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Betting Paradox



Here is a challenge for you guys:

Design three betting games where when played individually or in pairs are losing games but when all three are played in some alternating pattern the combined effect provides a positive result to the player.

**Make sure that the pair games are always result in a lost for player regardless of the way the player plays**

Edited by BMAD
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The casino keeps a running count, N, which is the number of games the player has played, modulo 3. So N can be 0, 1, or 2.

These are the games:

Game 0: if N=0, take action A, else take action B.

Game 1: if N=1, take action A, else take action B.

Game 2: if N=2, take action A, else take action B.

These are the actions:

Action A: pay $3, then roll a regular die and gain $ equal to your roll.

Action B: pay $599, then roll 100 regular dice and gain $ equal to your total roll.

It is obvious that action A is a slightly winning bet, while action B is a severely losing bet. If you can only play one game, you will be forced to take action B two games out of three. If you can only play two of the games, you will be forced to take action B one game out of three and still lose. If you can play any game, simply play game N every time and you will always take action A.

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so the games are as follows. if the number of games played is evenly divisable by 3, pay 3, roll a dice and take money equal to roll.

else pay 599 dollars, roll 100 dice and get money equal to roll. the other two games are the same, except for the slight differance of number of games being +1 evenly divisalbe by 3, and +2 evenly divisible by 3.

this system meets the nessicary criteria. that is any single game or any two games picked are definately losing, since at least 1/3 games will be severely losing. but if you can play all 3 in any order. you can simply play the first game then second game then third game and back to first and have a winning strat.

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