Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum of multiple dice rolls

  1. Oct 25, 2007 #1
    So, I'm having trouble seeing a pattern in this small analysis and was wondering if anyone might be able to lend a hand.

    Suppose I toss a die with a result of n points. Then, whatever those n points are determines the number of dice you throw a second time. If the sum of your second throw is under 12, you lose, if it's 12, you win nothing, and if it's higher than 12, you win.

    I have all the probabilities of 3 dice, but was wondering if there's a simpler way of finding the other probabilities? Since there are 18 possible numbers with 3 dice, I found that the triangular numbers starting with P(rolling a 3) for three dice is obviously 1/216, then P(rolling a 4) with 3 dice is 3 ways. Am I right to assume that there are then 6 ways for P(rolling a 5) and 10 ways for P(rolling a 6) until I reach the P(rolling a 9) because at that point the numbers are different as a result of having only six sided dice. Is there any way I might be able to see the others until I'm up to 6 dice.

    I realize its far easier to find the lower numbers as I use more dice. Am I right to conclude this works for all scenarios up till 6 dice?
  2. jcsd
  3. Oct 28, 2007 #2
    I have, I think a simple way. Given one die, we have [tex]\sum_1^6 j=21.[/tex] Dividing by 6 gives the average value of 3.5.

    We can just start with that. Thus tossing a die gives the likelyhood of 3.5, which means half the time it is 3 and half the time it is 4, probably. So that on the whole we arrive at [tex]\frac{3(3.5)+4(3.5)}{2} =(3.5)^2 =12.25.[/tex] (If in one case it was 2 and 5 rolls, it would average out to the same value above.)
    Last edited: Oct 28, 2007
  4. Oct 31, 2007 #3
    NM, this is actually incorrect. I found an neat and easy way to solve this problem. The thing is, I need all the probabilities of multiple dice in play at the same time, and expected value is simply the payout times whatever the various probabilities are. It's alright though, because I found a nice pattern with the probabilities with 1 dice and 2 dice, and I tested it for 3 dice and sure enough, it works. Thanks for the suggestion though!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook