Russian roulette: a game idolized in films and played around dark poker tables with the smell of tobacco and vodka thick in the air. Who would think such a dark and morbid game would yield surprising mathematical insights!

Before we begin, it is helpful to define a few terms. If each participant has an equal chance of winning, then the game is “fair”. Therefore, a coin toss would be “fair” since each person has a 50% chance of winning. We will define “winning” in roulette as not being shot, a very natural definition. “Losing” is defined as being shot. Each game is played until completion (someone loses). In Russian roulette, there are 2 participants, Alice as Player 1, and Bob as Player 2. Therefore, if each has a 50% chance of being shot then the game is fair. Enough with the introduction, let’s begin.

To define the game we will take a standard revolver with 6 chambers and 1 bullet. The bullet is loaded into a chamber and the barrel is spun to randomize the placement of the bullet. Alice picks up the gun and pulls the trigger. If the gun goes off, she loses. If not, she passes the gun to Bob. (At this point in the analysis, we will assume Bob grabs the gun directly with no further interaction. Later we will consider a different scenario.) Bob then pulls the trigger. If the gun goes off, he loses. If not, he passes the gun to Alice. This process repeats until someone loses the game. Each time someone pulls a trigger, we consider it a complete round. Below is a flow chart of a sample game.

As you can clearly see, this round was won by Alice because Bob lost by being hit with the bullet. Now we could play the game repeatedly with a non-lethal gun to see the different possibilities but this could take some time. We could run a computer simulation to play millions of games to try to determine a pattern. However, what if there was a systematic way to do this?

Well we are in luck. Mathematicians invented the concept of a probability tree to do just that. Below is a probability tree based on our version of Russian roulette. I have left out the probabilities for the time being so you can clearly see the potential outcomes. We will evaluate the chance of each option happening later on.

Enough with the chit chat, on with the tree.

I should probably explain why the nice pattern of having 2 branches off each box ends so suddenly. If you remember back when we defined the game, we stated that there were 6 chambers and 1 bullet. That means after 6 rounds of play the last turn, taken by Bob, will always end in a hit. The maximum number of rounds played is clearly 6.

Now we should add some numbers in to our tree. At the beginning, there is 1 bullet and 6 chambers total, so the chance of getting hit is 1/6. Let’s add this in on a new tree below (we will add the chances in place of the hit or miss entries so as not to clutter up the tree too much). This also means that the chance of missing will be 5/6 since there are 5 empty chambers out of the 6 chambers total.

Once the first round is over, we, as mathematicians, obtain some new information. If the bullet fired, the game is over and Bob won. However, if the bullet did not fire, this means that we now only have 5 chambers in question. There is still 1 bullet so the chance of a hit would be 1/5; 1 bullet in 5 remaining chambers. The chance of a miss would be 4/5; 4 empty chambers and 5 remaining chambers.

After the second round, we receive even more information. Only 4 chambers and 1 bullet remain. The chance of a hit would be 1/4; 1 bullet in 4 chambers. The chance of a miss would be 3/4; 3 empty chambers and 4 remaining chambers.

This process continues until round 6. This time it is Bob’s turn and there is 1 bullet with 1 remaining chamber. Hence, the chance of a hit is 1/1 and the chance of a miss is 0/1, or just 0.

This probability tree is ripe with information. But I am going to save the fruit for next time.

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