A Mathematical Analysis of Russian Roulette Part 2

When we left off, we were pondering the probability tree. This tree contains all the information we need to solve the game. We just need to decipher it. For example, if we want to know the probability that Alice will be hit on round 5 we can just read the value directly from the tree (I highlighted it in yellow for your convenience) and see that the chance is 1/2 or 50%. We must be very careful here though, just as we obtain new information from each round, the probabilities listed on the tree need to be understood in context. Alice has a 50% chance of being hit if she has made it to round 5. But what is the chance that Alice even makes it that far? Alternatively, what is the chance that the game progresses to round 5 and then Alice is hit?

tree 3

When dealing with probabilities that flow one after another, we use multiplication. If we follow the blue path to the yellow box, we can get the answer to the question. Therefore, the probability that Alice is hit on the 5th round is

\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} = \frac{1}{6}

Wow, that is fantastic! A quick way to calculate the chance of any eventuality, I love probability trees! Now let’s take it one step further. What is the overall chance of Alice winning? The tree tells us that 3 of the branches have Alice winning. Thus, by calculating the chance of each branch we can add them together to get the overall chance.

Branch one takes us to the end of round 2 where Bob gets hit and Alice wins.

\frac{5}{6} \times \frac{1}{5} \times = \frac{1}{6}

Branch two takes us to the end of round 4 where Bob gets hit and Alice wins.

\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{1}{6}

Branch three takes us to the end of round 6 where Bob gets hit and Alice wins.

\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2} \times 1 = \frac{1}{6}

Adding the probabilities together we get…

\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}

…the not so surprising result of 1/2 or 50%. If we did the same exercise for Bob we would find that he also has a 1/2 or 50% chance of winning. So there you have it, in a simple game of Russian roulette the outcome is fair, if rather gruesome.

One nagging question remains for me. What if you spun the gun after each round? Would that change the game? And how on earth would you calculate something like that.

Advertisements

1 Comment

Filed under Uncategorized

One response to “A Mathematical Analysis of Russian Roulette Part 2

  1. Pingback: A Mathematical Analysis of Russian Roulette Part 3 | the Math behind the Magic

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s