# Probability Question, Card Deals

Say I have a deck of 52 cards. A, K, Q, J are high cards. I shuffle this deck, and deal out 10 cards. Ideally, I want there to be at least 1 high card in amongst the 10. Sometimes though, I get deals with only low cards in them.

If I take out the aces, twos and threes from this deck, shuffle and deal the same as above, do I have a better chance of getting a deal with at least one high card in it?

In the first deck, high cards are 4/13 and in the second they are 3/10, so there is a small difference there, but I’d like to ignore that , if possible.

I’m not sure what you mean - if you take out the aces, twos and threes, the odds of getting a high card are worse, precisely because 3/10 is less than 4/13.

Because you are picking cards without replacement, you’ll need to use the hypergeometric distribution to figure out the exact odds. If you were doing the game with replacement, the odds of getting at least one high card in the first case is 1-(1-4/13)^10 = 0.9747, while the odds of getting at least one high card in the second case is 1-(1-3/10)^10 = 0.9718

If I took out all the suits bar one, in the second deck, I would have 10 cards, so if I deal out 10, I would definitely get a high card. In the first deck, I would have 13 cards left, so I would be highly likely to still get a high card, I think. I feel like if I am dealing from a smaller stack, I have a better chance of getting a high card, but I am not sure because probability is tricky.

In the first deck, the chance of getting a low card for the first one is 36/52. The conditional probability the second card is low is 35/51, etc. So the chances of getting all low cards is 36x35x…27/52x51…43 = 0.016.

For the second deck it is 28x27x…19/40x39…31 = 0.0154

It looks like @OldGuy is right, but if you want a more detailed explanation:

52\choose10 (pronounced “52 choose 10”) represents the number of different ways you could choose 10 cards from among 52 (where the order does not matter).

Since there are 36 “low cards” in the deck, the probability of getting all 10 low cards would be the number of ways that this could happen, divided by the overall number of ways of picking 10 cards. This is {36\choose 10} \div {52 \choose 10}, which (with the help of Wolfram Alpha) turns out to be about 0.016.

That means that the probability of not getting all 10 low cards—that is, of getting at least one high card—would be 1 – 0.016 = 0.984.

If you take out the aces, twos, and threes, now the deck has 40 cards of which 28 are “low cards” (the 4s, 5s, 6s, 7s, 8s, 9s, and 10s). So the analgous calculations would be {28\choose 10} \div {40 \choose 10}, which is about 0.0155; and 1 – 0.0155 = 0.9845.

Okay, so the smaller deck makes a difference, but not a huge one. Thanks, all.

Without getting into the details, this actually answers to the OP.

No this is not correct. Suppose you took out all but 10 cards, 1 King and 9 others. Then 1/10 < 4/13; however, when dealing 10 cards, you’ll deal them all so you are 100% sure of getting at least one high card. Clearly there is a trade-off between the fraction of cards that are high and the fraction of cards that you deal.

I’m sure it’s possible to set up examples when it isn’t 100% as well.

D’oh. That makes perfect sense. Your approach (above) is using the hypergeometric distribution, right?

Sort of. The method I described is the simplest way to determine the probability that no “events” happen. Then the probability of at least one event is 1 minus that number. It is a special case of the hypergeometric,

Thanks.