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?

The total combinations of three lit bulbs is:
15_C_3 = [(15!)/(15-3)!]/3! = 455

The total number of combinations of equilateral triangles is 35.

1.  1- 2- 3
2.  1- 4- 6
3.  1- 7-10
4.  1-11-15
5.  2- 4- 5
6.  2- 7- 9
7.  2-11-14
8.  2- 3- 5
9.  2-10-12
10.  2- 6- 8
11.  3- 5- 6
12.  3- 8-10
13.  3-12-15
14.  3- 7-14
15.  3- 4- 9
16.  4- 7- 8
17.  4-11-13
18.  4- 5- 8
19.  4- 6-13
20.  4- 9-12
21.  5- 8- 9
22.  5-12-14
23.  5- 6- 9
24.  5- 7-13
25.  5-10-13
26.  6- 9-10
27.  6-13-15
28.  6- 8-14
29.  7-11-12
30.  7- 8-12
31.  8-12-13
32.  8- 9-13
33.  9-13-14
34.  9-10-14
35. 10-14-15 

With all fifteen bulbs in the random cycle, the winning probability is 35/455, and, as a reduced fraction, is 1/13.


With one missing bulb, the total combinations of three lit bulbs is:
14_C_3 = [(14!)/(14-3)!]/3! = 364

The following shows the number of equilateral triangles that could be formed that includes one of the fifteen bulbs:

      4  
     7 7  
    8 9 8
   7 9 9 7
  4 7 8 7 4

With one missing bulb, the probability of an equilateral triangle formed is (35 - X)/364, where X is the total number of combinations that could be formed that would include the bulb. There are 4 different numbers of combinations -- 4, 7, 8 and 9. Respectively, the probabilities, in reduced fractions, are 31/364, 1/13, 27/364, and 1/14. Of these, only 1/14 succeeds as being a unary fraction and less than 1/13 (the original winning probability).

The new winning probability is then 1/14, with the one of the three center bulbs (5, 8 or 9) having been taken out of the random cycle.

Edited on February 5, 2008, 10:34 am

Posted by Dej Mar on 2008-02-04 23:12:33