## Finite sequence, infinite average?

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jseah
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### Finite sequence, infinite average?

I recalled a problem I met a few years ago that was never answered to my satisfaction. Perhaps you can help.

The problem:
I am playing a game in which I toss a fair coin. Each time I toss a coin I 'use' a coin toss. When I hit zero coin tosses remaining, the game is over.
Each time I toss a heads, I gain 2 coin tosses. Each time I toss a tail, nothing happens.
I start with 1 coin toss.

Qn: What is the average number of coin tosses in total I can expect to make?
Qn: Is there ever a time in which I will toss an infinite number of coins (and thus never finish)?

The weird part is this:
Average return (of number of coin tosses) per coin toss = P(heads) * 2 = 0.5 * 2 = 1
Therefore I break even on every toss and thus on average should toss an infinite number of coins.

But before any given coin toss, I have a number of 'tosses' remaining to use and a streak of tails equalling to the size of that number (ie. a probability of 2^number of tosses) that my coin tossing will come an end. The sum probability that I will run out of 'tosses' = 1. (0.5 + 0.25 + 0.125 + .... = 1)

Average number of coin tosses = infinite...
but at the same time P(finite number of coin tosses) = 1?
???
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Zamfir
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### Re: Finite sequence, infinite average?

jseah
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Joined: Tue Dec 27, 2011 6:18 pm UTC

### Re: Finite sequence, infinite average?

Hm! Recurrence equations seem powerful. They look like the math equivalent of recursive functions in programming. I should try to make sense of that.

Poor Stencil the flea though. =O
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brenok
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### Re: Finite sequence, infinite average?

jseah wrote:Average number of coin tosses = infinite...
but at the same time P(finite number of coin tosses) = 1?
???

Does that seem more unintuitive than every natural number being finite, but the "average" number being infinite?

jseah
Posts: 544
Joined: Tue Dec 27, 2011 6:18 pm UTC

### Re: Finite sequence, infinite average?

From what I understood of Zamfir's link, the thing that got me was that P(finite tosses) = 1. Because it's not actually P(finite tosses) = 1, but more like P(no more coins after X coins) approaches 1 as X approaches infinity (as per sum of exponential series).

So I was comparing a 'hard' 0 probability of infinite tosses against the average of infinite tosses, when it's really two infinities.
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jaap
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### Re: Finite sequence, infinite average?

See also almost surely. There are almost surely going to be only a finite number of coin tosses. Whilst going on forever is a possible outcome, the probability of that is nevertheless exactly zero, and the probability of finite tosses is exactly 1.

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