"dice probabilities" Topic

Hello

I need some help with some dice probabilities. Can't seem to wrap my head around this. What would the formula/method be for figuring out the following

What is the probability of rolling a 1 on at least 2 dice when the person rolls 3d6? or 4d6? or 3d8? 4d8?

What would the probability of the person rolling any double when rolling 3d6? 4d6? or 3d8? 4d8?

Thanks!

Sid

For the 1st, if you get at least two 1s on 3d6 you're getting 1-1-x, and there are three ways to arrange that and six possible values for x, so 18/216 or 1/2 or 8.25%.

On 4d6 it would be 1-1-x-x, there are six ways to arrange that, and 36 ways fill the two Xs, so 216/1296, or 1/6 which is 16.67%.

On 3d8 that would be 24/512 or 3/64 or just under 5%. On 4d8 it would be 384 out of 4096, whatever that calculates to, 9.5%?

Basically you just write out all the possibilities that satisfy your condition and divide by the total number of possibilities.

anydice.com is often helpful.

It's pretty easy to work out the probability of at least one of a particular number on any number of dice (1 – probability of no 1s), but what you want is a lot trickier unfortunately. As Andrew says, you have to work out all of the ways you can get the desired result. No simple formula I'm afraid.

IIRC there is a formula for "6 choose 1", but these questions are a compounded version of that. You could probably derive a formula, but you'd have to re-derive it every time you ask a new question. The math people have systematic ways to express these things, but they're really more notation that formula, that is you can't just plug in a couple numbers and get your answer.

And, bottom line, when you're designing a game, which I'm guessing is what's going on here, you need to get a really good sense of the probabilities because you're likely going to have DRMs or the like and if you haven't looked at a page full of numbers you won't know what is and is *not* a good idea.

So, yeah, make a list of possibilities and count them. You'll get better at it after a while.

Excel is your friend. Setup a spreadsheet with each column representing a single die. For a single d6, write 1 thru 6 down a row. For a second 6d, write 1 thru 6 in the second column. But the 2nd d6 can be 1 thru 6 if the 1st die is a 1, and again if the 1st die is a 2, rinse and repeat. You'll end up with 36 rows filled in. To add a 3rd d6, you'll need a 3rd column with 1 thru 6 for EACH of the current 36 rows (ie. combinations of die 1 and die 2), for a total of 216 rows. (see below) Then write an IF statement to quickly identify which rows have the combinations you are looking for. Add up the rows, divide by the total number of rows, and there's your percentage.

D1 D2 D3
1 1 1
1 1 2
1 1 3
…
6 6 4
6 6 5
6 6 6

It's easier than I made it sound…I think.

This website will help answer your question: anydice.com

Thanks. That is the approach I thought I probably had to take, but thought I must be doing something wrong by not recognizing a formula for it.

That's why I love this forum … always helfpul. :)