Flying Toupees!

An infinite number of men, all named Ronald Frump and all wearing toupees, line up in a row (a very long row!)

The first man's toupee flies off his head.

If the first man's toupee flies off his head, there is a 1 2 \frac{1}{2} probability that the second man's toupee will fly off his head.

If the second man's toupee flies off his head, there is a 1 3 \frac{1}{3} probability that the third man's toupee will fly off his head.

And so on...

In fact, in general rule is that...

If the ( n 1 ) th (n-1)^\text{th} man's toupee flies off his head, then there is a 1 n \frac{1}{n} probability that the n th n^\text{th} man's toupee will fly off his head.

So, now for the question...

What is the expected number of toupees that will fly off heads?

Give your answer to 5 decimal places.

Assumption: The only chance a man has of his toupee flying off his head is if the man before him had his toupee fly off as per the above probabilities. Otherwise, the toupee will stay firmly in place.


Image credit: http://www.solikeyouknow.com/


The answer is 1.71828.

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1 solution

Geoff Pilling
Oct 10, 2016

The expected number of flying toupees is given by:

E ( n ) = 1 P ( 1 ) + 2 P ( 2 ) + 3 P ( 3 ) + . . . = 1 n P ( n ) E(n) = 1P(1) + 2P(2) + 3P(3) + ... = \sum_{1}^{\infty}nP(n)

where P ( n ) = P(n) = The probability that exactly n n toupees fly off.

P ( n ) = n ( n + 1 ) ! P(n) = \frac{n}{(n+1)!}

So,

E ( n ) = 1 n P ( n ) = 1 n n ( n + 1 ) ! = e 1 = 1.71828 \large E(n) = \sum_{1}^{\infty}nP(n) = \sum_{1}^{\infty}n\frac{n}{(n+1)!} = e-1 = \boxed{1.71828}

I was guessing a value over 2 2 before making the calculation, so I was surprised it was only e 1 e - 1 .

And one "Frump" is one too many, let alone an infinite number of them. :P

Brian Charlesworth - 4 years, 8 months ago

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Hahaha... Definitely!

Geoff Pilling - 4 years, 8 months ago

I have not let learnt this advanced probability and I have never come across this expression to find the answer...but the only thing I do not understand is that the question is to find 'expected number' of toupees flying off...can we expect 1.71828 1.71828 number of toupees to fly off?

Skanda Prasad - 4 years, 8 months ago

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Yeah, its confusing... Even though 1.71828 toupees would never actually fly off, this is the average number of times you'd find over an ensemble of infinitely many tests.

Geoff Pilling - 4 years, 8 months ago

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I don't really understand what you meant by 'ensemble of infinitely many tests'.

Skanda Prasad - 4 years, 8 months ago

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@Skanda Prasad I mean, the 1st half of the comment made sense to me...but the other made it still more confusing...

Skanda Prasad - 4 years, 8 months ago

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@Skanda Prasad Yeah, just ignore the last part... Essentially its just an average number so it doesn't need to be an integer.

Geoff Pilling - 4 years, 8 months ago

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