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## Another Game of Dice

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #43905 Fun: (2.51) Difficulty: (2.59) Category: Probability Submitted By: javaguru

Your friend offers to play a game of dice with you. He explains the game to you.

"We each get one die, the highest die wins. If we tie, I win, but since you always lose when you roll a one, if you roll a one you can roll again. If you get a one the second time you have to keep it."

What is each person's probability of winning?

What are the probabilities of winning if you can keep rolling until you get something besides a one?

In the first game the probabilities of winning are 37/72 for your friend and 35/72 for you. In the second game each player has a 50% probability of winning.

There are 36 combinations (6 x 6) for your die and your friend's die on the first roll. In the first game if you don't roll a one on the first roll then that roll is equally likely to be a 2, 3, 4, 5 or 6, which can beat 1, 2, 3, 4, 5 different numbers, respectively. That gives a

(1+2+3+4+5)/36 = 15/36

probability of winning on the first roll. There is a 1/6 chance of rolling a one, in which case you have a 15/36 probability of winning on the second roll, which gives a

1/6 * 15/36 = 15/216 = 5/72

probability of winning on the second roll for a total probability of winning of

15/36 + 5/72 = 30/72 + 5/72 = 35/72 = .486111...

In the second game we never keep a one, so there are only 30 combinations (5 x 6) for the final roll. As shown above, there are 15 ways to win, which gives a probability of

15/30 = 1/2

of winning the second game.

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