Probabilities-Russian roulette

[Dassinia]

Homework Statement

Hello,
A player places a single round in a revolver leaving 5 empty emplacements.

a) What is the probability to stay alive after playing N times
b) What is the probability to stay alive after playing N-1 times and die the next shot ?
c) How many times can a player participate on average ?

Homework Equations

The Attempt at a Solution

a) (5/6)^N
b)(5/6)^(N-1)1/6
c) Really don't know how to solve this we saw basic law of probabilities to introduce the course of thermodynamics, the answer is given it is 6

Thanks !

 

[LCKurtz]

Homework Statement

Hello,
A player places a single round in a revolver leaving 5 empty emplacements.

a) What is the probability to stay alive after playing N times
b) What is the probability to stay alive after playing N-1 times and die the next shot ?
c) How many times can a player participate on average ?

Homework Equations

The Attempt at a Solution

a) (5/6)^N
b)(5/6)^(N-1)1/6
c) Really don't know how to solve this we saw basic law of probabilities to introduce the course of thermodynamics, the answer is given it is 6

Thanks !

I think for (b) you mean $\left(\frac 5 6\right )^{N-1}\left( \frac 1 6\right )$ don't you? In that case you have (a) and (b) correct. From part (b) you have that if $T$ = time of death then $P(T = n) = \left(\frac 5 6\right )^{n-1}\left( \frac 1 6\right )$. The average time of death is just the expected value of $T$. So you have two problems: What is the formula for $E(T)$ and can you calculate it? Can you take it from there? If not, come back with what you try.

 

[Dassinia]

What do you mean by E(T) ?

Thanks

 

[haruspex]

What do you mean by E(T) ?

Thanks

The expected value of T. (=average value.)

 

[Dassinia]

The formula to calculate the average is
E(T)=∑P(T)*T (sum n=0 to N)
=∑(5/6)^N-1*N/6
How can i get to E(T)=6 from here ?
Solved !

 

[haruspex]

The formula to calculate the average is
E(T)=∑P(T)*T (sum n=0 to N)
=∑(5/6)^N-1*N/6
How can i get to E(T)=6 from here ?
Solved !

Good.
There is a way to get the answer without summing a series.
Suppose the expected value is E. After pulling the trigger once, there's a 1 in 6 chance it's all over. Otherwise, the expected number of rounds remaining is still E:
E = 1 + (1/6)*0 + (5/6)*E

 

[PeroK]

Good.
There is a way to get the answer without summing a series.
Suppose the expected value is E. After pulling the trigger once, there's a 1 in 6 chance it's all over. Otherwise, the expected number of rounds remaining is still E:
E = 1 + (1/6)*0 + (5/6)*E

That's very clever!