Rainman

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• We should reject the first dowry.
• We should never accept a dowry unless it is larger than every dowry we have seen before.
• When we do accept, our chance of success depends only on total number of dowries seen.

With this in mind, I propose this strategy: start by seeing and rejecting K dowries, K being at least 1 and at most 99, then accept the first one that is larger than each of those. Now we need to find the chance of success for this strategy. We could succeed by correctly accepting the nth dowry, for any n from K+1 to 100. So our total chance of success is the sum of the chances of correctly accepting the nth dowry, for all n from K+1 to 100.

Note that our chance of refusing a dowry is K/(K+1), i.e. the chance that it is not larger than all of the K dowries we started by rejecting. Let x=K/(K+1) for simplicity.

P(correctly accepting dowry #K+1) = 0.01

P(correctly accepting dowry #K+2) = 0.01*P(refusing dowry #K+1) = 0.01*x

P(correctly accepting dowry #K+3) = 0.01*P(refusing dowries #K+1 and #K+2) = 0.01*x²

^...

P(correctly accepting dowry #100) = 0.01*P(refusing dowries #K+1 through #99) = 0.01*x^99-K

This is a geometric sequence, the sum of which is P(success) = (0.01-0.01*x^100-K)/(1-x) = (0.01-0.01*(K/(K+1))^100-K)*(K+1).

Now to maximize this with respect to K, anyone in the mood for some differentiation?

I threw some numbers into my computer calculator instead.

K = 37 yields P(success) = 0.3092

K = 44 yields P(success) = 0.3222

K = 45 yields P(success) = 0.3227

K = 46 yields P(success) = 0.3229

K = 47 yields P(success) = 0.3227

K = 48 yields P(success) = 0.3223

K = 55 yields P(success) = 0.3111

It seems the optimal strategy is to reject the first 46 dowries, then accept the first dowry that is larger than each of those.