Hard Prob/Stat Question! Please help!?
Posted by admin in Uncategorized on 05 21st, 2010 | one response
A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specied
number, you either win 35 if the roulette ball lands on that number or lose 1 if it does not.
If you continually make such bets, approximate the probability that
(a) you are winning after 34 bets;
(b) you are winning after 1,000 bets;
(c) you are winning after 100,000 bets.
Assume that each roll of the roulette ball is equally likely to land on any of the 38 numbers.
Can someone help please? Thanks!
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You can model this with a binomial distribution.
For (a) you are going to run 34 trials so n = 34
The probability of success is 1/38 = p
The probability of failure = 1-p = q = 37/38.
It is going to be easier to calculate the probability of loosing so I will figure that out and then subtract it from 1.
To loose after 34 bets, you have to have:
1) No winnings then loss is 34 and winnings is 0
After that you have at least won 35 so you will made more than 34 and be “winning”.
Now use the formula for calculating the binomial probability or you use and on-line calculator or a calculator. I used an on-line calculator.
n = 34, p = 1/38 = 0.0263 and x = 0
See reference below for a calculator.
P(losing) = 0.404. So P(winning in 34 bets) = 1 – 0.404 = 0.596
(b) After 1,000 bets you need to win more than 1000/35 times = 29.
Again it is easier to calculated the P(losing)
n = 1000, p = 0.0263 and x <=28
Using the on-line calculator:
P(losing) = 0.677, so P(winning) = 0.323
(c) After 100, bets you need to win more than 100,000/35 times = 28578.
Again it is easier to calculated the P(losing)
n = 100,000, p = 0.0263 and x <=28578
Using the on-line calculator, did not work, so I used Excel.
P(losing) = 1, so P(winning) = 0 . Not too surprising.