Expert: Josh - 2/28/2011

Question
Hello,

I was having difficulty understanding how to calculate some percentages dealing with combinations with decks of cards.

I want to be able to calculate the percent chance that, say, at least one ace would be in someone's hand after drawing seven cards from a 52 card deck.

This is what I tried:

percent chance to draw one ace + percent chance to draw two aces + 3 + 4 . . .  

[(1/13)*7]/[52 C 7] + [(1/13)*(1/13)*7]/[52 C 7] + [(1/13)*(1/13)*(1/13)*7]/[52 C 7] + [(1/13)*(1/13)*(1/13)*(1/13)*7]/[52 C 7]

But I still do not feel confident about my answer.

4.36*10^-9 That is way too small to be right.

Answer
Hi Stephen,

You are right in so far as "adding up the probabilities of drawing one/two/three and four aces" is concerned. However, I am not so sure that multiplying the number (1/13) repeatedly is the way to go. To me, it shows that you understand that the probability of getting a series of independent events is given by the product of the probability of individual selections. I think where it went wrong is how you came up with the numbers which represent individual selections.

Consider the argument used for the first case: the probability of drawing one ace.
The probability of picking an ace first up is 4*(1/52) = 1/13 because there are four suits.
However, our job is only half done at this point. Because we need to exclude the possibility of drawing any of the three aces remaining in the 51 cards. So, we need to consider the individual probabilities of selecting a non-ace card from the second pick, through to the seventh pick explicitly. For the 2nd pick, the probability of getting a non-ace is 48/51. For the 3rd pick, the probability is 47/50. This is done repeatedly up to the 7th pick. Finally, for this series of events to occur, the joint probability is given by the product of probabilities for each individual selection, multiplied by 7. The factor of seven accounts for the fact that the ace may appear in pick 1, pick 2,... or pick 7. The calculation is given by P_1=C(7,1)*(1/13)*(48/51)*(47/50)*(46/49)*(45/48)*(44/47)*(43/46). [Note: I use the notation C(N,k) to mean N!/((N-k)!*k!)]

Next, consider the argument used for the second case: the probability of drawing two aces.
For the time being, it is helpful to focus on the scenario where the two aces come from the first two of seven selections. The probability of picking an ace first up is (1/52). This may come from any of the four suits. So, the probability is 4*(1/52)=(1/13) as before. Now, for the scenario under consideration, we want to pick an ace again in the second pick. There are now only three aces remaining in a deck of 51. So, the probability is (3/51). Again, we need to exclude the possibility of picking an ace from pick 3 to 7. The reasoning is same as before, except we now have two aces remaining that we must avoid from this point onward. Fast forward to the joint probability which describes the pattern considered in this scenario, we get (4/52)*(3/51)*(48/50)*(47/49)*(46/48)*(45/47)*(44/46). Don't forget that this is only one of the scenario. It is possible to pick two aces from seven selections in C(7,2)=7!/[5!*2!]=7*6/2=21 ways. So, the probability associated with one scenario must be multiplied by the number of ways of selecting two aces from seven selections, to give us the case two probability: P_2= 21*(4/52)*(3/51)*(48/50)*(47/49)*(46/48)*(45/47)*(44/46).

I won't repeat the arguments for the third and fourth case. The respective probabilities are given by
P_3=C(7,3)*(4/52)*(3/51)*(2/50)*(48/49)*(47/48)*(46/47)*(45/46) and
P_4=C(7,4)*(4/52)*(3/51)*(2/50)*(1/49)*(48/48)*(47/47)*(46/46).

For case 4, I am just spelling this out aloud, of course, after the four aces are picked, we don't care what we pick in the other three picks (with absolute certainty, they cannot be an ace as they have all been removed).

The answer to the question is given by
P_1 + P_2 + P_3 + P_4
= 1.2928e-4 + 0.0058 + 0.0768 + 0.3669
= 0.4496

It's good that you are thinking about the answer and picked up that something is wrong.