Chances of rolling doubles on 2d6 vs 1d4 and 1d6.

Question

Trying to work out the probability of rolling doubles on two six sided dice (2d6) vs one four sided die (d4) and one six sided die. Maths isn't my strong suit. Little help?

'Kay, so the odds of rolling a double on 2d6 is nice and simple. There are 6 possible ways of rolling doubles on 2d6 out of a possible 36 combinations, so (1/(6*6))*6 = 0.167, or a 17ish percent chance of rolling a double.

There are 24 combinations of a d6 and a d4, and 4 ways of rolling doubles. So (1/(6*4))*4 = ... also 0.167. Exactly the same chance of rolling a double.

That doesn't seem right to me? I'd assumed that you'd be less likely to roll doubles on a d4 and a d6. So which is wrong, my maths or my intuition?

Answer 1

Your math is right and your intuition wrong.

I can't see inside your head to know your intuition but my guess is it's something like: "The $5$ and the $6$ are impossible to roll doubles with, so them being there reduces the odds of rolling doubles." And you know what, you're right, it does. The problem is you're comparing to the wrong baseline. You should compare to the odds of rolling doubles on two four-sided dice, which is $1/4.$ When you put the $5$ and $6$ on one of the dice, it reduces that to $1/6.$

If I got your reasoning wrong, at least be assured your answer is right.

Answer 2

I prefer this piece of intuition: throw one die first, then the other. In the second case, make sure that the d4 is the first die you throw. Then it's obvious that in both cases it's a $1/6$ probability that the second die lands on whatever the first die showed.