An odd(s) problem

Probability Level pending

A certain position in backgammon has odds of 8 to 1 against bearing off in a single move, but odds of 169 to 47 in favor of bearing off in 2 moves or fewer.

To three decimal places, what are the chances of successfully bearing off from this position in EXACTLY 2 moves?


The answer is 0.671.

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1 solution

Denton Young
Oct 19, 2018

Odds of 8 to 1 against mean a 1/9 chance. Odds of 169 to 47 in favor mean a 169/216 chance.

So the chance of bearing off in exactly 2 moves is 169/216 - 1/9, which is 145/216 or 0.671

I found that 169 216 1 9 = 169 24 216 = 145 216 0.671 \dfrac{169}{216} - \dfrac{1}{9} = \dfrac{169 - 24}{216} = \dfrac{145}{216} \approx 0.671 . Fortunately this was close enough to the posted answer of 0.657 0.657 to be accepted as correct.

Brian Charlesworth - 2 years, 7 months ago

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Fixed: now I just need to change the answer. It's been a long week.

Denton Young - 2 years, 7 months ago

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