Monday, May 22, 2006

Full House in Yahtzee

The game of Yahtzee is played with five dice. On the first turn, a player rolls all five dice, and then may decide to put aside any of the dice before rolling again. Each player has a maximum of three rolls to try to get a favorable combination of dice.

If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house in one of his/her next two rolls? In other words, what is the probability of getting either a 2 or a 4 in one of the next two rolls?



I haven't done a good stats problem in a while....

9 comments:

  1. The probability of getting a full house (2244 held, rolling the last) in two rolls is:

    1 - P(not getting it) =
    1 - (4/6)(4/6) =
    5/9

    If you start with 3 matching dice, what is the probability of getting a full house? (figure you've already got four of a kind and yahtzee)

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  2. That's right.

    On each roll, the probability of getting a 2 or a 4 is 1/3 and the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. Therefore, the probability of getting a full house is 1 - 4/9, or 5/9.

    As for your question, jonathan: the probability of getting a full house if you start with 3 matching dice is the same as saying what's the probability of getting a matching pair in two rolls of the die.

    I can say that because it doesn't matter what the three matching die are. If you have four 3's and roll two more threes, you can still put it in as a full house.

    Back to the question: P(getting a pair in two rolls) = P(getting a pair on 1st roll) + P(not getting a pair on first roll)*P(getting a pair on second roll) = (1/6) + (5/6)*(1/6) = 11/36.

    If I'm doing this right and if I remember correctly: this is the negative binomial distribution?

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  3. and the probability of getting a yahtzee in the first roll is ridiculess.

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  4. Mike,

    are you allowed to count 5 3's as a full house, or does it have to be with different numbers? If they have to be different, then the probability of starting with 3 of a kind and getting a full house needs to be a little lower.

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  5. The probability of getting a yahtzee in one roll is pretty bad... 6/6^5 = 6/7776 = .000772

    Jonathan, I would say you can use 5 threes to check off a full house. I wouldn't advise it since you get bonus points for yahtzee (5 of a kind), even if you already have a yahtzee.

    But you're right. If we disallowed that, the probability would be less.

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  6. I happened to be playing Yahtzee with my nephew this evening. He had three consecutive turns with the dice. In each of his three turns, he produced a Yahtzee, in which all five dice ended up showing the same number within the three throws allotted to him. Each turn consists of three throws of up to five dice. All of the Yahtzees were achieved within one to three throws of the dice -- i.e., after one to three throws of one to five dice, all five dice were showing the same number. I tried to calculate the range of probabilities for this occurence, but failed. My understanding is that the range should be between an upper and lower limit. The higher probability would be that each of his three turns would require all three throws of the dice and each of the throws would produce at least one and not more than two dice with the same number as the previous throw showing. The lowest probability would be that each turn would result in all five dice showing the same number in the first throw.

    Can anyone help me with this? Oh, and by the way, my nephew is very, very intelligent ... but he's only eight, and I need to find a way to explain this to him in terms that he can understand (of course, the fact that I'm much less intelligent but much older means I'd like to understand the explanation as well).

    Thank-you --

    Roger Miller

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  7. well, seeing as this was posted a while ago and I just stumbled upon it, I don't know if anyone will see this, but here goes...

    There will never be a "range of probabilities" for an event like dice throwing. There is always a singular probability. To solve a problem like this, find the total number of outcomes that could possibly happen, then find the number of outcomes that result in what you want. The simplest (and longest, often) way to do this is to make an event tree (which would be the size of a small European country in your case). Write out the initial outcomes of the event, then branching off from each one of those, write the outcome of the next event given those circumstances. At the end you can just count them up.

    However, in this case, the cases of "two 2s, one 3, one 1, and one 5" and "two 3s, two 4s, and one 6" are indistinguishable to you, since all you care about is that you have two of a kind that you will keep. In this case, the branches from the first event would look something like 1) five of a kind (1/6^4), 2) four of a kind (5^2/6^4) 3) three of a kind (5^2*10/6^4) 4) two of a kind (1-P(everything else), 75/108, by far the largest one) and 5) all different numbers (5/54)). From each of these branches, you carry out more branches until everything ends in "three turns, no yahtzee" or "yahtzee". The probability of getting to the end of each branch is the product of all probabilities along the way.

    How to calculate, for example, the probability of rolling 3 of a kind of the first roll: there are 6^5 ways the dice can land. There are 6 different numbers that can be your "three of a kind". There are 5^2 ways for the other two to land. There are 10 different unique ways for the dice to be arranged at that point (10*5^2*6 ways divided by 6^5 different outcomes).

    I left out some details here, but you can find a good stats book or google it if you have any more problems.

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  8. That has to be the most complete explanation I've seen on this site. Thanks, Abe.

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  9. Actually, I disagree with the answers given above about the probability of getting a full house in the last two rolls being 5/9. If you draw a tree diagram of the problem, you roll the single die once and have the possibility of getting: 1, 2, 3, 4, 5, or 6. If you got a 2 or a 4, you would stop, and not roll again. If you got a 1, 3, 5 or 6 you would roll again, with the same 6 possibilities as the outcome. Therefore, there are a total of 26 possible outcomes, 10 of which would result in a full house. So... the probability of getting a full house would be 10/26, or about 0.385.

    ReplyDelete

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