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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Simulation of differing distribution | Comment 6 of 9 |
On the possibility that the distribution of probable prices is more centralized than a uniform distribution between the lowest possible price and highest possible price in each of the three price ranges, I simulated a more central distribution by using a program:

RANDOMIZE TIMER
FOR trial = 1 TO 1000000
  FOR item = 1 TO 3
    amt = 500 * item
    FOR i = 1 TO 3
      amt = amt + 500 * RND(1)
    NEXT
    amount(item) = amt
    bin = INT(amt / 500)
    totCt(item, bin) = totCt(item, bin) + 1
  NEXT
  ct = ct + 1
  IF amount(3) > amount(2) AND amount(2) > amount(1) THEN winCt = winCt + 1
NEXT
FOR item = 1 TO 3
  FOR bin = 1 TO 5
    PRINT USING "########"; totCt(item, bin);
  NEXT
  PRINT
NEXT
PRINT
avg = winCt / ct
PRINT USING "######### #.##### #.#####"; ct; avg; SQR(ct * avg * (1 - avg)) / ct

As this program uses three uniform distributions with a range of 500, added together to produce a number from 0 to 1500 to add to the lowest value in each range, and the variance of a uniform distribution is r²/12, the variance of each price is 3(500²/12)=62500, making the s.d. 250. The results of one run are:

166601 666750 166649 0 0
0 166565 666642 166793 0
0 0 167612 666078 166310

1000000 0.83892 0.00037

-----------
showing how many of each prize were in each bin of prices. In this instance, no longer is the probability of each of 2 or 3 orders within matching bins either 1/2 or 1/6.

The average shows as 83.8% or 83.9% (the standard error of the mean showing as .00037), so the centralized tendency does help raise the odds of a win.
  Posted by Charlie on 2003-10-16 09:33:12
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