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Easy as 1 2 3 (Posted on 2003-10-15) Difficulty: 3 of 5
In a game show, there is a game in which you have to order the value of three prizes in order of least expensive to most expensive. You have to get all three right in order to win.

The only problem is you have your spouse do all the shopping, so you only know that the first prize is between 500 and 2000, the second prize is between 1000 and 2500, and the third prize is between 1500 and 3000.

Which order should you put them in so that you have the highest probability of winning, and what is the probability that you will win using this arrangement?

See The Solution Submitted by Gamer    
Rating: 3.8333 (6 votes)

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Different distribution(s) | Comment 3 of 9 |
The probabilities will depend on the probability distributions within the three ranges specified. The preceding solutions assumed a uniform distribution within these ranges. Other possibilities exist. An example would be if the median of each range were not the arithmetic mean of the endpoints but rather the geometric mean. This would occur if the logarithm of the value were distributed uniformly rather than the value itself. This has the advantage that, for example, 500-2000 is described as probably 1000 but possibly off by a factor of 2.

In that instance the probability of winning increases to about .609962587. This is computed by the following program:
FOR i = 1 TO 6
  cp(i) = LOG(500 * i)

FOR k = 6 TO 4 STEP -1
  maxJ = 5: IF k < 5 THEN maxJ = k
  FOR j = maxJ TO 3 STEP -1
    maxI = 4: IF j < 4 THEN maxI = j
    FOR i = maxI TO 2 STEP -1
      PRINT k, j, i,
      p1 = (cp(k) - cp(k - 1)) / (cp(6) - cp(3))
      p2 = (cp(j) - cp(j - 1)) / (cp(5) - cp(2))
      p3 = (cp(i) - cp(i - 1)) / (cp(4) - cp(1))
      totP = p1 * p2 * p3
      IF j = k THEN
        totP = totP / 2
        IF i = k THEN
          totP = totP / 3
        END IF
        IF i = j THEN totP = totP / 2
      END IF
      totPr = totPr + totP
      PRINT USING "#.#######"; totP

It relies on the fact that since, within the same band (of the six bands discussed) given that any two or three are in the same band the probability distribution of each is the same, so each ordering possibility still has 1/2 or 1/6 probability of being the case.

More complicated, but more likely, distributions would require numerical integrations. Particularly more likely distributions would concentrate probability near the center of the ranges, and trail off at the edges of each of the three ranges. They would presumably make success even more probable, as in the extreme case, pinpointing the prices would guarantee success.
  Posted by Charlie on 2003-10-15 22:18:06
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