 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  An Olympic trial (Posted on 2022-03-01) In some Olympic sports, the athletes compete for the highest score on a trick. The judges score the trick based on how well it is completed as well as its level of difficulty. As a result, the competitors try the hardest trick they think they can successfully complete. How hard should they try?

Let's call the difficulty of the attempted trick a number 0<d<1.

The attempt is a randomly chosen from the uniform distribution 0<a<1.

If a<d the trick fails and no points are scored.

If a>d the trick succeeds and scores d*a.

What difficulty should be attempted for the highest expected score?

 No Solution Yet Submitted by Jer Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Hope I did not make a mistake (spoiler) Comment 1 of 1
The probability of a trick suceeding is (1-d).
If it suceeds, a is between d and 1, so the expected value of a given that the trick suceeded is (1+d)/2.

Then the expected score is (1-d)(d)(1+d)/2 = (d-d^3)/2

The derivative with respect to d is (1-3d^2)/2

Setting this equal to 0 gives (1-3d^2) = 0.
d^2 = 1/3
d = 1/sqrt(3)  = .57735  (final answer)

Expected score = sqrt(3)/9 = .19245

 Posted by Steve Herman on 2022-03-01 12:27:20 Please log in:

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