Two persons engage in a game of chance. The game is to nominate a sequence of three consecutive coin tosses [H or T].

Player one firstly nominates a sequence and then player two makes a nomination. The game finishes when the last three tosses match either one of the players' nominations.

How can player two be assured of winning most of the time?

Given the choices that can be made by player one, what are the odds of player two winning?

Oh, it doesn't matter who tosses the coin.

Based on Charlie's table, here is a set of rules for player 2 to follow to optimize his/her results based on player 1's choice of a pattern:
If player 1 chooses "xyz" then player 2's best choice is  "wxy". 
If x = y     then w is the opposite.   (so "not x", x, x)
If x <> y   then w is the same as x. (so x, x, y).

 

It makes sense that the last 2 of player 2's pattern would be xy, since in a series of coin flips, for player 1 to win, there has to be a wxyz (unless he wins right from the beginning), so if player 2 has some wxy pattern, he wins before player 1 has the chance.

But I can't explain why the 'w' should be either heads or tails except by looking at Charlie's table

Posted by Larry on 2013-06-21 16:34:35