I see a similar question in Russian roulette. Now if instead of two bullets being placed side by side, two bullets are randomly put in the chamber. Your opponent played the first and he was alive after the first trigger pull. You are given the option whether to spin the barrel. Should you spin the barrel?

My approach: I felt this would be case dependent. As in first if they are side by side, then it is 1/4 probability of losing in not spinning the barrel vs 1/3 in spinning the barrel, so I go for not spinning the barrel. If not, if one space between them, then I have 1/2 p of losing which is bad so I go for spinning the barrel, now the answer given is 2/5. And I have no idea why. Help please!

My idea of russian roulette: When we don't spin the barrel and we have pulled the trigger, every piece moves clockwise. As in if we number the pieces, where bullet was in 1 and 3, rest 2,3,4,5,6 are all empty, then after first trigger pull (trigger was at position 1), bullet in 3rd position moves to 4th position. Am I right?

  • $\begingroup$ There are 15 different ways 2 bullets can be placed: 6 adjacent, 6 one-apart, 3 opposite. For adjacent, 3/4 chance next is empty as per the classic problem. For the others, 50% (2/4) chance the next is empty (imagine going clockwise and count how many empty spots have another empty spot as its successor). Then just calculate with (6/15*3/4) + (6/15*2/4) + (3/15*2/4) = 60% chance of surviving the next round. Spinning just gives you the "standard" 66.7% (4/6) chance of surviving, so you should spin. $\endgroup$ Commented Jul 8, 2022 at 19:51
  • $\begingroup$ Where is this problem from? $\endgroup$
    – bobble
    Commented Jul 8, 2022 at 23:42
  • $\begingroup$ Is my idea of russian roulette correct? $\endgroup$
    – Charlie
    Commented Jul 9, 2022 at 5:21
  • $\begingroup$ It is from Practical Guide to Quant interview $\endgroup$
    – Charlie
    Commented Jul 9, 2022 at 5:22

1 Answer 1


It is better by $6\frac23\%$ to

spin the barrel

because I think we can assume that all $15$ possible positions are equally likely. Your chance of survival as an average of each of those $15$ cases is $3/5 = 60$%. Your chance of survival of spinning the barrel is $2/3$

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  • 2
    $\begingroup$ Hey, the explanation given in the book is something quite different. > "If you spin the barrel, the probability that you will lose in this round is 2/6. If you don't spin the barrel, there are only 5 chambers left and your probability of losing in this round (conditioned on that your opponent survived) is 2/5. So you should spin the barrel" Quote from the book A Practical Guide to Quantitative Interviews $\endgroup$
    – Charlie
    Commented Jul 8, 2022 at 20:00
  • $\begingroup$ I just realized that that answer agrees with my answer. Change your 2/6 to 1/3 and you will see that it lines up with my answer. $\endgroup$
    – JLee
    Commented Jul 8, 2022 at 20:06
  • $\begingroup$ thank you @msh210 for the edit $\endgroup$
    – JLee
    Commented Jul 8, 2022 at 20:46
  • $\begingroup$ Yess, it agrees with your answer but is there any shortcut like in the book it seems it's a one liner. $\endgroup$
    – Charlie
    Commented Jul 9, 2022 at 15:07
  • 1
    $\begingroup$ You've got the book answer, my post, and the answer in the comment of @LukasRotter, all correct. Not sure how many more ways we can slice this. $\endgroup$
    – JLee
    Commented Jul 9, 2022 at 16:32

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