A very interesting probability/ strategy puzzle for children, coming straight out of the movie Sholay, one of the most popular Bollywood movies of all time. All the Indians reading this blog I am sure will vividly remember the scene where Gabbar (the villain) has a revolver in his hand which has a cylinder that can hold 6 slots, he puts bullet into 3, and then rotates the cylinder to a random position and there are three people to the shot.

I got the puzzle from my elder daughter, Anisha, who got this from Alok Mittal’s Mathematical Circles class yesterday.

Lets look at 3 variations of this situation. In all of them assume that two people are playing a game, lets call them A and B. The game is that one person shoots the other starting from a random position of the cylinder. If there is actually a bullet at that position, the other person is dead. If not, then the other person gets the revolver and shoots…and the game goes on until one is dead. Objective, of course, is to be alive. The revolver cylinder has 6 slots.

1. If there is only one bullet in the revolver and we start from a random position, what should be the strategy for A…should he/she shoot first or let the other person shoot first. What are the chances of being alive or winning with this strategy?

2. If instead of 1, there are 2 bullets in continuous slots, what would be A’s strategy?

3. If there are two bullets, but both inserted in random locations, what would the strategy be?

Happy shooting….oops, I meant only for a playful cause!

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