I will post a challenging math problem each day. The level of difficulty will vary, but most problems should not require any specialized knowledge beyond calculus. Problems are also posted on Twitter using the hashtag #mathpotd.
Sunday, September 20, 2009
Triple 6
You throw three dice n times. How big does n have to be for you to have greater than a 50% chance of throwing three 6's?
If three dice are thrown, the probability of NOT getting triple sixes is 1-(1/6)^3 = 215/216. So the probability of not getting triple sixes in n throws is (215/216)^n
Now we solve an inequality. (215/216)^n < 0.5 log[(215/216)^n] < log(0.5) n * log(215/216) < log(0.5) n > log(0.5)/log(215/216) = 149.37... n >= 150
The answer is 150.
ReplyDeleteIf three dice are thrown, the probability of NOT getting triple sixes is 1-(1/6)^3 = 215/216. So the probability of not getting triple sixes in n throws is (215/216)^n
Now we solve an inequality.
(215/216)^n < 0.5
log[(215/216)^n] < log(0.5)
n * log(215/216) < log(0.5)
n > log(0.5)/log(215/216) = 149.37...
n >= 150