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 hardest part about this for me was making sure I had all the possible equilateral triangles accounted for when using all 15 lights. I think I have everything now.

I’ll first start with equilateral triangles that have one edge that is horizontal. I think we can agree that these are the "easy" triangles to find so I don’t think anyone needs me to call out each triangle by vertices. There are:
16 with an edge length of one unit,
7 with an edge length of two units,
3 with an edge length of three units, and
1 with an edge length of four units.

So that is 27 so far.

Now we have the "off angle" triangles (that I forgot about during my first pass of the puzzle). We have 6 triangles with an edge length of approximately two units. For clarity, they are:

7-13-5
8-14-6
10-13-5
9-12-4
4-9-3
6-8-2

And we have 2 triangles with an edge length of approximately three units. For clarity they are:

7-14-3
10-12-2

So that brings our total up to 35 possible equilateral triangles.

The total number of 3-light combinations that are possible is 15C3 (not to be confused with 15P3 which calculates the number of permutation of 15 choose 3). So that is 15!/[3!*(15-3)!] = 15!/(3!*12!) = 13*14*15/3! = 13*7*5 = 455.

So the current probability of winning a fuzzy animal is 35/455 = 1/13 = 7.6923%

Now lets consider the situation when a bulb burns out. Technically one of 15 bulbs could go out, but due to symmetry many bulbs are equivalent to 2 or 5 others. It turns out there are only 4 "representative" bulbs:
1: which is equivalent to 11 and 15
2: which is equivalent to 3, 7, 10, 12 and 14
4: which is equivalent to 6 and 13
5: which is equivalent to 8 and 9.

Alrighty, well when bulb 1 goes out:
From the horizontal triangles we lose:
1 with an edge length of one unit (1-2-3),
1 with an edge length of two units (1-4-6),
1 with an edge length of three units (1-7-10), and
1 with an edge length of four units (1-11-15).
And from the off-angle triangles we lose:
0 with an edge length of approximately two units, and
0 with an edge length of approximately three units.
So we lose 4 equilateral triangles, which means there are 35-4 = 31 equilateral triangles which can win.

When bulb 2 goes out:
From the horizontal triangles we lose:
3 with an edge length of one unit (2-1-3, 2-3-5, 2-4-5),
1 with an edge length of two units (2-7-9),
1 with an edge length of three units (2-11-14), and
0 with an edge length of four units.
And from the off-angle triangles we lose:
1 with an edge length of approximately two units (2-8-6), and
1 with an edge length of approximately three units (2-12-10).
So we lose 7 equilateral triangles, which means there are 35-7 = 28 = 2²*7 equilateral triangles.

When bulb 4 goes out:
From the horizontal triangles we lose:
3 with an edge length of one unit (4-2-5, 4-5-8, 4-7-8),
3 with an edge length of two units (4-1-6, 4-6-13, 4-11-13),
0 with an edge length of three units, and
0 with an edge length of four units.
And from the off-angle triangles we lose:
2 with an edge length of approximately two units (4-3-9, 4-12-9), and
0 with an edge length of approximately three units.
So we lose 8 equilateral triangles, which means there are 35-8 = 27 = 3³ equilateral triangles.

When bulb 5 goes out:
From the horizontal triangles we lose:
6 with an edge length of one unit (5-4-2, 5-2-3, 5-3-6, 5-4-8, 5-8-8, 5-9-6),
1 with an edge length of two units (5-12-14),
0 with an edge length of three units, and
0 with an edge length of four units.
And from the off-angle triangles we lose:
2 with an edge length of approximately two units (5-7-13, 5-13-10), and
0 with an edge length of approximately three units.
So we lose 9 equilateral triangles, which means there are 35-9 = 26 = 2*13 equilateral triangles.

The total number of 3-light combinations that are possible now is 14C3. So that is 14!/[3!*(14-3)!] = 14!/(3!*11!) = 12*13*14/3! = 2*13*14 = 2²*7*13 = 364.

So the four different probabilities of getting an equilateral triangle are now:

If bulb 1 goes out, the probability is 31/364 and this is far as we can reduce that fraction. By the way, this ends up being 8.5165%, which is greater than what our probability was before the bulb went out. So we know this is not the bulb that went out since we didn’t get 1/n for the fraction, and the probability increased (which would have given Carni Val incentive to replace the bulb).

If bulb 2 goes out, the probability is 28/364 = 1/13. Well, we did end up with the form of 1/n for the fraction, but this is the same probability as the original game (35/455 = 1/13). So while Carni Val still might not have incentive to replace the bulb, in this case, we were told that the probability to win did indeed decrease. So we know this is not the bulb that went out.

If bulb 4 goes out, the probability is 27/364 and this is as far as we can reduce that fraction. So while this ends up being 7.4176%, which is indeed less than the original game, we know this is not the bulb that went out because we didn’t get 1/n for the fraction.

If bulb 5 goes out, the probability is 26/364 = 1/14 = 7.1429%. This is indeed less than the original game, and it is in the form of 1/n.

So the answer is, bulb 5, 8 or 9 went out and the probability is 1/14 = 7.1429%

Posted by nikki on 2008-02-04 16:26:20