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The following Python program runs in 51ms.

Solution:He would expect to pay out on 1 out of 14 games.At first I thought we have an incorrect solution.

When all 15 bulbs are working, there are 15!/(12! 3!) = 455 different configurations of three lit bulbs. The possible equilateral triangles are sixteen of side 1, seven of side 2, three of side 3, one of side 4; as well as six of side root(3) perpendicular to the sides of the array, and also — easy to miss — two more of side root(7) at about 19°:

That makes 35 in all, so the suckers have a 1-in-13 chance of winning.

I think we have to assume that a bulb does not blow, for then people would notice that sometimes only two come on. Rather, the random mechanism just ignores one of them. There are 14!/(11! 3!) = 364 different configurations of the remaining 14 bulbs. Which equilateral triangles don’t appear depends on which bulb is out of action: if it’s one of the corner ones, then obviously the triangle of side 4 is one of them, for example.

As you say, we have to distinguish four cases of crippled bulb:

corner (3 possibilities): 31 triangles still possible

next to corner (6 possibilities): 28

mid-side (3 possibilities): 27

inner (3 possibilities): 26.

The (weighted) mean is 28, representing a 1-in-13 chance as before.

So on average it makes no difference that one bulb is not working.

After thinking about it I realized that the sideshow owner of course knew it was one of the inner bulbs and worked out the improved 1-in-14 chance.

If it had been one of the corner ones we can bet he’d have fixed it double quick!

Oh, how I hate eMail gateway that suppress leading spaces and tab characters!

My wee diagram actually looked quite good when I sent it.

I think I’ve fixed it up for you with some

<pre>...</pre>tags.