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I would like to understand the math behind a basic racing game.

2d6 are rolled and the result advances the corresponding horse along the track (a roll of '7' would advance the #7 Horse). To win the race, the #7 Horse needs to advance a total of 8 times - while the # 2 and #12 horses only need to advance three times.

How do I calculate the probability that the #2 horse will win the race?

[NOTE: The 3/11 Horses must advance 4 times, the 4/10: 5 times, the 5/9: 6 times, 6/8 horses: 7 times. These do not need to be included in the answer, but are included for completeness.]

I understand the basic probability of rolling two fair six-sided dice.

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    Just answering your title as a hint: You need to add the probability of your event happening given $n$ rolls, then you need to sum the result over all $n$, I believe. A very interesting question. – 2017-01-19

1 Answers 1

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Start with the question in the title.

The probability that you throw 3 twos before 7 sixes is: that on the $3+7$ tries that matter, the first $3+7-1$ contain 3 twos and 6 sixes in some order, and try $3+7$ contains the seventh six.

$$\mathsf P(~\uparrow~) = \dbinom{3+7-1}{3}\dfrac{2^3\,5^7}{(2+5)^{3+7}}$$

... because the probability of obtaining a two given you obtain a two or a six is:$\frac{2/36}{2/36+5/36}$

Moving on, the probability that you throw 3 twos before 7 sixes or 3 twelves is: that on the tries that matter you do not throw all the twelves or all the sevens before the last two. Thus by the rule of complements, and the principle of inclusion and exclusion:

$$\mathsf P(~\uparrow~) = {1-\dbinom{3+7-1}{7}\dfrac{2^3\,5^7}{(2+5)^{3+7}}\\-\dbinom{3+3-1}{3}\dfrac{2^3\,2^3}{(2+2)^{3+3}}\\+\dbinom{3+7+3-1}{7+3}\dfrac{2^3(5+2)^{7+3}}{(2+5+2)^{3+7+3}}}$$

And, as you see, the probability that twos win the race will be rather complicated.