# Probabilities-Russian roulette

1. Jan 20, 2014

### Dassinia

1. The problem statement, all variables and given/known data
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 ?

2. Relevant equations

3. 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 !

2. Jan 20, 2014

### LCKurtz

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.

3. Jan 20, 2014

### Dassinia

What do you mean by E(T) ?

Thanks

4. Jan 21, 2014

### haruspex

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

5. Jan 21, 2014

### 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 !!!

Last edited: Jan 21, 2014
6. Jan 21, 2014

### haruspex

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

7. Jan 21, 2014

### PeroK

That's very clever!