# 2 card hand probability question

1. Oct 6, 2012

### CAF123

1. The problem statement, all variables and given/known data
This is just a sort of clarification question: suppose I wanted to know the probability of getting at least one ace in a 2card hand. My question is : can this be calculated directly or would I need to use the complement?

Calculating directly, I said for the first card of the two there are 4 possibilities and for the second there are 51 possibilities so the required amount of hands is 4x51. I think this needs to be divided by 2! because I am not considering order.

The probability would then be $$\frac{\frac{4×51}{2!}}{(52 choose 2)},$$ could someone tell me whether this is correct or not. When I use it as part of a later question, I get the wrong answer.
Thanks

2. Oct 6, 2012

### LCKurtz

You could calculate it directly but why would you want to? It is much easier to calculate the probability of getting 0 aces and subtracting that from 1. And no, your answer isn't correct.

3. Oct 6, 2012

### Ray Vickson

Note that {1 ace} = {AN or NA}, so P{1 ace} = P{AN} + P{NA}; here A = Ace, N = non-ace. Of course,
P{AN} = P{A first}*P{N second|A first}, etc.
P{2 aces} = P{AA}.
Or, as LCKurtz has suggested, compute 1 - P{NN}, which is a bit easier.

RGV

Last edited: Oct 6, 2012
4. Oct 7, 2012

### CAF123

What exactly does my method fail to consider? Given that the question is probability that at least one ace is in a 2 card hand we have 4 possibilities for the first card (one of them must be ace) and 51 for the other card. Divide by 2! to neglect order and divide by sample space.

5. Oct 7, 2012

### Dick

Let's call the ace of spades AS, ace of hearts AH and one of spades 1S. Then your 4*51 count includes {AS,AH} and {AH,AS}, so you want to divide that case by two. On the other hand, it only contains {AS,1S} NOT {1S,AS}. So dividing by two in that case is a mistake. You are bunching too many cases together.

Last edited: Oct 7, 2012