Running order
Say there are 333 marathon runners. Each is randomly handed a number (1 to 333) to wear. What is the probability that the first three to finish will be wearing numbers that are in decreasing (not necessarily sequential) order?
Labels: logic
posted by Chris at
2:31 PM
8 Comments:
My Answer: 166/662
Ignoring the myriad of variations, the 2nd guy has exactly 50% chance of being less than 1st. Of all the possible ways to be 3rd, the guy has almost a 50% chance, but since 1st place runner is already greater than 2nd, it's a 166/331 chance. Compound the events = 1/2 * 166/331
Hi Nathan. All of that is wrong, sorry :(
16.6% of the time the order will be as stated.
That's the right answer. How did you do it?
hi Chris
i'm 46 years old and 277 pounds.i smoke two packs a day and live on my couch. what make you think i know ANYTHING about marathons!?
LOL. Good post Knightmare, sounds like a suitable name too. I smoke two packs myself. Until I found this problem, I thought a Marathon was a chocolate bar (now called Snickers). Greetz.
The three numbers are different (shocked?). Call the runners A, B and C. They can be arranged as ABC, CAB, BCA, BAC, CBA and ACB. That's 6 ways. Only one these ways gives the right order for the numbers they wear. So the probability of coming in in decreasing order is 1/6.
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