And the answer is…

Hairy, you are seriously the man. No other person I’ve told this puzzle to even came close to a solution.

For those who don’t get what I’m saying here. Do not read further until you have read the previous post and its comments.

“i’m sticking by my answer, C, htt will have the higher average. But not by ten times.

This answer hinges on the ’semi-assumption’ that the robots do not ‘clear their memory’ after the coins do in fact concatenate their names. (If the robots did clear their memories and start a fresh list, it really looks like a statistical dead heat.)

If, instead, the next value is tacked onto the two that preceded it, in essence allowing for a ‘frameshift,’ statistical deviation appears. (If this assumption is incorrect, i’m totally in over my head here! toss life preserver here)

Instead of a ten-fold-longer average, i’m thinking that htt will have an average list length 2 times as long as hth’s list, as ‘tt’ must go through three subsequent coin flips to have a single chance in eight to be born again, whereas ‘th’, with it’s self-effacing symmetry needs only two subsequent coin flips to regenerate itself.”

Hairy, you were just one more deduction away from the final answer.

Your assumption is incorrect. Each experiment until the achievement of their target pattern constitutes one “event” and the robots do clear their memory after noting down the value of “number of tosses” for each event.

The answer is B, in fact.

The last observation you couldn’t make was the fact that the robots do not clear their memory unless they achieve their target pattern for one “event”. So if they achieve an incorrect pattern, the next toss is concatenated to the previous incorrect pattern. Therein lies the key.

Solution to the puzzle:

Suppose one particular event is in progress for the robot hth.

If the last toss was a T, He has a 1 in 8 probability of achieving his target pattern in the next three tosses.

_ _ _ T

Now suppose he gets a H in the next toss. Now, the probability of winning in the next two tosses is 1 in 4.

_ _ _ T H

Now he gets a T in the next toss. Now the probability of winning in the next one toss is 1 in 2.

_ _ _ T H T

Now he can get a H or a T. If he gets an H, he wins, and reports the number of tosses for that event as 7. If he gets a T, however, he loses and is back where he started – He has a 1 in 8 probability of achieving his target pattern over the next three tosses.

_ _ _ T H T T

.

.

Now suppose a similar scenario for the robot htt. The last toss was a T, and he has a 1 in 8 probability of achieving his target over the next three tosses.

_ _ _ T

He gets an H. Now there’s a 1 in 4 probability of winning in the next two tosses.

_ _ _ T H

He gets a T. Probability is now 1 in 2.

_ _ _ T H T

And here things are different fron hth. If he gets a T, he wins, and reports the number of tosses for that event as 7. But if he gets an H, his situation is better off than hth, because he has a 1 in 4 probability of winning over the next two tosses (unlike hth, who had a choice between winning and falling back to 1 in 8).

_ _ _ T H T H

This improvement in probability translates to the fact that htt will report a smaller average value of “number of tosses”. The average number reported by htt is 8, and that reported by hth is 10.

I found a formal mathematical proof here. I’m not that good with probability mathematics so I’m not sure how valid it is, but looking through it, it appears to be simply a mathematification of the same principles I’ve explained above.