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You are challenged to a game of Russian roulette. Your opponent places two bullets side by side in a six chamber revolver and spins the chamber. She explains that the game is single turn-based and MUST be played through to its conclusion. She then offers that you can choose who goes first. Should you go first, or let her?

Edit: In my envisioning of the scenario the cylinder is spun once and once only, at the beginning of the game.

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  • $\begingroup$ dont be first if its rotating $\endgroup$ – balazs.com Nov 8 at 20:39
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After the chamber is spun, there are six possible bullet configurations (B = bullet, E = empty):

BBEEEE: player 1 dies
EBBEEE: player 2 dies
EEBBEE: player 1 dies
EEEBBE: player 2 dies
EEEEBB: player 1 dies
BEEEEB: player 1 dies

You have a 2/3 probability of surviving if you elect to shoot second.

I had posted a variation of this puzzle on my blog in 2009: Russian Roulette

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There seem to be two variants of the game.
In the first, the cylinder is spun just once, after the rounds are loaded; players then pull the trigger in turns until a live round fires.
In the second, the rounds are loaded and then each player spins the cylinder and pulls the trigger at each turn until a live round fires.

In both variants, play second.
In variant one, as noted by @Penguino, you have a $\frac23$ chance of survival playing second.
In variant two, the odds are still in your favor, but at the slightly lower odds of $\frac6{10}$.

Penguino explains the odds for variant one here.
The odds for variant two are trickier, and were determined by simulation.

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  • $\begingroup$ For the exact calculation of probabilities in the second variant of the game see the answer by @masonk and my comment there. $\endgroup$ – CiaPan Mar 6 '17 at 14:27
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Assume the bullets are in chambers 5 and 6 (you can do this because the number of the cylinders is arbitrary). Then the first player dies on their first turn if the cylinder is at 5 or 6 after the initial random spin, or after their second turn if the cylinder starts on 3, or after their third turn if the cylinder starts on 1. So best to start second with 2/3 survival rate.

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Cool answer.

Rubio's second variant ("spin chamber after every trigger pull") is the more difficult one to answer, but it can be solved exactly with a Markov matrix.

This matrix, call it M, has four states: Alice's Turn, Alice Died, Bob's Turn, Bob Died

0     0    4/6    0
2/6   1    0      0
4/6   0    0      0
0     0    2/6    1

Solve for the eigenvalues of this matrix and you will find closed form solution for the equilibrium probability between Alice Died and Bob Died. The answer is exactly .6, .4.

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    $\begingroup$ Thanks to the symmetry this specific case can be easily solved without general Markov description. If Alice starts, her chance to die in the first move is $1/3$. Let $p$ denote her overall chance to die in the game. If she survives, then Bob, the second player, appears in the same situation as Alice was when starting. That means Bob has chance $p$ to die in the following game. So the overal chance of Bob's death is $2/3\cdot p$, and that equals $(1-p)$, hence $p=3/5$ – the first player dies with probability of $60\%$. $\endgroup$ – CiaPan Mar 6 '17 at 13:58
  • $\begingroup$ @CiaPan: And that is exactly the same as the second part of my answer. $\endgroup$ – Peregrine Rook Mar 6 '17 at 20:27
  • $\begingroup$ @PeregrineRook You're right. I just tried to show the same with less math formalism, but exposing more 'nature of the problem' instead. $\endgroup$ – CiaPan Mar 14 '17 at 9:35
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As Rubio points out, the question is slightly ambiguous. If we assume that a player spins the cylinder on each turn, then

In round 1, Player 1 dies with probability $\frac13$.

So there is a $\frac23$ probability of there being a round 2. Then, Player 2 dies with probability $\frac13$. Thus, the absolute probability of Player 2 dying in round 2 is $\frac23\times\frac13$, following the rule that the probability of event A happening, and then event B, is $$P(A)\times P(B\text{, given that }A\text{ has happened})$$
For example, in this pinball machine (where the ball is dropped in at the top, and takes branches with equal probability):
pinball machine
the probability that the ball will go down the leftmost chute is $\frac{1}{10}$ — because $P(A)$, taking the left branch at the upper fork, is $\frac12$, and $P(B\text{ given }A)$, taking the leftmost branch at the lower division point, is $\frac15$.

And so it goes: the probability of Player 1 dying in round 3 is $\left(\frac23\right)^2\times\frac13$ because this requires Player 1 to survive the first round ($\frac23$), Player 2 to survive the second round ($\frac23$), and Player 1 to run out of luck in round 3 ($\frac13$).  Similarly, the probability of Player 2 dying in round 4 is $\left(\frac23\right)^3\times\frac13$, etc.  (If it helps you to understand the exponential factor, consider flipping a fair coin ten times in a row.  Most people find it “obvious” that the probability of getting heads 10 times in a row is $\big(\frac12\big)^{10}$.)

Thus, the total probability of Player 1 dying is \begin{align}\qquad\qquad P_1&=\frac13~~+~~~\left(\frac23\right)^2\,\times\frac13~~+~~\left(\frac23\right)^4\times\frac13~~+~~\dots~\end{align} Solving this is simple algebra: \begin{align}P_1&=\frac13~~+~~\,\left(\frac23\right)^2\,\times\frac13~~+~~\left(\frac23\right)^4\times\frac13~~+~~\dots\\[1ex]&=\frac13~~+~~\left(\frac49\right)\phantom{^1}\times\frac13~~+~~\left(\frac49\right)^2\times\frac13~~+~~\dots\\[1ex]-\qquad\frac49~P_1&=\phantom{\frac13~~+~~}\left(\frac49\right)\phantom{^1}\times\frac13~~+~~\left(\frac49\right)^2\times\frac13~~+~~\dots\\[1ex]\hline\frac59~P_1&=\frac13\\[2ex]P_1&={\frac13\over\frac59}~~=~~\frac13\times\frac95~~=~~\frac{9}{15}=\frac35~~=~~0.6\end{align}

Another way of getting to the same result is to note that

\begin{align}P_2&=\left(\frac23\right)\times\frac13~~+~~\left(\frac23\right)^3\times\frac13~~+~~\left(\frac23\right)^5\times\frac13~~+~~\dots\end{align} and then note that $P_2=\frac23P_1$.  But $P_1+P_2=1$,  so $P_1+\frac23P_1=\frac53P_1=1$,  so $P_1=\frac35$.

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  • $\begingroup$ That makes some sense to me. To be clear, my version of the game involved only one spin of the chamber at the beginning. I have updated the question and am trying to calculate these odds following the same algebra, but I am not very good at algebra. Can you explain why the probabilities of there being a subsequent round are being raised by an escalating power? $\endgroup$ – Robert Mar 7 '17 at 21:40
  • $\begingroup$ @Robert OK, I've expanded my answer. $\endgroup$ – Peregrine Rook Mar 8 '17 at 1:17

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