Probability Question from Frederick Mosteller, #7

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?

Thanks in advance!

Mean Joe

TS Contributor
When Mosteller does his calculation he does not include the Gambler's original stake
Actually Mosteller does. In a winning situation, the Gambler plays $1 and gets back $36 ($35 + the stake). Mosteller's calculation uses the net, and in a winning situation that net is 36-1=35. In a losing situation, the net is -1, on which you agree with Mosteller.