Q: A swindler showed an honest man a six sided die. If the man rolled a ONE, he wins, and gets back twice the amount of his bet. If not, the swindler keeps 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 if I have 3 tries, each try has a 1/6 chance of winning, so my chances of winning are 3/6 or 1/2. Is the bet really fair? If not, what are the chances of the man winning?
You cannot just add 1/6 + 1/6 +1/6 so 1/2 is incorrect. The probability of not getting a 1 is 5/6 (there are 6 sides and the other possible outcomes are 2, 3, 4, 5 or 6). The probability of no 1s in 3 throws is 5/6 x 5/6 x 5/6 = 125/216 which is the probability of the swindler winning. So the probability of the man winning is 1 – 125/216 = 216/216 – 125/216 = 91/216.