Mark Gritter (markgritter) wrote,
Mark Gritter
markgritter

Modelling 1 vs. 100

The game show "1 vs. 100" (on NBC in the U.S. though many variants exist throughout the world) is a quiz show pitting a lone constestant against "the mob".

The player accumulates a prize pool by answering multiple-choice questions correctly. He gets a fixed amount per member of the mob that answers the question incorrectly. Those mob members are then eliminated from future rounds. If the player gets a question wrong, the mob members who got the question correct split his prize pool. In the U.S. version, a player may choose to take his winnings and leave after answering a question correctly. He or she also receives a $1 million prize if all of the members of the mob are eliminated; it's not clear to me how many questions are allowed.

When should a rational player pack up and leave? How much does the nonlinear payoff weight the results?

For example, the maximum value per question is $10,000. Suppose there is just one mob member left, and that you and the mob member have just 1/3 chance of getting the question right. Then, ignoring the $1M jackpot:

1/9th of the time you both get it right and go on
2/9th of the time you get it right and your opponent does not, so you win $10K
6/9th of the time you get it wrong and you lose your prize pool (and your opponent may or may not get the money.)

Let your prize pool = Y and the expectation of playing this game = X. Then X = (1/9)X + (2/9)(10000) + (7/9)(-Y). X=(20000-7Y)/8, so if your prize pool is greater than $2857 you shouldn't play.

With the jackpot, X = (1/9)X + (2/9)(1000000-Y) + (7/9)(-Y), so X = (2000000-9Y)/8. If your prize pool is greater than $222,222 you shouldn't play.

Now suppose there are two mob members remaining, again with 1/3rd chance of getting the question right. Then

1/27 of the time everybody gets it right and goes on
4/27 of the time you get it right but both other players are wrong and you win the $1M
4/27 of the time you get it right and so does just one mob member, so you get $10K and the opportunity to play the game above.
18/27 of the time you get the question wrong

X=(1/27)X + (4/27)(1000000-Y) + (4/27)*(Y+10000-MAX(0, (2000000 - 9*(Y+10000))/8)) + (18/27)(-Y)

Left as an exercise for the reader... I think I need to go off and write a spreadsheet or something.
Tags: games, math
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