## Russian Roulette: An Optimal Strategy

OK now we’ve got ourselves a game
– Mike (Robert De Nero), The Deer Hunter

You may have seen the 1978 classic (and Academy Awards winning) Vietnam War film, The Deer Hunter.  It features a particularly harrowing scene involving two prisoners of war, Mike (played by Robert De Nero) and Nick (played by Christopher Walken), being forced by their captors into playing Russian Roulette.  In the scene Mike significantly ups the ante of the “game” by having three bullets placed in the barrel, rather than the more traditional single round.  Crucially, after Mike survives his turn the barrel is not re-spun before the pistol is handed over to Nick.  I won’t spoil the ending, but it is often a topic of debate among viewers of this scene whether Nick’s chances of survival were improved or reduced by not spinning the barrel again.

In a game of Russian Roulette, assuming that the bullets are placed in sequence, and given that Player 1 survives the first turn, is the optimal strategy to spin again or not spin again?

Good question.  The answer, it seems, depends on how many bullets.

The Deer Hunter scene featured three bullets, giving Mike even odds of surviving his turn.  What were Nick’s chances?  Before answering that question let’s consider, for the sake of completeness, all the combinations of one through to six bullets placed in sequence.  What are the chances of surviving a second turn either with re-spin or without re-spin?

The table below represents the different combinations (combn. 1 to 6) of having 1 to 6 bullets placed in sequence in the barrel of a pistol.  Imagine the circular barrel of a pistol with its 6 chambers laid out flat.  “X” denotes a bullet.  “O” denotes an empty chamber.  Also imagine that each chamber has a number.  So for example combination OOXXXO means that chambers 1, 2, and 6 are empty and chambers 3, 4 & 5 have bullets in them.  The chamber numbers are simply arbitrary.  What is important is that Chamber “1” represents the live chamber that exists on Player 1’s turn.  Chamber 2 represents the live chamber that exists on Player 2’s turn.  OOXXXO represents both Player 1 and Player 2 surviving their respective turns.

 combn. 1 combn. 2 combn. 3 combn. 4 combn. 5 combn. 6 1 bullet XOOOOO OXOOOO OOXOOO OOOXOO OOOOXO OOOOOX 2 bullets XXOOOO OXXOOO OOXXOO OOOXXO OOOOXX XOOOOX 3 bullets XXXOOO OXXXOO OOXXXO OOOXXX XOOOXX XXOOOX 4 bullets XXXXOO OXXXXO OOXXXX XOOXXX XXOOXX XXXOOX 5 bullets XXXXXO OXXXXX XOXXXX XXOXXX XXXOXX XXXXOX 6 bullets XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX XXXXXX

Consider the first, single bullet scenario.  We discount combination 1 as this would result in the death of Player 1 and the game would end.  Given that Player 1 survives their turn then the bullet has to be in either chamber 2, 3, 4, 5 or 6.  So without a re-spin Player 2 now has a 1 in 5 (20.0%) shot (sorry, couldn’t resist) of killing themselves.  This compares to “only” a 1 in 6 (16.7%) chance of dying with a re-spin.  Conclusion?  In a single bullet game of Russian Roulette it is (very slightly) better to re-spin.

In the two bullet scenario we discount combinations 1 and 6.  So the chance of dying without a re-spin reduces to 1 in 4 (25.0%) vs. 2 in 6 (33.3%) with a re-spin.  In a two bullet game of Russian Roulette it is actually better not to re-spin.

And so on…

3 bullet (Deer Hunter) scenario:  Chance of dying with no re-spin=1 in 3 (33.3%) vs. spin=3 in 6 (50%).  Significantly better to not re-spin.

4 bullets?  No re-spin=1 in 2 (50%) vs. re-spin=4 in 6 (66.7%).  Don’t re-spin.

5 bullets?  No re-spin=1 in 1 (100%) vs. re-spin=5 in 6 (83.3%).  Definitely re-spin!

6 bullets!  Umm… I don’t think it counts as Russian Roulette with 6 bullets.  Obviously Player 1 is guaranteed to die on the first turn…

Next time you’re watching The Deer Hunter with your mates (which I highly recommend as it’s an awesome film) tell them confidently that Nick did the right thing by not re-spinning the barrel of the pistol.