# Russian Roulette probability distribution

• SpY]
In summary: In fact, the continuous solution is often preferred as it allows for more precise calculations and can be used in cases where the discrete solution is not applicable. However, it is important to note that the two solutions will not always give the exact same result and the continuous solution is an approximation of the discrete one.
SpY]
Hi. This isn't exactly like the previous thread and 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.
$$P_D(n)={\frac{5}{6}}^{n-1}\frac{1}{6}$$

Finding the mean for the discrete case is easy -
$$\overline{n}= \sum_{n=1}^{\infty}n{\frac{5}{6}}^{n-1}\frac{1}{6}$$
Let $$q=\frac{5}{6}$$ and rewrite the argument of the sum as a partial derivative:
$$\frac{1}{6} \sum_{n=1}^{\infty}\frac{\partial }{\partial q}q^n$$
Which can be taken out of the sum (linear) and we use the formula for geometric series (q<1):

$$\frac{1}{6} \frac{\partial }{\partial q}\frac{q}{1-q} = \frac{1}{6} \frac{q}{(1-q)^2}$$

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:
$$p(x)={\frac{5}{6}}^{x-1}\frac{1}{6}$$

Now to try this as a continuous problem through an integral:
$$\overline{x} =\int_{0}^{\infty}x \frac{1}{6} \big( \frac{5}{6} \big)^{x-1} dx$$
$$= \frac{1}{6} \frac{6}{5} \int_{0}^{\infty} x (\frac{5}{6})^x dx$$

Through IBP (or wolframalpha) where u=x and dv= dx q^x and q=5/6
$$= \frac{1}{5} \big( \frac{x\cdot q^x}{ln(q)} - \frac{q^x}{(ln(q))^2} \big) \right|_0^\infty$$

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

$$= \frac{1}{5\cdot(ln(q))^2}$$

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)?

Last edited:
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)?
The answer to the first question is yes for q < 1, qx goes to 0 a lot faster than x becoming infinite. The easiest way to see it is use L'hopital's rule on x/q-x.

Using the integral as an approximation is valid.

## 1. What is the Russian Roulette probability distribution?

The Russian Roulette probability distribution is a mathematical concept that describes the likelihood or chance of an event occurring in a game of Russian Roulette. It is used to calculate the probability of a player surviving or dying in a game where a single bullet is placed in a revolver and the cylinder is spun before each player takes a turn firing the gun at their own head.

## 2. How is the Russian Roulette probability distribution calculated?

The Russian Roulette probability distribution is calculated by dividing the number of desirable outcomes (surviving) by the total number of possible outcomes. For example, if there are 6 chambers in the revolver and only 1 bullet, the probability of surviving would be 1/6 or approximately 16.67%.

## 3. What factors affect the Russian Roulette probability distribution?

The main factor that affects the Russian Roulette probability distribution is the number of chambers in the revolver and the number of bullets in the chamber. Other factors that can affect the probability include the skill of the player in spinning the cylinder and the quality of the revolver.

## 4. Is the Russian Roulette probability distribution the same for every player?

No, the Russian Roulette probability distribution may vary for each player depending on their individual circumstances, such as their level of fear or risk-taking behavior. Additionally, the probability may change with each turn as the number of bullets in the chamber decreases.

## 5. Can the Russian Roulette probability distribution be used to predict the outcome of a game?

No, the Russian Roulette probability distribution is based on mathematical calculations and does not take into account external factors such as luck or chance. It cannot be used to accurately predict the outcome of a game as each spin of the cylinder is independent of the previous spin.

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