the probability distribution is symmetric so probability of getting 0 heads = probability of getting 3 heads = (1/2)^3 so you have to sum it twice

The answer is no: In each round, there are 6 coins in play, so the sample space is 2^6. Count the ways you can both get 0 heads, 1 head, 2 heads, and 3 heads. 0 --> There is 1 way to get no heads. 1 --> My first is head and any of your three is a head also. Similarly, I do the same if my second or third flip is a head. This leads to 3*3 = 9 ways. 2 --> Count the coin that is tail (which means that 2 are heads). This means we have to calculate it the same way as before, again leading to 9 ways. 3 --> finally, there is only 1 way for us to both have three heads. Add this up: 1 + 9 + 9 + 1 = 20. The sample space is 2^6=64. Thus, the probability we both have the same number of heads is 20/64 = 5/16. Expected value = (5/16)($1) + (1 - 5/16)($2) < 0. Thus, we should not play the game.