Aaron plays a game in which in round 1, he automatically wins $1. Each successive round, he wins or loses twice the amount of the previous round, with the win or loss randomly chosen.

Question

He may declare in advance of play that he will drop out after a certain number of rounds. How much should he expect to win on the average, and how many rounds should he choose (in advance) to maximize his expected winnings? Possible multiple choice answers are: A) $0, don't play at all B) $1, play 1 round C) $2, play two rounds D) $6, play 3 rounds E) no limit to expected winnings, play as long as possible

Answer 1

Expected value = a predicted value of a variable, calculated as the sum of all possible values each multiplied by the probability of its occurrence.

At "Each successive round." the expected value is +0.5*$x - 0.5*$x = $0, where $x is "twice the amount of the previous round" and the +/- 0.5 is the probability of win/lose.

Now, "in round 1, he automatically wins $1," so the expected value is: 1.0*$1 = $1 (probability = 1.0 to win $1)

Thus, this is the maximum value of his expected winnings. Aaron should play one round and claim his $1.

Note, however, that actual winnings or losses may be large amounts (with various probabilities).

Answer 2

Strategy: (B). There is NO expected increase on earnings, by playing longer, because the wins and losses are random. Of course, in actual gambling situations, “the house” takes a percentage, and so, the longer one plays and/or the more one bets, the more (on the average) one loses. If, as is customary at hotels with associated casinos, one has a no-cash-value voucher for an amount of free chips, one might opt to play the chips to convert them to possible winnings, and then stop play. Human brain chemistry being what it is, one is more stimulated by winning than losing, and one tends to keep playing until the casino wins all -- theirs as well as yours.

-- Cheers, -- Mr. d.