Requesting assistance on the following. I am under the beliefthat the properties

Question

Requesting assistance on the following. I am under the beliefthat the properties of expectations for constant values is tobe used on this, but I am not sure how to plug the numbers in.


You and a friend play the following game: You pay yourfriend $3 each turn and then flip a fair coin. If it’s tails,your friend pays you $ (2^n), where n is the number of timesyou’ve flipped the coin, and the game ends. If it’sheads, you have the choice of stopping or continuing. If you have mdollars to start with, and you play the game either until you winor until you have no money left, what will you win on theaverage?

m = amount of money n = number of flips

E(cX) = cE(X)

m = 3X so E(m) = E(3X) = 3E(X) = 3 x 2 = 6

So if this is correct, then the amount that I should win onaverage would be $6. If this is not correct, please offer yourguidance. Thanks in advance! Requesting assistance on the following. I am under the beliefthat the properties of expectations for constant values is tobe used on this, but I am not sure how to plug the numbers in.


You and a friend play the following game: You pay yourfriend $3 each turn and then flip a fair coin. If it’s tails,your friend pays you $ (2^n), where n is the number of timesyou’ve flipped the coin, and the game ends. If it’sheads, you have the choice of stopping or continuing. If you have mdollars to start with, and you play the game either until you winor until you have no money left, what will you win on theaverage? m = amount of money n = number of flips E(cX) = cE(X) m = 3X so E(m) = E(3X) = 3E(X) = 3 x 2 = 6 So if this is correct, then the amount that I should win onaverage would be $6. If this is not correct, please offer yourguidance. Thanks in advance!

Explanation / Answer

I'm afraid I can't follow your reasoning here. Forstarters, you don't know what m is, so your answer should be afunction of m. Second, the amount you win is 2^n, not 2*n,which means you have an exponent, and you can't pull exponents outof expectations. (Expectations follow all the same rules assums do.) So I don't think you're goign to be able to use therules about constants. The only way I see to approach thisproblem is to construct the expectation out of the definition of anexpectation. I assume your book covered the definition ofexpected value. . Let's think through a couple of possibilities. . Suppose m = $3. Then you can afford only flip of thecoin. There's a probability p=0.5 that you win 2^1=$2 (afterpaying $3!) and a probability p=0.5 that you lose and cannotcontinue. Those are the only 2 possible states. By thedefinition of expectations as the sum over all possible states of(the value in that state times in the sum in that state), thisgives us E = 0.5(-1) + 0.5(-3) = -0.5 - 1.5 = -2. So if m=3,E(X) = -2, where X is the amount of money you have at theend. . Suppose m=$6. Then you can afford two flips of thecoin. This gives us 3 possible states:     probability 0.5 you win the first toss, win$2, and pay $3, for value -$1     probability 0.25 you lose the first toss,win the second, pay $6, win $4, for value -$2     probability 0.25 you lose both tosses, pay$6, for value -$6 . E(X if m=6) = 0.5*(-1) + 0.25*(-2) + 0.25*(-6) = -0.5 -0.5 -1.5 = -2.5 . I'm starting to wonder if this person should really beconsidered a "friend"! This does not look like a game Iwant to play! . Can you use the definition to calculate the expected value ifm=9, so you can make up to 3 flips? . Can you use the definition to calculate the probability ofwinning on flip n times the payoff if you win on flip n? Thisis NOT conditioned on first having lost the first n-1 flips. The probability of losing on flip 2 times the payoff if you win onflip 2 is 0.5*0.5*($4-$6). . From your last problem we worked on together, I know youdon't like to be given the answer straight off (which is good,because I don't give answers straight off). I'm going toleave you for now with this, but I'll be happy to comment some moreafter you've worked on this for a bit.

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