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?
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I found that 2 1 6 1 6 9 − 9 1 = 2 1 6 1 6 9 − 2 4 = 2 1 6 1 4 5 ≈ 0 . 6 7 1 . Fortunately this was close enough to the posted answer of 0 . 6 5 7 to be accepted as correct.
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Fixed: now I just need to change the answer. It's been a long week.
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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