# Probability Question from Frederick Mosteller, #7

#### barrelbowl

##### New Member
Hey all,

I'm trying to improve my probability skills so hopefully the questions I'm asking aren't too amateur-ish.

The following is a question from Frederick Mosteller's "Fifty Challenging Problems in Probability", #7 Curing the Compulsive Gambler.

A summary of the question is as follows:
• Gambler always plays \$1 on the number 13 in a game of American Roulette, which has 38 equally likely numbers
• Winner of roulette receives initial stake + \$35*initial stake, and loser is out initial stake
• Kind Friend wants to cure the Gambler and bets him \$20 at even money that he will be down at the end of 36 plays
• How does this cure work?

So my difference with Mosteller is in calculating the expected value of the game. The game has an expected value of:

(\$35+\$1)*(1/38)-(\$1)*(37/38) = -\$1/38*(36 trials) = -\$36/38 is expected value at the end of 36 trials.

When Mosteller does his calculation he does not include the Gambler's original stake, so he does \$35*(1/38)-\$1(37/38)=-\$2/38*(36 trials) = =-\$72/38 at the end of 36 trials.

For the bet with his friend he loses \$20 if wins at least once during his 36 games, because he will break-even. So his expected winnings on his off-bet with his friend are...((37/38)^36)*\$20-(1-(37/38)^36)*\$20=\$4.65.

So he gets \$4.65-\$36/38=\$3.73 according to me at the end of 36 trials if he wins once, and \$4.65-\$72/38=\$2.79 according to Mosteller. Who is right and why?