Hi. This isn't exactly like the previous thread and(adsbygoogle = window.adsbygoogle || []).push({}); nota homework problem. I'd just like to check the validity of my solution. It concerns the relation between a discrete and continuous probability distribution.

The problem:

A player inserts a bullet into a 6-chamber revolver. He then spins the drum and points it to his head. If he lives, he re-spins the drum and tries again. Find the mean number of trials before he shoots himself.

Discrete solution

So this is clearly a discrete problem and statistically independent (re-spinning) so I'll start defining the discrete probability distribution for him dying in the n'th game.

[tex]P_D(n)={\frac{5}{6}}^{n-1}\frac{1}{6}[/tex]

Finding the mean for the discrete case is easy -

[tex]\overline{n}= \sum_{n=1}^{\infty}n{\frac{5}{6}}^{n-1}\frac{1}{6}

[/tex]

Let [tex]q=\frac{5}{6}[/tex] and rewrite the argument of the sum as a partial derivative:

[tex]\frac{1}{6} \sum_{n=1}^{\infty}\frac{\partial }{\partial q}q^n[/tex]

Which can be taken out of the sum (linear) and we use the formula for geometric series (q<1):

[tex]\frac{1}{6} \frac{\partial }{\partial q}\frac{q}{1-q}

= \frac{1}{6} \frac{q}{(1-q)^2}[/tex]

Then substituting q=5/6 we get the mean to be6.

Continuous solution

Though not technically allowed, define the probability density function for dying in the x'th game:

[tex]

p(x)={\frac{5}{6}}^{x-1}\frac{1}{6}[/tex]

Now to try this as a continuous problem through an integral:

[tex]\overline{x} =\int_{0}^{\infty}x \frac{1}{6} \big( \frac{5}{6} \big)^{x-1} dx[/tex]

[tex]= \frac{1}{6} \frac{6}{5} \int_{0}^{\infty} x (\frac{5}{6})^x dx[/tex]

Through IBP (or wolframalpha) where u=x and dv= dx q^x and q=5/6

[tex]= \frac{1}{5} \big( \frac{x\cdot q^x}{ln(q)} - \frac{q^x}{(ln(q))^2} \big) \right|_0^\infty[/tex]

For these limits I'm taking a leap of faith, particularly the first term. As a product of a linear (x) and exponential (q^x), to zero they both go to zero - but to infinity, one goes to infinity and the other to zero. The limit of a product is the product of the limits only when they are not infinite (I'm sure). Otherwise I'm saying ∞*0 = 0 ...

By some numerics, putting in x=1000 the result is to the order 10^-77. The graph of p(x) also tends to zero as x goes to infinity. So assuming the limits for this term go to zero, I then have

[tex]= \frac{1}{5\cdot(ln(q))^2}[/tex]

Substituting q I get the mean to be 6.02... sufficiently close to the 6 - as I got previously. Not sure of the significance of the error though...

My question is what I did with an integral mathematically valid? One of those cases where getting the correct answer doesn't necessarily mean your method is ok..

Also, can we use a continuous distribution (integral) as an approximation for a discrete distribution (riemann sum)?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Russian Roulette probability distribution

**Physics Forums | Science Articles, Homework Help, Discussion**