Probability-Random Variables?

Question

For a game of roulette, a gambler bets $1 on red with the probability 12/26 to win. He repeats this bet four times. What is the probability that her profit is zero?

Answer 1

If the profit is zero, i would assum that that means you neither won nor lost money, so for 4 trials you would need two wins and two losses.  Just caculate the probibility of that.  (ignoring the fact that sometimes a zero/dzero returns 1/2 your initial wager)

if 12/26 is the probibility of a win, and you **assume** the reamiaining 14/26 is a loss you would just multiply the ods of two consectutive wins and two consecutive losses.

14/26 may not be the actual odds of a loss depending on the rules offered at the particualr roullette table as suggested above.

It comes to 6.17% if done in that manor.

Answer 2

If you have to use your numbers given, from a statistical stand point over time you should hit the numbers given thus each time you don't hit red, your statistical odds go up. To find out your odds of not hitting a red number over 4 rolls you divide the 12 by your 26 and get 0.461 or 46% Subtract from 100 and that gives you 0.539 or a 54% chance of not winning. Multiply the .539 x .539 x .539 x .539 for the second, third and fourth spin. This gives you 0.084 which is 8.4%. Thus you would have a 8.4% chance of not hitting a red number and showing a zero profit.

Keep in mind though, on a standard roulette table there are 36 regular numbers not 26 and 18 are red and 18 are black. Then there is either 1 or 2 green numbers depending on the roulette verions. Thus you are looking at 18/37 or 18/38 as your true odds on red. With 37 numbers your odds are 7% and for 38 numbers your odds are 7.7% to not turn a profit.

Answer 3

you need two wins and two losses to break even.

Let X be the number of wins.  X has the binomial distribution with n = 4 trials and success probability p = 18/38 = 0.4736842



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.

The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.

X ~ Binomial( n = 4 , p = 0.4736842 )

the mean of the binomial distribution is n * p = 1.894737

the variance of the binomial distribution is n * p * (1 - p) = 0.99723

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.998614

The Probability Mass Function, PMF,

f(X) = P(X = x) is:

P( X =  0 ) =  0.0767336

P( X =  1 ) =  0.276241

P( X =  2 ) =  0.3729253 <<<<< ANSWER

P( X =  3 ) =  0.2237552

P( X =  4 ) =  0.05034492