Carnie Val ran a game of chance on the boardwalk at the Jersey Shore. I won't mention what beach. The player would plunk down his dollar and an array of 15 lights arranged in an equilateral triangle would start to flash. After a few seconds, three chosen at random would remain lit. If the three lit bulbs formed the vertices of an equilateral triangle, the lucky player would win a fuzzy stuffed animal. The game was on the up-and-up in the sense that any combination of lights was as likely to turn up as any other. For convenience of discussion, the bulbs are numbered as follows:

               1
               
             2   3
           
           4   5   6
          
         7   8   9  10
         
       11  12  13  14 15

One day, one of the lights failed to work. It was taken out of the random cycle, so that at the end three of the remaining 14 lights would stay lit, again with equal likelihood of any of the possible arrangements.

Val has no incentive to fix the broken light, as the new arrangement gives the player a lower probability of winning. That probability is the reciprocal of an integer, that is 1 over a whole number.

What is that probability?

(In reply to Solution by Michael)

If 5, 8, or 9 were out, 13 possible winning combinations would be removed.

What are the 13 possible winning combinations that would be removed?  Following along with the original 27 triangles you accounted for, I would have expected you to call out 7 combinations that were removed.

Like it did with me, this result may lead you to realize that there are more triangles to account for.

Edited on February 4, 2008, 4:40 pm

Posted by nikki on 2008-02-04 16:40:01