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Mutually Exclusive Events

Probability of two events that cannot occur at the same time P(A or B) = P(A) + P(B)

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Russian Roulette

Consider a game in which one of the six chambers of a revolver is loaded with a bullet. The cylinder of the gun is spun once, and the gun is passed between two players until it fires; whoever fires it loses. Would you prefer to go first or second?

How Probability Helps

This is a famous game known as Russian roulette (though it’s not much of a game if the gun is pointed towards the players). Many people tend to think that going second is better, since going first exposes the player to the risk of firing first. The second player has a chance of winning “without even having to try” if the first person fires. Others think that the first shot has a one-sixth chance of firing, while the second has a one-fifth chance, so they’d choose to go first. Interestingly, however, both players actually have an equal chance of firing.

Let’s say Player 1 goes first, followed by Player 2. Player 1 will lose if the bullet is in the first, third, or fifth chamber. If the bullet is in the second, fourth, or sixth chamber, then Player 2 will lose. The event that it’s in Chamber 1, the event that it’s in Chamber 2, the event that it’s in Chamber 3, etc. are all mutually exclusive events. If you do a little math, you’ll see that the probability of losing is \frac{1}{2} for both players.

There are many variations of the game: single-player versus multiplayer, several chambers containing bullets instead of one, having players spin the cylinder before each shot, etc.

Read more here: http://mindyourdecisions.com/blog/2009/07/28/the-russian-roulette-puzzle/

Explore More

Here’s one of the variants of Russian roulette. Assume that there are three bullets in consecutive chambers and, as before, whoever fires the first shot loses. Who has the higher probability of winning: the player who goes first or the player who goes second?

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