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This seems so simple, but I'm not sure how to calculate it.

I have one six-sided die and one 12-sided die.

What is the probability that, on a roll of both dice, that the six-sided die will win?

I'm not overly familiar with mathematical notation. Thank you!

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    $\begingroup$ Just one poor little six-sided? I assume wins means greater. Then the probability is $(1/6)(0/12)+(1/6)(1/12)+(1/6)(2/12)+(1/6)(3/12)+(1/6)(4/12)+(1/6)(5/12)$. This simplifies to $5/24$. $\endgroup$ – André Nicolas Apr 14 '15 at 9:16
  • $\begingroup$ Similarly, the probability of a tie is $6(1/6)(1/12)=2/24$ and the probability of a loss is $17/24$, so if you roll again after a tie then the probability of the six-sided die eventually winning is $5/22$ $\endgroup$ – Henry May 14 '15 at 19:09
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The number of possibilities is 6 * 12 (every combination of a six sided dice and a 12 sided dice).

The general way to do this, particularly for such a small problem space is to list out every possible combination, make rectangular table with 12 entries on one side and 6 entries on the other and fill in who "wins," sum up the winners for each side and divide by 72 (the total number of possibilities).

In a comment above it's shown how to do this more simply, but this is the general solution to these sorts of problems.

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