Question

If two fair dice are tossed, what is the smallest number of throws, n, for which the probability of getting at least one double 6 exceeds 0.5? (note: this was one of the first problems that de méré communicated to pascal in 1654.)

Answer 1

This is a rather famous probability problem.

The easiest way to solve this is to calculate the probability that you WON'T roll a "double 6" (or a twelve) each time you roll the dice.  There are 36 ways in which dice rols can appear and only one is a twelve.  So, for one roll, the probability that you will NOT get a twelve is (35/36)^n where 35/36 is about .97222222 and n would equal 1 for the first trial.  So for your first roll the odds that you WON'T get a 12 is .97222222.

For the second roll we calculate (35/36) to the second power or (35/36)^2 which equals about .945216.

When we get to the 24th roll we calculate (.97222222)^24 which equals 0.508596.  

For the 25th roll, we calculate (.97222222)^25 which equals  0.494468.  For the first time we have reached a probability which is lower than 50 per cent.  That is to say, after 25 rolls, we have reached a point in which the probability is less than 50 per cent that we will NOT roll a twelve.

To phrase this more clearly, after 25 rolls we reach a point where the probability is greater then 50 per cent that you will roll a 12 at least once.

Please go to this page 1728.com/puzzle3.htm and look at puzzle 48. (The last puzzle on the page).  An intersting story associated with this probability problem is that in 1952, a gambler named Fat the Butch bet someone $1,000 that he could roll a 12 after 21 throws.  (He miscalculated the odds [as we know you need 25 throws] and after several HOURS, he lost $49,000!!!)

Please go that page and it has a link to the Fat the Butch story.