# Probability + Gambling = Interesting Question

1. Sep 10, 2008

### His_Dudeness3

Hey everyone, I was given this fun little probability question from my tutor after I finished early in one of my classes about three weeks back, and I just can't seem to crack it!! Something about gambling and probability makes my brain go haywire (or maybe its some other, deeper problem :uhh:). Anyway,here it is, and have fun!!

A roulette wheel is numbered from 0 to 36. 0 is Green. Half of numbers are
Red and half are Black. The game has an entrance fee $1. The player then stakes$10 and must choose the parity (Odd or Even) and the color (Red or Black). If
he gets right parity or color $12 is returned, that is a gain is$1. If he get right
both parity and color $20 is returned, , that is a gain is$9. If he does guesses
neither correct color nor parity, and the number is not 0, then the entrance fee \$1
is returned. If 0 comes up, the player gets nothing.

(a) If X is the gain on a single game, complete the table of the probability distribution
of random variable X:

(b) Find E(X) and standard deviation of X

(c) If player plays twice, what is the probability that he comes out ahead (i.e.
positive net gain).

(d) If player plays this game fifty times, find the mean and standard deviation of
his overall net gain.

(e) Use your answer to part (d) and a suitable approximation to calculate the
probability of coming out ahead after playing fifty games.

(f) Given a roulette wheel where the half of odd numbers are Red and half are
Black, and similarly for even numbers, check that color and parity appear
independently.

2. Sep 10, 2008

### Focus

Doesn't look that interesting to me.
a)b) Analysing the question (pretty simple)
c)$$Pr(X_1+X_2>0)$$ use conditional probability to solve that (Law of Total probability)
d) Use linearity of expectation and the indipendence for the variance.
e)Depends on d) but my hunch is that you can work out how many s.d. the mean is away from zero and use that.
f)no idea what that is asking for

3. Sep 10, 2008

### His_Dudeness3

For some reason, I can't figure out if the distribution I get for part (a) is correct (as pretty much the rest of your answers would be wrong if you don't get the right distribution):

x -11 -10 1 9
Pr(X) (1/37) (32/37) (2/37) (1/37)

Can anyone clarify if this is correct?

4. Sep 11, 2008

### rbeale98

this is not interesting at all, nor is this even close to the payout scheme of roulette

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