# Thread: "Exploding" Dice Probability (Talk to me like I'm an idiot)

1. ## "Exploding" Dice Probability (Talk to me like I'm an idiot)

Hi All,

First off, let me just start by saying that my mathematical savvy leaves a lot to be desired. I didn't take a lot of math in school, and while I'm good with numbers, some of what may seem like basic terminology is completely foreign to me. Please be patient if I ask some fairly basic questions to broaden my understanding.

That said, I'm working with some friends on designing a game system that relies on rolling ten-sided dice. I've already sussed out how to calculate the odds of rolling N number on a ten-sided die if I'm rolling D number of dice. Fairly basic stuff.

Then, we started talking exploding dice. For the uninitiated, an "exploding" die is a die that, upon rolling a certain number (typically the highest number on the die, a 10 in the case of a 10-sided die), allows you to reroll and add the die to your total. Thus, if you rolled an "exploding" 10 on a 10-sided die, you could roll again, continuing to re-roll 10's, until you roll a non-10, at which point you total your amount. So you roll a 10, re-roll, get another 10, re-roll and get a 6, your total is 26.

I found a fairly straightforward, if slightly dense (not a stats pro), explanation on how to determine the odds of rolling N number with a die if you're using one die. However, there are three complications that I need assistance with:
1. How do you determine the odds of rolling N number when rolling D number of "exploding" 10-sided dice?
2. We're talking about introducing "exploding" 1's - you can re-roll and add together 1's instead of 10's. How do you determine the odds of rolling N number when rolling D number of 10-sided dice with exploding 1's?
3. We're talking about also adding "re-rolls" on 1's - you can re-roll any 1, and replace it with the next roll. How do you determine the odds of rolling N number when rolling D number of 10-sided dice, re-rolling any 1's (future 1's could also be re-rolled)?

Sorry for the three-part question, feel free to answer any part of it you feel you've got a grasp on.

Thanks!
Rob

2. ## Re: "Exploding" Dice Probability (Talk to me like I'm an idiot)

I give an answer to part 3) first as it is only replacement and will be the easiest one. Essentially you have a truncated discrete uniform, which is also a discrete uniform - the final outcome of any dice can only contain 9 numbers (assume you always replace) and they have probability of 1/9 each. So you can proceed as usual.

For part 2), this "exploding" feature means one need to add the number of 1's before reaching the final outcome as in part 3. So the total will be the sum in part 3), plus i.i.d. geometric random variables (negative Binomial). But, of course the sum of these random variables could be very messy.

Part 1) is similar to part 2), but now you end with 1-9 and now multiply the negative Binomial by 10. Let see if this is what you want.

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