Since subtlety is a primary characteristic of combinatorics, that makes this a really nice question to dig into!Ĭards are discrete objects, so that there is only one of each this makes permutations and combinations (which involve selecting and/or arranging distinct items) appropriate. And I think the ultimate answer is going to be that you are being fooled by a superficial similarity between them. I’ll start out by saying that my first impression was that cards and dice behave very differently (because there’s only one of each card, while dice are independent), so I didn’t think the work would be anything like the same. He demonstrated the overcounting with cards is there any with dice? To correct my mistake, one would have to remove the 3!, but how then do you account for double-counting? If you could help me understand why I should ignore double counting in Yahtzee, but not in Poker, I would be grateful. Then I divide with the total number of dice-rolls which is 6^5. Then, the same way I did in poker, I divide with 3!, since a composition of 5-5-1-2-3 is the same as 5-5-3-2-1. I choose one value out of six, choose two dice then pick the last three dice. I want to use the same method I did when I calculated the poker-version. Finally I divide with the total number of hands which is C(52,5) and get approximately 42%. These three cards can be arranged in 3! ways, and since order does not matter, I divide by 3! to remove duplicates. For example, if my third card is six of spades, my fourth is seven of spades and my fifth is eight of spades, that is the same hand as if my third card was eight of spades, my fourth card was seven of spades and my fifth card was six of spades. To account for double-counting, I divide by 3!. Then I choose the third, fourth and fifth cards, all of which can be combined with 4 suits, thus 4^3. (C(13,1) x C(4,2) x C(12,1) x C(11,1) x C(10,1) x 4^3)/3!įirst, I take 1 of the 13 values in which I have a pair, and their suit can be combined in C(4,2) ways. Poker (An ordinary deck of cards with 5 cards in a hand): Now, Alexander carefully showed his thinking about each problem: The key is to recognize whether you’ve done it! It means that we have counted the same outcome more than once, and need to either compensate, or find a way to avoid it. To my knowledge, “one pair” plays no role in Yahtzee, but would mean exactly two of one number (5 again), with all other dice having different numbers:Īs for double-counting (or overcounting), we’ve mentioned it, for example, in Permutations and Combinations: Undercounts and Overcounts, Arranging Letters with Duplicates, and Interpreting and Solving a Counting Problem. In poker, one pair means a hand containing two cards with the same number (here, 5), all other cards having different numbers: It will just be assumed that in poker, 5 distinct cards are chosen randomly from a deck of 52, and in Yahtzee, 5 six-sided dice are rolled once. Therefore I will first calculate the probability of getting a one-pair in poker, then do the same thing for Yahtzee and hopefully, you can observe where I went wrong.īefore I show the rest of the question, I want to mention that he is not concerned with how one actually gets the cards in poker or the numbers in Yahtzee, which in both cases can involve more than one step and complicate the probabilities. However, when I try to calculate the probability of getting a one-pair in Yahtzee, I get an incorrect answer which I believe has to do with my confusion with double counting. I know how to calculate the probability of getting a one-pair in poker. More precisely, I am having trouble with the phenomenon called double-counting, when the same event is counted more than once. I have a question regarding the probability of getting one pair in poker and Yahtzee. This very detailed question came from Alexander in early July: Both were about apparent conflicts between similar problems involving cards and dice. A couple recent questions involved related subtleties in probability and combinatorics.
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