Question

Suppose any one round of a gambling game pays 3 to 1 and the odds are 4 to 1 against you. Suppose you will play for 100 rounds and bet one dollar on each round. We want the probability you will come out ahead.
a. Determine what the possible net gains are. (Not all integers from −100 to 300 are possible.)

b. What is the smallest possible positive net gain?

c. What is the chance you will come out ahead?

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

The odds are 4 to 1 against, so we can estimate the probability of success (p) as

$\frac{p}{q}=\frac{p}{1-p}=\frac{1}{4}\\\\4p=1-p\\\\5p=1\\\\p=0.2$

The expected pay for every success is 3 to 1, so we lose $1 for every lose and we gain $3 for every win.

The number of winnings in the 100 rounds to be even can be calculated as:

$W+L=100\\\\L=100-W\\\\\\Payoff=0=3*W-1*L=3W-1*(100-W)=3W+W-100\\\\0=4W-100\\\\W=25$

We have to win at least 25 rounds to have a positive payoff.

As the number of rounds is big, we will approximate the binomial distribution to a normal distribution with parameters:

$\mu=np=100*0.2=20\\\\\ \sigma=\sqrt{npq}=\sqrt{100*0.2*0.8}=4$

The z-value for x=25 is

$z=\frac{X-\mu}{\sigma}=\frac{25-20}{4}=1.25$

The probability of z>1.25 is

P(X>25)=P(z>1.25)=0.10565

There is a 10.5% chance of having a positive payoff.

NOTE: if we do all the calculations for the binomial distribution, the chances of having a net payoff are 13.1%.