In the game of roulette, a player can place a $9 bet on the number 15 and have a

Question

In the game of roulette, a player can place a

$9

bet on the number

15

and have a

1/ 38

probability of winning. If the metal ball lands on

15,

the player gets to keep the

$9

paid to play the game and the player is awarded an additional

$315.

Otherwise, the player is awarded nothing and the casino takes the player's

$9.

A) What is the expected value of the game to the player?

B) If you played the game 1000 times, how much would you expect to lose? Note that the expected value is the amount, onaverage, one would expect to gain or lose each game.

Explanation / Answer

A) The expected value of the game to the player here is computed as:

= Probability of winning the bet * 315 - Probability of losing the game * 9

= (1/38)*315 - (37/38)*9

= - 0.47

Therefore the expected value to the player here is - $0.47 ( that is a loss of 47 cents )

B) If we play the game 1000 times we are expected to lose

= 1000*0.47

= $470

Therefore we are expected to lose approximately $470

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