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. But sometimes they know what score they need to beat to win the competition. How hard a trick should they try?
Model the situation as before: the difficulty, d, of the trick to be attempted is chosen as a number from 0 to 1.
The attempt, a, is a randomly chosen from the uniform distribution (0,1)
If a<d the trick fails and scores 0.
If a>d the trick is successful and scores a*d.
Now suppose our competitor is going for the gold and has one final attempt to beat the current highest score of 0.6
What difficulty should be attempted to maximize the chance of winning?
In order for the trick to succeed, a must be greater than d.
In order for the score (a*d) to beat 0.6, a must be greater than .6/d
The probability of winning the gold is therefore 1-max(d, .6/d).
The probability of winning the gold is maximized when d = .6/d
Solving, the best d = sqrt(.6) = approx 77.5%
Chance of winning the gold = 22.5%
Winning the gold (spoiler)
