In the game of roulette, when a player gives the casino $21 for a bet on the number 33, the player has a 37/38 probability of

Question

3 years ago

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In the game of roulette, when a player gives the casino $21 for a bet on the number 33, the player has a 37/38 probability of

losing $21 and a 1/38 probability of making a net gain of $735.(The prize is $756, but the player's $21 bet is not returned, so the net gain is $735.) If a player bets $21 that the outcome is an odd number, the probability of losing $21 is 20/38 and the probability of making a net gain of $21 is 18/38. (If a player bets $21 on an odd number and win, the player is given $42 that includes the bet, so the net gain is $21.)
If a player bets $21 on the number 33, what is the player's expected value?

The expected value is _______ dollars. (round to the nearest cent as needed)

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Mathematics

1 answer:

ohaa [14] · Answer 1 · 2022-02-16T11:19:48+03:00

ohaa [14]3 years ago

4 0

Answer:

Step-by-step explanation:

In the game of​ roulette, when a player gives the casino $21 for a bet on the number 33​, the player has a 37/38 probability of

In the game of roulette, when a player gives the casino $21 for a bet on the number 33, the player has a 37/38 probability of