In a game of roulette, Jorge places 170 bets of each on the number 3. A win pays off with odds 35:1 and on any one spin there is a probability that 3 will be the winning number. Among the 170 bets, what is the minimum number of wins needed for Jorge to make a profit? Estimate the probability that Jorge will make a profit.

probability that Jorge makes a profit is = 0.46412  

Step-by-step explanation:

Solution:-

- The number of bets made on number "3", N = 170

- He bets on each number "3", k = $1

- The winning pay-off odds : $ ( 35 : 1 )

- The probability of getting number "3" on a spin, p = 1/38

- The total amount paid (C) for n = 170 bets on number "3" are:

                 C = N*k

                 C = (170)*($1)

                 C = $170

- The probability of getting a number "3" on a spin is independent for each trial.

Denote:

- The amount received per win = $ 35

- The number of wins = r

- So the minimum "N" number of wins must be enough to match loss.

                 Amount Win = Amount Loss

                 r*$35 = C

                 r*$36 = C

                 r = $170 / 36

                 r = 4.7222 ≈ 5 wins

- So the minimum amount of wins required by r = 10 to make a profit.

- Let a random variable "X" denote the number of times Jorge spins to get number "3" - Number of wins. The probability to get a number "3" on each spin is independent for each trial. Therefore X follows Binomial distribution.

- So,                         X ~ B ( N , p )

                               X ~ B ( 170 , 1/38 )  

                               1 - p = 37 / 38

- So we need to determine that Jorge get number "3" at-least r = 5 times. Where the probability mass function for binomial distribution is given below:  

So,

- So the probability that Jorge makes a profit is = 0.46412                                        

Note:- The normal approximation to Binomial distribution may be a less cumbersome choice; however, care must be taken to verify the conditions for normal approximation i.e

                          N*p ≥ 10

With the given data, N = 170 , p = 1/38:

                          N*p = 170/38 = 4.4737 ≤ 10

Hence, the normal approximation is an invalid choice for the data given.

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