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# Probability: equal sequences

Started by BMAD, Apr 14 2014 02:58 AM

5 replies to this topic

### #1

Posted 14 April 2014 - 02:58 AM

Try out (or imagine) the following simple game with a friend:

You each toss a fair coin as many times as necessary to get a required sequence.

Your required sequence is H T H

Your friend's required sequence is H T T.

Do this several times and each time write down the number of tosses you needed

The winner is the player whose average number of tosses is the lowest. For example:

Game 1:

You toss: H T T H T H

Friend tosses: H T H H H T H H T T

Your 'score' is 6 and your friend's is 10

Game 2:

You toss: T T H T T H H T H

Friend tosses: T T H H T H T T

Your 'score' is 9 and your friend's is 8

Game 3:

You toss: T T H H T H

Friend tosses: H H T H T T

You both 'score' 6.

At this point your average is 7 and your friend's average is 8.

The conundrum: Assuming you played the game many times, which of the following outcomes would you expect?

A) You win (i.e. your average is lower than your friend's).

B) Your friend wins (your average is higher than your friend's).

C) You tie (your average is about the same)

### #2

Posted 14 April 2014 - 05:58 AM

Spoiler for Looks like

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #3

Posted 14 April 2014 - 02:49 PM

agreed

### #4

Posted 14 April 2014 - 08:31 PM Best Answer

Spoiler for I would say

**Edited by Rob_G, 14 April 2014 - 08:31 PM.**

### #5

Posted 15 April 2014 - 07:10 AM

Spoiler for I would say

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #6

Posted 15 April 2014 - 12:15 PM

Hmmm. I did not consider a single coin

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