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This is going to seem pretty random but I know DakkaDakka has a lot of math wizards especially when it comes to dice probabilities.

I was wondering if anyone could give me the probability of rolling triples on 6D6 compared to two doubles on 6D6.

Might be helpful Type in 6d6 and get a break down of the bell curve for the outcomes

Short answer: 32.15% and 8.1%, respectively. Do you want the long answer?

Always show your working

 Dheneb wrote:
Short answer: 32.15% and 8.1%, respectively. Do you want the long answer?

I'm surprised triples is more likely. Wouldn't mind long answer if you've got time. If not, don't worry about it.

 KamikazeCanuck wrote:

 Dheneb wrote:
Short answer: 32.15% and 8.1%, respectively. Do you want the long answer?

I'm surprised triples is more likely. Wouldn't mind long answer if you've got time. If not, don't worry about it.

Lets start with the chance of rolling 3 one's with the six dice. The probability of a successful event (rolling a one) on one die is 1/6; an unsuccessful event has probability 5/6. The formula to use is nCk * p^k * (1-p)^(n-k), where nCk is the binomial coefficient (on my calculator it's listed as nCr), p is the chance of success from one attempt (1/6), n is the number of dice (6) and k is the number of successes you want (3).
To get the chance of any double you multiply that by 6, which gives 0.3215, or 32.15%.

Part two was wrong on my part, the right answer is the probability of getting the same double twice (the same formula as above, but with k=4) plus the chance of getting two different doubles (which uses a generalisation called a multinomial). Right answer is 97.4%.

 Dheneb wrote:

 KamikazeCanuck wrote:

 Dheneb wrote:
Short answer: 32.15% and 8.1%, respectively. Do you want the long answer?

I'm surprised triples is more likely. Wouldn't mind long answer if you've got time. If not, don't worry about it.

Lets start with the chance of rolling 3 one's with the six dice. The probability of a successful event (rolling a one) on one die is 1/6; an unsuccessful event has probability 5/6. The formula to use is nCk * p^k * (1-p)^(n-k), where nCk is the binomial coefficient (on my calculator it's listed as nCr), p is the chance of success from one attempt (1/6), n is the number of dice (6) and k is the number of successes you want (3).
To get the chance of any double you multiply that by 6, which gives 0.3215, or 32.15%.

Part two was wrong on my part, the right answer is the probability of getting the same double twice (the same formula as above, but with k=4) plus the chance of getting two different doubles (which uses a generalisation called a multinomial). Right answer is 97.4%.

97.4%!? The reason I asked for long answer is because I suspected I didn't explain the second scenario very well. What I mean by two doubles is, for example: two 1s, two 2s and then a 3 and a 4. Is that really 97.4%?

 KamikazeCanuck wrote:


97.4%!? The reason I asked for long answer is because I suspected I didn't explain the second scenario very well. What I mean by two doubles is, for example: two 1s, two 2s and then a 3 and a 4. Is that really 97.4%?

A multinomial will describe the chance of rolling different types of doubles (that is, double 1 and double 2). So the chance of rolling double 1 and double 2 is, according to its formula, about 3%. Although, I'd like to revise it again ( ) since I was double-counting most combinations. Different doubles are 46%, if you allow the same ones add 5%.

I wonder if you are also counting the doubles that occur within triples and quads and so on. Oh well, turns out this was harder than I thought. Thanks for your effort.