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Swindler's Dice

Braingle.com on

Published in Brain Teasers

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 is 1/2. Why not give it a try?

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


Solution:

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 chances of the man winning.

 

Today's brain teaser courtesy of Braingle.com.


 

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