I had to work out a sample game the long way to understand the strategy for a 3-sequence game, maybe others will find it helpful:
Suppose some games where the players choose a sequence of 3 and the game goes up to 4 tosses max. For any sequence of 3 there are going to be 4 possible ways to win. Let's suppose player one chose THT, the ways of winning are:
HTHT - game ends after 4 tosses
TTHT - game ends after 4 tosses
THTH - game ends after 3 tosses
THTT - game ends after 3 tosses
Now, as player two, you know that any combination you chose will also have 4 ways of winning, so in order to beat player one you need to rob them of one of their ways of winning by picking a sequence that appears before their winning sequence.
You have no options in the latter two cases, player one wins after only 3 tosses, but you can beat them in the first two cases. Both HTH and TTH appear before THT, but in the third case THT appears before HTH. So HTH won't work, you both take one of each other's winning conditions leaving you both with 4. But there are no cases where TTH appears after THT, so you have 4 ways of winning while player one only has 3.
I've never seen this comprehensive an exposition of intransitive coin games; that's pretty cool. But I must admit some surprise at the lack of Rosenkrantz and Guildenstern references (http://www.youtube.com/watch?v=RjOqaD5tWB0).
Suppose some games where the players choose a sequence of 3 and the game goes up to 4 tosses max. For any sequence of 3 there are going to be 4 possible ways to win. Let's suppose player one chose THT, the ways of winning are:
HTHT - game ends after 4 tosses TTHT - game ends after 4 tosses THTH - game ends after 3 tosses THTT - game ends after 3 tosses
Now, as player two, you know that any combination you chose will also have 4 ways of winning, so in order to beat player one you need to rob them of one of their ways of winning by picking a sequence that appears before their winning sequence.
You have no options in the latter two cases, player one wins after only 3 tosses, but you can beat them in the first two cases. Both HTH and TTH appear before THT, but in the third case THT appears before HTH. So HTH won't work, you both take one of each other's winning conditions leaving you both with 4. But there are no cases where TTH appears after THT, so you have 4 ways of winning while player one only has 3.