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Guys, try to solve this interesting probability question

Assume that a beauty contest is being judged by the following rules: (1) there are N contestants not seen by the judges before the contest, and (2) the contestants are individually presented to the judges in a random sequence. Only one contestant appears before the judges at any one time. (3) The judges must decide on the spot whether the contestant appearing before them is the most beautiful. If they decide in the affirmative the contest is over but the risk is that a still more beautiful is in the group as yet not displayed. In that case the judges have made the wrong decision. On the other hand, if they pass over the candidate, the contestant is disqualified from further consideration even if it turns out that all subsequent candidates are less beautiful. What is the probability of correctly choosing the most beautiful contestant?

Assume A,B,C are the candidates and A happens to be the most beautiful
In order to make a proper judgement I need to see A at the end.
A,B,C can be arranged in 3! ways. If A comes at the end then other two come before A in 2! ways
P = 2!/3! = 1/3
In case of N P = (N-1)!/N! = 1/N

In order to make a proper judgement I need to see A at the end.

Well, if A is first or second, does that not count if she is chosen outright and second or third person is just ignored? Hence. does A need to be third for there to be a valid judgement?

I'll give it a try.
Let's say there are 3 contestants {A,B,C}
We have to multiply the probability of the most beautiful person {A} taking a given position by the remaining probability of having her win the competition given that the other persons before are rejected.

Let's say A is the first person to show up.
Probability of that happening: 1/3
Probability of her being chosen: 1/1 because she is most beautiful and competition stops there

Let's say A is the second person to show up.
Probability of that happening is 1/3
Probability of her being chosen: 1/2 since this is her chance to win if the person before her is also beautiful and is chosen before her

Let's say A is the third person to show up.
Probability of that happening: 1/3
Probability of her being chosen: 1/3 since this is her chance to win if either of the 2 persons before her are chosen

Therefore, for a 3 persons competition, the probability of the most beautiful person winning would be:
=1/3*1/1 + 1/3*1/2 + 1/3*1/3
=1/3 (1 + 1/2 + 1/3) = 11/18

The general formula would simplify to:
1/n (1 + 1/2 + 1/3 + ... 1/n)
I'm not sure how to simplify this though.

Acctually, the probability will depend on a. It means on how many contestants we will ignore first, and choose the most beautiful among them.
So, Paul's general formula is wrong.

Acctually, the probability will depend on a. It means on how many contestants we will ignore first, and choose the most beautiful among them. So, Paul's general formula is wrong.

Hmmm, can you then confirm what the answer would be for a 4 person contest then? I believe that "a" was taken into consideration when I allocated the "remaining probability of having the most beautiful person win the competition given that the other persons before were rejected".