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Yahtzee Trouble

  1. Dec 3, 2004 #1
    Hello,

    I'm confused as to what is the probability of rolling 4 of a kind in one roll, in Yahtzee. I'm thinking its simply C(6,2) * C(2,1), which gives a probability of [tex]\frac{30}{6^5} = \frac{5}{1296}[/tex] . However, most of the sources I have seen regarding this outcome give a probability of [tex]\frac{25}{1296}[/tex] , which is five times greater than what I have. What is the correct way to go about this?

    Thanks a ton
     
  2. jcsd
  3. Dec 5, 2004 #2
    You have consideerd only one order of dice. There are several orders of dice, so the probability is greater C(5,1) greater.

    The probability of gettting 2,3 is the same as getting 5,5 if you consider order. If you don't the probability is greater than getting 5,5.
     
    Last edited: Dec 5, 2004
  4. Dec 5, 2004 #3
    first of all, thank you very much for replying

    now, thats exactly what i thought at first. However, my teacher told me to consider the calculations without order, hence utilising combinations, not permutations. So I found out that in total, there are [tex]C(6,1)^5 = 6^5 [/tex] possible combinations of faces. Proceeding in the same manner, there should be [tex] C(6,2)*C(2,1)*C(1,1)*C(1,1) = 30 [/tex] combinations for 4 of a kind. If we consider permutations, then the total number of ways one can get 4 of a kind is [tex] P(6,2)*P(2,1) = 60 [/tex] correct?

    Please point out any flaws in the calculations above
     
  5. Dec 5, 2004 #4
    No actually there are 150 ways of getting a four of a kind. You HAVE to consider order. Take the following simple example:

    The probability of getting a 2 and a 3 is greater than getting two fives if you DO NOT consider order. This is because the either die#1 can be a 2 and die#2 can be a 3 or die#2 can be a 3 and die#1 can be a 2.

    So, for a four of a kind, it will be C(6,1) * C(5,4) * C(5,1) * C(1,1) = 150

    If the calculation is not self-explanatory please let me know and I will elaborate.
     
  6. Dec 6, 2004 #5
    I see why we must consider order in this case, as the total number of outcomes possible is taken to be [tex]|S| = 6^5 [/tex]. And this number is inclusive of distinct arrangements, taking each arrangement to be a different element of set [tex] S [/tex]. Your example clearly shows why order is necessary.
    However, I don't follow the mathematical product itself. I see the [tex] C(6,1) [/tex] as "pick any one of the six elements of the die-face-set for the 1st position", and [tex] C(5,1) [/tex] as "pick any element distinct from that of the 1st position for the 2nd position". Now, where does the [tex] C(5,4) [/tex] come from? I'm thinking that if you have two distinct elements for the first two positions, then you have a choice of two elements for the third position and are forced to fill the rest of the positions with the same element as that of the third position. Nevertheless, I do see how there are 150 possibilities in all, since I like to see it as the different positions which could hold the single (5 in all) multiplied by [tex] C(6,2)*C(2,1) [/tex], which yields 150.

    Pardon my inability to connect the mathematics to the game, but the only part I don't understand is the mathematical product.
     
  7. Dec 6, 2004 #6
    C(6,1) - choose a number from one to six
    C(5,4) - pick 4 dice from your five and assign the number to them
    C(5,1) - pick another number different from your initial choice
    C(1,1) - pick the remaining die and assign the number to it.

    Hope this clarifies.
     
  8. Dec 6, 2004 #7
    Thank you so much himurakenshin!!!
    The muddy waters of my mind have thus become crystalline.

    I am going to use your method to calculate the probability of getting a pair (and only a pair; a full house counts as a triple, and a double pair is still a pair, for my purposes). Once I have that, I can calculate the probability of getting a Yahtzee in 3 rolls, which was the original objective of my project.
     
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