An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you will lose $1, Suppose that in one evening you play this game n=2500 times in a row.
(a) Estimate the probability that by the end of the evening you will not have lost any money.
(b) Estimate the probability that the number of "even rolls" (roll a 2, 4, or 6) will fall between 1250 and 1300.
(c) Estimate the probability that you will win $100 or more.

According to the honest coin principle, if the random variable X denotes the number of heads in n tosses of an honest coin (n ≥ 30), then X has an approximately normal distribution with mean, [tex]\mu=\frac{n}{2}[/tex] and standard deviation, [tex]\sigma=\frac{\sqrt{n}}{2}[/tex].

Here the number of tosses is, n = 2500.

Since n is too large, i.e. n = 2500 > 30, the random variable X follows a normal distribution.

The mean and standard deviation are:

[tex]\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25[/tex]

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, μ = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

= μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

= 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

= 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.