Answer to Question #318878 in Statistics and Probability for Rachel

Question

A game called “It Pays to Wait” is played in which you roll a single die. If you roll a six on the first try you lose. If you roll a six on the second try you win $5 and the game is over. If you roll a six on the third try you win $10 and the game is over. If you roll a six on the fourth try you win $15 and the game is over. The maximum number of rolls a player can take is four. It costs $4.00 to play the game. Determine the expected profit for the operator of this game after 200 games have been played.

Answer 1

The expected value is the sum of each outcome multiplied by the probability of that outcome,

For the person betting, the four outcomes are winning 5, 10,15$ and losing 4$ .

To win 5$ (5*1/6*5/6)

To win 10$ (10*5/6*5/6*1/6)

To win 15$ (15*5/6*5/6*5/6*1/6)

To lose 4$ (-4*1/6), if you roll 6 on the first try. Or (-4*5/6)⁴, if you do not roll 6 at any round.

The expected value is then 5 * 1/6*5/6 + 10*5/6*5/6*1/6+15*5/6*5/6*5/6*1/6+ (-4) * 1/6+(-4)*(5/6)⁴=0.694+1.157+1.447-0.667-1.929 which gives an expected value of 0.7 meaning you can on average expect to win 0.7 $ every time you play.

So operator will lose 0.7 $ every time.

For 200 games operator will lose 140.4$