Question

Game of dice - fair dice:

If I throw a number >= 5 I win. If he throws a number =< 4 he wins.

I throw the first dice.

Given that he loses(throws a number >4) what is the probability of me winning ?

What is the expected number of throws before either of us wins?

Answer 1

We know a dice contain 6 sides having numbers 1-6

Let's take

Winning questioner =W1

Losing questioner=L1

Winning opponent=W2

Losing opponent= L2

If the questioner wins he should get 5 or 6 in his throw means the probability of winning P(W1) =2/6=1/3

Probability of lose P(L1) =4/6=2/3

Same way probability of winning his opponent P(W2)=4/6=2/3

Probability of Losing opponent P(L2)=2/6=1/3

Here given that th questioner started to throw and also said he wins in that then

Chance is like he wins in his 1st throw, or he lose his first also his opponent lose his 1st throw and he wins his 2nd throw so on

Then probability can be written as =W1+L1L2W1+L1L2L1L2W1+.......

P(winning questioner)=(1/3)+(2/3*1/3*1/3)+(2/3*1/3*2/3*1/3*1/3)+......

=(1/3){1+(2/3*1/3)+(2/3*1/3*2/3*1/3)+.......}

=(1/3){1÷(1-2/9)} For infinite GP sum=first term/(1-common ratio)

=1/3*9/7

=3/7

The number of throws before either of them win can be said as 0(when questioner wins in his 1st throw ,1(when opposite wins in his 1st throw means total 2nd throw so on we get ,2,3,4,5,6...........

Hence totalling we can say it's expected value can varie from zero to infinite.