Question

Alice is a good gambler and always knows when to stop. Suppose in a game, in each round she has a winning probability of 0.8 and she will stop at the first loss. Try to find the upper bound on the probability that she can win at least 10 rounds using Markov's inequality.

Answer 1

The answer is "0.500".

Explanation:

The chance that round will be won,
`P (win) =0.8`

The likelihood of losing round by complementary law

`P(loss) = 1 -P(win)\\\\`

`=1-0.8\\\\=0.2\\\\`

Let X be the number of rounds to the first loss played.

Here,
`X \sim Geometric (p=0.2)`

The PMF of X is the following:

`P(X=x)=p(1-p)^x\\\\`

`=(0.2)(1-0.2)^x\\\\=(0.2)(0.8)^x, \ \ \ \ \ x=1,2,.....`

The mean of X is the geometric random variable property,
`E(X) =(1)/(P)= (1)/(0.2)=5\\\\`

Markov's Inequality:

Suppose X is any random variable that has no negatives. So,

`P(X \geq s)\leq (E(X))/(s) for s >0\\\\`

Find the upper bound of
`P (X \geq 10)`

`P(X \geq 10) \leq (E(X))/(10)`

`=(5)/(10)\\\\=(1)/(2)\\\\=0.5\\\\`

The highest probability is, therefore, that Markov's inequality will win at least ten rounds: 0.500