In a game of chance, I bet x dollars with probability p, I win y dollars...

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Question

In a game of chance, I bet x dollars with probability p, I win y dollars...

In a game of chance, I bet x dollars with probability p, I win y dollars (considering y is the profit). What should x be for this to be a fair game? What is the value of x when y=$10 and p=0.2?

math Statistics-And-Probability

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Answer 1

Answer #1

Answer:

Given,

P(success) = p

P(failure) = q = 1 - p

Consider,

If the game is fair, then the expected value = 0

Mean = 0

xp(x) = 0

p(y-x) - (1-p)x = 0

py - px - x + px = 0

x = py

It is given y = 10

p = 0.2

substitute in x

x = 0.2*10

= 2

x = 2

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Answer 2

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In a game of chance, I bet x dollars with probability p, I win y dollars...

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