Probability Question from Frederick Mosteller, #7

#1
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
#2
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.