Question

Suppose two players play a traditional Russian Roulettes game. One bullet is put into a 6-revolver and the barrel is randomly spun so that there is equal chance for each chamber to be under the hammer. Two players take turns to pull the trigger against themselves until one kills him or herself. Under such rules, would you rather go first or second? What’s the probability of survival to go first?

b) Now the rule is changed, the barrel gets spun after each shot. Now would you rather go first or second? What’s the probability of survival for each choice?

c) Suppose two bullets instead of one are randomly put into the chamber. Your opponent went first and survived the shot. Now you are given the chance to spin the barrel. Should you do it or not?

d) What if the two bullets are consecutively put into the chamber and the barrel is spun (i.e. two bullets are next to each other), should you spin again after your opponent survived the first round?

Answer 1

a) One bullet is put into a 6-revolver and the barrel is randomly spun.Two players take turns to pull the trigger against themselves until one kills him or herself. Under such rules, would you rather go first or second? What’s the probability of survival to go first?

The problem can be solved by calculating the probability of survival for the choices.

First, consider the odds of survival if the cylinder is spun. The cylinder is equally likely to stop at any of the six chambers. One of the chambers contains the bullet and is unsafe. The other five chambers are empty and you would survive. Consequently, the probability of survival is 5/6, or about 83 percent.

Next, consider the odds if the cylinder is not spun. As the trigger was already pulled, there are five possible chambers remaining. Additionally, one of these chambers contains the bullet. That leaves four empty or safe chambers out of five. Thus the probability of survival is 4/5, or 80 percent.

Next As the trigger was already pulled twice, there are four possible chambers remaining. Additionally, one of these chambers contains the bullet. That leaves three empty or safe chambers out of four. Thus the probability of survival is 3/4, or 75 percent.

Next time the probability of survival will be 2/3 or 66.66 percent.

5th shot will have survival probability of 1/2 or 50 percent.

6th will have survival probability 0 percent.

In case of  to pull the trigger against themselves until one kills him or herself. The best choice is to have higher survival probability.

If you start first, survival Probability for each shot you will have 3 shots to survive : 83\75\50

If you start second, survival Probability for each shot you will have 3 shots to survive : 80\66\0.

I would go first as it has higher probability of survival.

b) Now the rule is changed, the barrel gets spun after each shot. Now would you rather go first or second? What’s the probability of survival for each choice?

As cylinder is spun. The cylinder is equally likely to stop at any of the six chambers. One of the chambers contains the bullet and is unsafe. The other five chambers are empty and you would survive. Consequently, the probability of survival is 5/6, or about 83 percent.

Next, consider the odds if the cylinder is spun. It brings us back to same condition that is it can stop at any of 6 chambers, one of the chamber has bullet. survival probability would be 5/6 again.

Each time the survival probability is same.

Going first or second doesn't matter.

c) Suppose two bullets instead of one are randomly put into the chamber. Your opponent went first and survived the shot. Now you are given the chance to spin the barrel. Should you do it or not?

First shot, the cylinder is equally likely to stop in any of the six chambers, four of which are empty. Consequently, the probability of survival is 4/6, or about 67 percent.

first shot went by and nothing happened. we have option to spin or not lets check survival probability.

Next, consider the odds if the cylinder is not spun. This situation again requires careful accounting. We know the gun is loaded with two bullets so it has two loaded chambers (let’s label them 4 and 6) and four empty chambers (let’s label these 1, 2, 3, and 5). When the game starts, the trigger is pulled and nothing happens, so we know the cylinder started on an empty chamber. Then, the cylinder advances one chamber forward, meaning the cylinder must now be in one of the positions 2, 3, 4, or 6. If you don’t spin, you will survive in two of these four cases (4 and 6 are loaded). Therefore, not spinning the gun has a probability of survival of 2/4, or 50 percent.

if we spin we have survival probability of 67 percent.

In conclusion, the best choice in the non-adjacent case is to spin.

d) What if the two bullets are consecutively put into the chamber and the barrel is spun (i.e. two bullets are next to each other), should you spin again after your opponent survived the first round?

First, the odds when the cylinder is spun. The cylinder is equally likely to stop at at any of the six chambers. Two of the chambers contain bullets and are unsafe. The other four chambers are empty and you would survive. Consequently, the probability of survival is 4/6, or about 67 percent.

Next, consider the odds if the cylinder is not spun. This situation is a bit trickier so careful accounting is helpful. We know the gun is loaded with two bullets so it has two loaded chambers (let’s label them 5 and 6) and four empty chambers (let’s label these 1, 2, 3, and 4). When the game starts, the trigger is pulled and nothing happens, so we know the cylinder started on an empty chamber. Then, the cylinder advances one chamber forward, meaning the cylinder must now be in one of the positions 2, 3, 4, or 5. If you don’t spin, you will survive in three of these four positions (all but 5). Therefore, not spinning the gun has a probability of survival of 3/4, or 75 percent.

It is now better not to spin, unlike the single bullet case.