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SpY]

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Hi. This isn't exactly like the previous thread and

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

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 be

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)?

*not*a 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 be

**6**.__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)?

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