Probability: equal sequences

[BMAD]

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)

[Rob_G]

that my friend has a greater chance of winning. Lets examine what can happen at any given point assuming we haven't already completed the sequence. (If last flip was tails start at 1 if heads start at 2)

1) If either of us flip tails here nothing changes, we stay at step one and we both still need 3 correct flips to finish. If we get heads we continue.

2) If either of us flip heads here we stay on this step. If we get tails continue. (Me 3A my friend 3B)

3A) If I flip tails here I have to start over at step one until I flip heads (at least 3 flips to finish). If heads I complete my sequence.

3B) If my friend flips heads here he goes back to step 2 to try for a tails (at least 2 flips to finish). If tails he completes his sequence.

Edited April 14, 2014 by Rob_G

[bonanova]

If there is an average number of tosses to get a particular pattern, then the pattern that happens more quickly on average will win if the game is played many times. So the questions are [1] does each pattern have an average number of tosses to appear and [2] how would we determine those numbers?

[BMAD]

agreed

[bonanova]

I learned a trick once, how to calculate waiting times for coin sequences. For HTH it's 10 and for HTT it's only 8. So the person flipping for HTT would win. But that applies only when each contestant is flipping his own coin. Curiously, if only a single coin is flipped, the patterns have an equal chance of occurring first: they will appear only after HT has appeared, and of course the third and deciding outcome is, as they say, a coin toss.

[BMAD]

Hmmm. I did not consider a single coin