perplexus.info :: Probability : Carny Game

Carny Game (Posted on 2008-02-04)

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?

Submitted by Charlie

Rating: 3.5000 (2 votes)

Solution: (Hide)

There are C(15,3) = 455 ways the lights could originally have remained lit.
After the malfunction, there are C(14,3) = 364 ways the lights can remain lit.
When the full 15 are working, there are 35 possible equilateral triangles that could remain lit at the end: 16 with side length 1 (10 "rightside up", 6 "upside down"), 7 with side length 2, 3 with side length 3, 1 with side length 4, 6 with side length sqrt(3) (such as bulbs 3,4,9) and 2 with side length sqrt(7) (such as bulbs 3,7,14).
Any given original vertex bulb (such as bulb 1) being out would reduce the possible triangles by 4 to 31, making the probability 31/364, but that's not the reciprocal of an integer.
Any given bulb next to one of these (such as bulb 2) being out would reduce the possible triangles by 7, to 28, making the probability 28/364 = 1/13 -- indeed the reciprocal of an integer, but not lower than the original probability of 35/455, which is also 1/13.
A bulb from the middle of a side of the original triangle, such as 4, would reduce the possible winning triangles by 8, to 27/364, but again, that's not the reciprocal of an integer.
But a bulb out that was not on the original perimeter, such as bulb 5, would reduce the possible winning triangles by 9, to 26/364 = 1/14, which is the reciprocal of an integer, and is also lower than the original probability of 1/13.
So 1/14 is the new probability of winning this game.
From Enigma number 1473 by Susan Denham, New Scientist, 15 December 2007.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Answer	K Sengupta	2008-10-13 02:42:54
Solution	Dej Mar	2008-02-04 23:12:33
re(2): Solution	Michael	2008-02-04 18:52:47
re: Solution	nikki	2008-02-04 16:40:01
Solution	nikki	2008-02-04 16:26:20
Solution	Michael	2008-02-04 15:54:22

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