A Mathematical Analysis of Russian Roulette Part 3

Warning: the following is not for the faint of heart. It will involve some higher-level math. As my old calculus teacher used to say, “Here be Dragons”. If you want the answer, skip to the end to read the conclusion and my best attempt an intuitive explanation as to why the game changes.

This is the conclusion to a series (part 1 and part 2). The new rules of the game are as follows: 1 gun, 6 chambers, 1 bullet and 2 Players. After each round the gun is spun and then passed to the opposing Player. Below is an example of how a game could progress with the new rules.

flow

There is a crucial difference between this game and the previous version of Russian roulette. In our previous version, the game had a maximum of 6 rounds since after 6 trigger pulls it was guaranteed that the bullet would have been fired. Now, it is possible to envision a scenario where the game could continue for 7 rounds, 8 rounds, possibly 100 rounds or more. How do we draw a probability tree with an undefined number of branches? Below is my best attempt at it, but as you can see it gets complicated.

fading tree

I drew the tree with each branch fading out since, although it is possible for the game to continue for 100 rounds, the branches are not as likely the further away you travel from the start. This raises another interesting question, how do we account for an infinite number of possibilities? There are many ways to do this but I am going to use an infinite series. I am going to assume you are familiar with these. If not, you can refer to any intro Calculus textbook.

Let’s start with the first branch where Alice wins. We considered this before, but now the odds are different. The chance of a miss is 5/6 and the chance of a hit is 1/6.

\frac{5}{6} \times \frac{1}{6} = \frac{5}{36} =\frac{1}{6} \times \frac{5}{6} = \frac{1}{6} \times \left(\frac{5}{6}\right)^1

The reason for writing the result as a power of 5/6 will become apparent later on. Now we calculate the second branch where Alice wins. This must continue until round 4, so there must be 3 misses and end with a hit.

\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{125}{1296} = \frac{1}{6} \times \left(\frac{5}{6}\right)^3

Continuing with rounds 6, 8 and 10 we start to see a pattern.

\frac{5}{6} \times\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{3125}{46656} = \frac{1}{6} \times \left(\frac{5}{6}\right)^5

\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{78125}{1679616} = \frac{1}{6} \times \left(\frac{5}{6}\right)^7

\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{1953125}{10077696} = \frac{1}{6} \times \left(\frac{5}{6}\right)^9

The pattern should be apparent to most. Each time we go up a round we add 2 to the power of the ratio 5/6. If we forget about the 1/6 factor in front for the time being we should realize we have a geometric series! To simplify a little bit we will label the quantity 5/6 as “r” (r since it was originally a ratio of empty chambers to total chambers). So to find the probability of Alice winning after 10 rounds we just add up the probabilities we calculated.

As an aside, the ∑ symbol is a compact way of expressing addition. The number on the bottom is the index from which you start counting. The number on the top is the index at which you stop counting. The content to the right of the symbol is what you count. For example,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = \sum_{i=0}^{4} \left(i \right)

Now, on with the show.

\frac{1}{6} \times \left(\frac{5}{6}\right)^1 + \frac{1}{6} \times \left(\frac{5}{6}\right)^3 + \frac{1}{6} \times \left(\frac{5}{6}\right)^5 + \frac{1}{6} \times \left(\frac{5}{6}\right)^7 + \frac{1}{6} \times \left(\frac{5}{6}\right)^9

= \frac{1}{6} \times \left\{\left(\frac{5}{6}\right)^1 + \left(\frac{5}{6}\right)^3 + \left(\frac{5}{6}\right)^5 + \left(\frac{5}{6}\right)^7 + \left(\frac{5}{6}\right)^9\right\}

= \frac{1}{6} \sum_{i=0}^4 \left(\frac{5}{6}\right)^{2i+1}

= \frac{1}{6} \sum_{i=0}^4 \left(r \right)^{2i+1}

\approx 0.381

To calculate the full probability that Alice wins we need to play the game until someone loses. This can be thought of as add up all of the probabilities To calculate the full probability that Alice wins we need to for an infinite number of rounds, since in theory there is a chance that the game will go on forever. Using our handy friend, the limit, this is easily accomplished. The series is a power series, so the result is available in any Calculus textbook. Also since r = 5/6 we can rewrite 1/6 = 1 – r. Using our formula for power series and some algebra the answer is as clear as day.

\lim_{n \to \infty} (1-r) \sum_{i=0}^n \left(r \right)^{2i+1}

= (1-r) \frac{r}{1-r^2} = \frac{r}{1+r}

Just take a minute to absorb this formula. It is beautiful. Since we used an arbitrary variable “r” we can now calculate the probability that Alice will win for any exotic configuration of bullets. We could imagine a gun with millions of chambers, hundreds of bullets and get an exact answer. This closed form solution to the game is truly remarkable.

Using our original fraction of 5/6 we get:

\frac{\frac{5}{6}}{1 + \frac{5}{6}} = \frac{\frac{5}{6}}{\frac{11}{6}} = \frac{5}{11} \approx 45.5\%

We can approach the probability of Bob winning the same way. Skipping most of the details, we get the following:

\lim_{n \to \infty} (1-r) \sum_{i=0}^n \left(r \right)^{2i}

= (1-r) \frac{1}{1-r^2} = \frac{1}{1+r}

Using our original fraction of 5/6 we get:

\frac{1}{1 + \frac{5}{6}} = \frac{1}{\frac{11}{6}} = \frac{6}{11} \approx 54.5\%

As a quick check, the probability that Alice or Bob wins should be 1, since someone has to win the game eventually. The above formulas confirm this:

\frac{r}{1+r} + \frac{1}{1+r} = \frac{1+r}{1+r} = 1

But wait, 45.5% ≠ 54.5%! This means the game is not fair! Bob has an unfair advantage. How could a small change, such as spinning the barrel between rounds, affect the outcome so drastically? In fact, before I crunched the numbers, I believed that the first game was unfair and that spinning the gun each time would make the game fair by “resetting” the game after each round. Boy was I wrong!

This unfairness becomes more and more pronounced as the number of bullets in the gun increases. We can model this by decreasing our value for “r”. Below is a graph for different values of “r”, from very high (1 bullet, many chambers) to low (many bullets, few chambers).

graph

We can see clearly from the graph that as the chance of being hit goes up so does Bob’s chance of winning. This makes sense since in the extreme case of a fully loaded gun, where r = 0, Bob has a 100% chance of winning. In other words, Alice doesn’t stand a “chance”. The above graph also clearly shows that as the number of empty chambers increases the game approaches a fair game. Therefore, in theory, the only fair game is a game with no bullets!

Conclusion (Welcome back non-calculus readers)

The conclusion of the matter, from a calculus point of view, is as follows:  if we spin the barrel between rounds, Bob will always have an advantage of about 10%. Therefore, Bob will win 55% of the games and Alice will only win 45% of the games. For those of you who have not taken calculus, I will now offer an intuitive explanation as to why Bob has such an advantage.

The key to understanding the enigma is to imagine a different configuration of bullets. Take an extreme case where the gun is fully loaded. Here Alice gets hit 100% of the time. The reason Bob wins is that he never has to play a round. He is able to get a free walk since Alice will take the bullet every single time. This same logic can be applied to our situation of 6 chambers and 1 bullet. Now we have 5 empty chambers out of 6 available chambers. Again, it would be possible that the bullet would hit Alice in the first round and Bob would get a free walk. This inconsistency was not in our original Russian roulette because the game was limited to a maximum of 6 rounds. This fixed probability made sure that no Player would win by default so to speak.

Another way of understanding the paradox is to consider the probability that Bob has of being hit on round 2. In the original game this probability was 1/5, 1 bullet in 5 remaining chambers. However, in the new game where the barrel is spun, this probability decreases to 1/6 since there is 1 bullet and 6 chambers. Since 1/5 > 1/6, it should be clear that, at least for round 2, the resetting of the probability favours Bob. This same logic can be applied each round and Bob will always be lowering his chance of being hit by resetting between rounds.

So there you have it; if you are in a locked room, with no way out, a gun on the table and your turn as Player 2 (Bob), ask if you can spin the barrel so that the both of you have a “fair chance”; it just might save your life.

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2 Comments

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2 responses to “A Mathematical Analysis of Russian Roulette Part 3

  1. Nihilist

    Can you explain how you obtained 1/(1-r^2)?

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