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ken vh
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11 months ago
in Interview Questions: Two Bowling Balls | 20bits on 20bits
I couldnt help myself and went ahead and found the optimal strategy under the (not unreasonable) assumption of uniform distribution. The 14 strategy outlined above is the best worst case, but it will always take 14 tries. The following sequence of skips has the optimal expected tries of around 12.84--not much of an improvement--with worst case 19:
10, 9, 9, 8 ,8 ... 3, 3, 2, 2, 1, 1
10, 9, 9, 8 ,8 ... 3, 3, 2, 2, 1, 1
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Can you share the math that gave you those numbers? I'd be interested in taking a look.