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7 years, 1 month ago
Our good friends that like games, Alice and Bob, play a game called "Say 100 and win".
The rules are:
In the beginning they use one regular 6 faced die to get a random number between 1 and 6. Let's call this number $n$
Alice goes first and says a number $k$ where $n < k \leq n+10$.
Bob goes second and says a number $m$ where $k < m \leq k+10$.
They take turns following the rules above. Each one has to say a number bigger than what the previous said but not bigger than 10+ previous number.
The person that gets to say 100 wins the game.
As we all know Alice and Bob, they are smart and play the optimal strategy.
What are the chances of each winning the game and what's the strategy?
45.1k 10 10 gold badges 146 146 silver badges 378 378 bronze badges
Apr 6, 2016 at 10:23
17.8k 4 4 gold badges 53 53 silver badges 100 100 bronze badges
alice wins with probability 5/6
whoever gets to say 89 wins, because the other guy then has to add 1-10 to that number, making it possible to say 100. Therefore whoever gets to say 78 wins, by the same reasoning. Keep subtracting 11 in the same way until you get to 12. Whoever gets to say 12 wins, therefore as long as the dice rolls >1, alice gets to say 12 and wins. theres a 5/6 chance of this happening. If the dice rolls 1, bob wins.
Apr 6, 2016 at 10:38
2,839 14 14 silver badges 14 14 bronze badges
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