We have seen that, in playing roulette at Monte Carlo (Example 6.13), betting 1

Question

We have seen that, in playing roulette at Monte Carlo (Example 6.13), betting 1 dollar on red or 1 dollar on 17 amounts to choosing between the distributions m_x = (-1 18/37 -1/2 1/37 1 18/37) or m_x = (-1 36/37 35 1/37) You plan to choose one of these methods and use it to make 100 1-dollar bets using the method chosen. Using the Central Limit Theorem, estimate the probability of winning any money for each of the two games. Compare your estimates with the actual probabilities, which can be shown, from exact calculations, to equal.437 and.509 to three decimal places.

Explanation / Answer

Let us work on each of the bets separately

Bet1 : $1 on red

Let x be the amount won or lost, then probability distribution of x is given as

x p(x)

-1 18/37

-1/2 1/37

1 18/37

Step I: Find mean.

We can find the mean or expected value of x, E(x) = x*p(x)

µ = E(x) = -1*(18/37) + -1/2*(1/37) + 1*(18/37)

µ = E(x) = -0.01351

Step II: find standard deviation:

Variance = E(x^2) - µ^2

E(x^2) = x^2 * p(x)

E(x^2) = (-1)^2 * 18/37 + (-1/2)^2 * 1/37 + (1)^2 * 18/37

E(X^2) = 0.97973

Hence Variance = 0.97973 - (-0.01351)^2

Variance = 0.979547

Standard deviation() = root of variance = (0.979547)

= 0.9897

Step III: Find probability of winning. For winning x>0

n = 100 bets

Applying central limit theorem we get:

P(Z> (0-100µ)/10) = 1 - P(Z < (0-100µ)/10)

Hence P(Z < (0-100µ)/10) = P(Z < (0 + 1.351)/9.897) = P(Z < 0.1365) = 0.5557

Hence P( Z > (0-100µ)/10) = 1 0.5557 = 0.4443

Bet II: $1 on 17

Let x be the amount won or lost, then probability distribution of x is given as

x p(x)

-1 36/37

35 1/37

On repaeting the same steps as used for bet I, we get the following result:

Step I : mean = µ = -0.02703

Step II: Standard deviation = = 5.8378

Step III: Find probability of winning, follow the same steps as above, for winning x>0

Given n =100

Apply central limit theorem

P(Z > (0-100µ)/10) = 1 - P(Z <(0-100µ)/10)

P(Z < (0-100µ)/10) = P(Z < (0+2.703)/58.378) = P(Z < 0.046) = 0.5185

Hence P(Z> (0-100µ)/10) = 1 - 0.5185 = 0.4815

Hence we get estimated probability for bet 1 = 0.4443 & for bet 2 = 0.4815

Hence we find that bet 2 has more chances of winning

Compare estimates with actual probability

For bet 1 actual probability (0.437) is less than the estimated probability (0.444)

For bet 2 actual probability (0.509) is more than the estimated probability (0.4815)

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