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A game

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A and B played a game with some amount. They agreed that whoever will lose have to give half his amount to the winner.

They played 100 games and a result was that A won First game B won next two games A won next three games and so on

A B B A A A B B B B A A A A A...............

Who will have graeter amount after the 100th game?

Edited by Utkrisht123
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Posted (edited) · Report post

Unless they both start with zero (which would make the game pointless), in which case that both A and B would always have the same amount, then whoever takes the 100th turn will have the most, as they'll have 50% of their opponent's points plus the points they already had.


As B has the 100th turn (which is in the 14th set of turns taken between the two of them), B will have the most points at the end of turn 100.
Edited by David_Walker
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Posted (edited) · Report post

After 91 games, A will hace won 13 in a row (1+2+..+13=91) and B will be virtually penniless. Howevere, B wins the next 9 games, decreasing A's stack to 1/2^9, leaving A now virtually broke. Assuming both started with equal money

But david walker has easier solution

Edited by jhawk
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Posted · Report post

A and B played a game with some amount. They agreed that whoever will lose have to give half his amount to the winner.

They played 100 games and a result was that A won First game B won next two games A won next three games and so on

A B B A A A B B B B A A A A A...............

Who will have graeter amount after the 100th game?

Whosoever wins the latest game will have more money, so B.

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Posted · Report post

B will win with 51 to 49

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Posted (edited) · Report post

Without cognitive bias called "Gambler's fallacies" 3 : 2 in favor of A chances is presumable in the future..it is unlikely to happen for 6 consecutive win for B . Why is it that in NBA no team has ever recovered and win from 3:1 deficit in the finals

Edited by TimeSpaceLightForce
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