Question Heip In the game of roulette, a player can place a $8 bet on the number

Question

Question Heip In the game of roulette, a player can place a $8 bet on the number 27 and have a an additional $280 Oherwise, the player is awarded nothing and the casino takes the 38 probabity of winning Ithe metal bai lands on 27, the player gels to keep the $8 paid to pay the game and he player awarded player's $8 What is the expected value of the game to the player? if you played the game 1000 bmes, how much would you expect to lose? The expected value is Round to the nearest cent as needed) The player would expect to lose about S Round to the nearest cent as needed)

Explanation / Answer

Solution:-

Winning probability = 1/38 = 0.026316

Expected value = Winning probability × Winning amount - Lossing probability × Amount paid

Expected value = 0.0263158 × 280 - 0.973684 × 8

Expected value = 7.3684 - 7.78947

E(X) = - 0.42107

Expected value in playing 1000 times = 1000 × - 0.42107 = - 421.07

The player would expect to loose about $421.07.

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---