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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?