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Statistics Help Please...?
In the game of roulette, a player can place a $7 bet on the number 2 and have a 1/38 probability of winning. If the metal ball lands on 2, the player gets to keep the $7 and is rewarded $245. Otherwise, the player is awarded nothing and the casino takes the $7. What is the expected value of the game to the player? If you played the game 1,000 times, how much would you expect to lose?
The expected value is:
The player would expect to lose about:
Please explain your answer also. I don't just want the answer. i want to understand how you got it. Thanks.
3 Answers
- cidyahLv 71 decade agoFavorite Answer
status---x----p(x)
Win ----252--- 1/38
lose ....-7....37/38
Expected value = 252(1/38) -7 (37/38) = (252-259) /38 =-7/38
Loss per game : -7/38
Loss per 1000 games = -7000/38 = $184.21
- TeresaLv 45 years ago
Having taken a number of university level statistics courses, I would tend to question any statistical results posted anywhere. There are so many vairables that for the most part, statistics are irrelevant in serving as an accurate predictor of much of anything. BB, Raji the Green Witch
- 1 decade ago
1000 spin he is expected to win
1000/38 times
It’s 26.3
Therefore the player will get
26.3x$252=$6631
And he will place at the table $7000
The difference is $368
You can simply multiply $7000 he played with 0.
$368 is 5.26% (the house advantage)
I am preparing to play roulette.
Source(s): roulette forums
