Enigmatics |
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Posted on: Friday 14/12, 2007; 1:36 PM This week's enigma is unusually trivial to solve. Compute the positions of the bulbs.
Display the triangular array of bulbs.
Create all the alternative lists of bulb positions where a single bulb is omitted (i.e. it has failed). Don't bother to use symmetry to reduce the number of cases.
Pick all possible triples of bulbs from each of these lists.
Determine whether each triple of bulbs is an equilateral triangle.
For each of the possible bulbs that failed determine what fraction of cases are equilateral triangles (i.e. probability of payout). Compute the corresponding fraction for the original full triangular array of bulbs.
Pick the cases where a bulb has failed, and has reduced the probability of payout, and is a fraction of the form The payout now occurs 1 in every n games where n is given by |
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