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Uh. I totally forgot everything about statistics and probabilities (and I'm too lazy to remember those, so I won't check numbers 4 and 6), but I think the core
by drdaeman 11y ago
Uh. I totally forgot everything about statistics and probabilities (and I'm too lazy to remember those, so I won't check numbers 4 and 6), but I think the core idea how it works is that Bob's option just has lower chances. Say, we toss a coin up to 3 times.
Possible outcomes are: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH
Alice successes are: HT, THT, and HHT, but Bob has less options: HH and THH. That's why he needs more tosses on average.
- deleted 11y ago[deleted]
- knughit 11y agoYes but why is Bob's harder? Answer without enumerating the full list of possible flip sequences
- jastr 11y agoIf they both fail to get what they need on their second tosses, it's possible for Alice to win on the next flip, but Bob needs two flips to win! If Alice has failed to win on her second toss, her flip sequence was "HH", so she can win on the next toss, by flipping a T! If Bob has failed to win on his second toss, his flip sequence was "HT", so he needs two more flips to win!