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The entry fee is $11, and the entrants make the recipes at home and vote for their favorite, which cannot be their own recipe. The top 2 finishers win money. First place wins $58, and second place wins $48. You give yourself an X chance of finishing in the top 2, and if you finish top 2, then you have an equal chance of finishing first or second.

Required:
What is x such that entering this tournament is a fair gamble for you as an expected value maximizer?

1 Answer

1 vote

Answer:

The value is
X =  0.1719

Explanation:

From the question we are told that

The entry fee is
a =  \$ 11

The outcome for first place is
b =  \$ 58

The outcome for second place is
c =  \$ 48

The outcome for losing is
d =  - \$ 11

The chance of finishing in the top two is X

Given that there are equal chance of finishing first or second, then the chance of finishing first is
(X)/(2)

and the chance of finishing second is
(X)/(2)

Then the chance of losing(i.e not finishing in the first two) is
1 - X

given that the game is a fair gamble for the player as an expected maximizer then it mean is that the expected value of entering the tournament E(X) = 0

So


E(X) =  b *  (X)/(2)  +  c  *  (X)/(2) *  d  *  ( 1 - X)

=>
0 =  58 *  (X)/(2)  +  48  *  (X)/(2) *  -11  *  ( 1 - X)

=>
0 =  (58X)/(2)  +   (48X)/(2) *  -11  + 11X)

=>
X =  (64)/(11)

=>
X =  0.1719

User Ludesign
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