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## Swindler's Dice

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

 Puzzle ID: #16401 Fun: (2.71) Difficulty: (2.05) Category: Probability Submitted By: silver_sword Corrected By: Winner4600

A swindler once approached an honest man with a die. He handed him the die and told him about the bet. The die had six sides. If the man rolled a ONE, he wins, and gets back twice the amount of his bet. If not, the swindler would keep the bet.

"But...my chances are only one out of six," retorted the man.

"True," grinned the swindler, "But I'll give you three tries to get a one."

The man considered. Three tries, with each try having a 1/6 chance of winning. So his chances of winning are 1/2. Why not give it a try?

Is the bet really fair? If not, what are the chances of the man winning?

As you have guessed...the bet is not fair. He had calculated the probability wrongly. Probability does not accumulate, like 1/6 x 3.

The probability of the man not getting a ONE in three throws is: 5/6 x 5/6 x 5/6, which is 125/216. This is the probability of the swindler winning.

Hence, the remaining fraction, 91/216, is the actual chance of the man winning.

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