Your mother did tell you to stay away from gambling, but now it's far, far too late to head her advice. You're sitting in a smoke-filled room, your single opponent staring you down. Between you, on a table, there is six-shot revolver, containing exactly one live bullet. Someone has spun the cylinder so neither of you knows whether the next shot will be blank or lethal. Each of you will take one shot at the opponent in turn, and the cylinder will be spun again after each shot. The game ends, obviously, when one of you fires the lethal shot, and the other person is dead. It's your choice whether to go first or second — which should you choose? Does it matter?
This puzzle was inspired by a problem in the book Number Story, which is reviewed in this issue of Plus.
If you are stumped by last issue's puzzle, here is the solution.
For some challenging mathematical puzzles, see the NRICH puzzles from this month or last month.
Go for the first shot.
In the second worst case you get to fire 3 shots first and you have more opportunity..
Surely it's best to go second! Whether you go first, second, third, fourth, etc, when the gun is in your hand, the chance of the shot being a killer is exactly the same. But...
During the first go, the shooter has a 1/6 chance of dying but the opponent (the second shooter), who doesn't have the gun, has a 0 chance of dying.. because they don't have the gun! Therefore your chance of survival is massively greater if you go second!
It doesn't matter
The probability is reset every time you spin the barrel so it is a 1/6 chance every time you pull the trigger, so it doesn't matter which you choose.
Third, because in the second, it can give you the bullet in the head, and therefore, in the third, it cannot give you because when it shoots, it turns and when you reload, it turns again, that's why the third