Finding Patterns in Multiple Dice Rolls: A Simple Solution

[rbzima]

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?

 

[robert Ihnot]

I have, I think a simple way. Given one die, we have \sum_1^6 j=21. 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 \frac{3(3.5)+4(3.5)}{2} =(3.5)^2 =12.25. (If in one case it was 2 and 5 rolls, it would average out to the same value above.)

 

[rbzima]

I have, I think a simple way. Given one die, we have \sum_1^6 j=21. 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 \frac{3(3.5)+4(3.5)}{2} =(3.5)^2 =12.25. (If in one case it was 2 and 5 rolls, it would average out to the same value above.)

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!