Description
Alter/Nate
Two friends, Alter and Nate, have a conversation:
Alter: Nate, let’s play a game. I’ll pick an integer between 1 and 10 (inclusive), then you’ll pick an integer between 1 and 10 (inclusive), and then I’ll go again, then you’ll go again, and so on and so forth. We’ll keep adding our numbers together to make a running total. And whoever makes the running total be greater than or equal to 100 loses. You go first.
Nate: That’s not fair! Whenever I pick a number X, you’ll just pick 11-X, and then I’ll always get stuck with 99 and I’ll make the total go greater than 100.
Alter: OK fine. New rule then, no one can pick a number that would make the sum of that number and the previous number equal to 11. You still go first. Now can we play?
Nate: Um… sure.
Who wins, and what is their strategy?
Solution
The first player, Nate, is guaranteed to win in this case.
Our goal is to let Nate get the sum of all , which would eventually let him get , a guaranteed win.
Therefore, our strategy for Nate is to let first pick number , which contributes to the offset. Later, for him to increment by each round, if Alter picks any number , Nate can pick the number . Otherwise if Alter picks , Nate then picks . Given the rule that two consecutive number cannot add up to the sum of , Alter can only pick in this case. Such that in these three rounds (Alter, Nate, Alter). Nate will then pick to reach .
If Nate can guarantee he reaches a sum of , he wins the game by reaching .
More
For more puzzles, checkout
- Jane Street Puzzles