Coin Flip Poser

You are given a choice of two coin flip games to play. One choice is to flip 3 coins at once, and if exactly 2 of them come up heads, you win. The other choice is to flip 4 coins at once, and if exactly 2 of them come up heads, you win. Which game do you choose to play, and why?

1st game chance to win = 3/8
2nd game chance to win = 6/16 = 3/8

Seems like the same odds to me. Did I miscalculate?

three coins:

HHH
HHT *
HTH *
HTT
THH *
THT
TTH
TTT

Three winners out of 8

4 coins:
HHHH HHHT HHTH HHTT* HTHH HTHT* HTTH* HTTT
THHH THHT* THTH* THTT TTHH* TTHT TTTH TTTT

6 winners out of 16 = 3/8

Yep, no gain either way

Originally posted by iamatiger
three coins:

HHH
HHT *
HTH *
HTT
THH *
THT
TTH
TTT

Three winners out of 8

4 coins:
HHHH HHHT HHTH HHTT* HTHH HTHT* HTTH* HTTT
THHH THHT* THTH* THTT TTHH* TTHT TTTH TTTT

6 winners out of 16 = 3/8

Yep, no gain either way

That's what I got. It's easy, but a little counter-intuitive. Intuitively we can sense that the odds of exactly 2 heads start to drop off as N increases further, certainly, say, to 10 or 20.

I pick 4 coins. That way if one of the coins lands on it's edge my odds don't go down.