Date: 02/18/97 at 13:58:35 From: Christopher K. Subject: Card probability Hello, I have a problem. I found that if you take a deck of cards, name any two cards, and then look for them in the deck, they will be by each other or only one card away from each other. I tried to find the probability of this occurring and have asked a lot of people about it, but I can't find the answer. So far, when trying to simulate this, I have only failed once.
Date: 02/27/97 at 21:10:09 From: Doctor Ken Subject: Re: Card probability Hi Chris - That's some pretty good luck! Here are some pointers when trying to find the probability. When computing ANY probability, you must remember the probability rule of gold: probability is the number of favorable possibilities divided by the total number of possibilities. Often it's a good idea to break the problem into pieces. In your case, we can break the problem into two pieces: the probability that the two cards are next to each other and the probability that they are separated by a single card. We can compute the two pieces separately, and then add the probabilities together to find the final probability. In your case, the total number of possibilities (for both pieces of the problem) is the number of different arrangements of a deck of cards. So in order to calculate the probability that the conditions you named are satisfied, we have to find this number. So let's do it. There are 52 cards in a deck. If we have them in a certain order, then there are 52 choices for the first card, 51 for the second, 50 for the third, and so on. The total number of different arrangements is thus 52 * 51 * 50 * ... * 2 * 1, which you can write as 52! (pronounced 52 factorial). So that's the denominator in our probability fraction. The harder part is to find the numerator. But it's not too bad either. Let's concentrate on the first piece first. Given two cards, how many different arrangements of the deck have those two cards next to each other? Well, let's start by writing a few down. If the two cards are called A and B, then we could have these arrangements (I'm using x to just mean any old card here): A B x x x ..... x B A x x x ..... x x A B x x ..... x x B A x x ..... x and so on. Since there are 50 x's in each line, each line actually represents lots of different arrangements; you could order the cards that fill in the x's in lots of different ways. How many? Well, 50 * 49 * 48 * ... * 2 * 1, which is 50!. Now how many lines are there? There are 51 different places we could put the "A B" pair (convince yourself of that), and for each place there are two lines (one for "A B" and one for "B A"), so in total there are 2 * 51 * 50! different arrangements that have A and B next to each other. So what's the probability? Well, we divide: 2 * 51 * 50! 2 1 Probability = -------------- = ---- = ---- 52! 52 26 So that's the first piece of the problem. I'll leave it to you to figure out the second piece, the probability of having A and B separated by a single card. Once you have that, you can add the two probabilities together to find the total probability. Good luck! -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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