly computed $P(A\cap B\cap C\cap D\cap E)$.

However, as was pointed out in detail by Ahmed Masud, if our sample space consists of ordered sequences of $5$ cards, then the event "full house" can happen in other ways. For example, we could get an Ace, then a King, then a King, then an Ace, then an Ace. This pattern is not accounted for by your calculation.

We can define new events that deal with this sort of pattern. It can even be done in a style that is almost exactly the same as yours. The new $B$ could be the event that the fourth card picked is of the same kind as the first card picked. Your new $C$ would be the event that the fifth card picked is of the same kind as the first and fourth. The probabilities would be computed in the same way, and the answer for this pattern is exactly the same as the one for the pattern you analyzed.

It turns out that there are $\binom{5}{3}$ different patterns. This is because a pattern is uniquely identified by identifying the places where we get our three cards of the same kind. For each such pattern, the probability is the same as the one you found, and one simply multiplies your number by $\binom{5}{3}$.

  • $\begingroup$ Thanks very much, I think that makes more sense :) . I just failed to see that I'd implicitly started computing a particular permutation of the cards in considering the events A, B, C, D, E. For some reason, I've always had difficulty with this - it seems that sometimes multiplication implies order (e.g. 5! arrangements) and sometimes it doesn't (e.g. the combinatorial method I gave above) :S thanks again though $\endgroup$ – tom Mar 8 '12 at 1:24

Your scenario looks at:


But there other sequences of events that can give you a full house:


In particular there are 10 sequences in total.

To choose 3 cards of one type and 2 cards of another type you worked out the probability. In particular,

$$ {3 \over 51} \times {2 \over 50} \times {48 \over 49} \times {3 \over 48} = 0.0001440576 $$

So total number of ways to do it is $10 \times 0.0001440576 = 0.001440576$. Which is the same as above.

  • $\begingroup$ Thanks, but I'm still a little confused. Apparently, what I have calculated is the probability that $A\cap B\cap C\cap D\cap E$ occured. Is this correct? It's what the formula clearly states. But there's absolutely no difference between that event and, e.g. $A\cap C\cap B\cap D\cap E$ or any other permutation - intersections are commutative. I don't see why we have to take into account order here... thanks $\endgroup$ – tom Mar 7 '12 at 12:05
  • 2
    $\begingroup$ Intersections are commutative but the particular event sequences are not. It's a counting problem, you have to count all the ways you can get a full house not just one. The reason you can simply multiply the number of event permutations with the probability of one of them to get the final result is due to the commutative property, i.e. changing the sequence does not change the probability of one of them, however you still have 10 sequences. So $P = P(E_1) + P(E_2) + ... + P(E_{10})$ $\endgroup$ – Ahmed Masud Mar 8 '12 at 14:13

after taking first card,in factt,you may not be able to get the second card you want.Totally there are $5C3$ conbination and hence the answer concide with the other one calculations.You are just calculating the special case of getting that set of symbol.


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