A game is played starting with 6 fair coins laid out with heads face up.
Each round consists of flipping all of the coins showing heads.
If fewer than half of the flipped coins come up heads the player loses.
Rounds continue until the player either loses or has one heads remaining.
The player wins by getting to one heads without losing.
What is the probability of winning this game?
Examples:
6→4→1 would be a loss. (1 is less than half of 4.)
6→5→3→2→2→2→1 would be a win.
I think you need to work backwards.
If there are 2 heads (and you haven't already lost),
then on the next flip the probabilities are as follows:
1/4 -- 0 heads -- you lose
2/4 -- 1 head -- you win
1/4 -- all heads -- flip again
The third case (all heads) leaves your situation unchanged, so we can ignore it and just figure out the relative probabilities of the other cases, which become:
1/3 -- 0 heads -- you lose
2/3 -- 1 head -- you win.
So, if there are 2 heads (and you haven't already lost), then your probability of winning is 2/3 (i.e, 1/3 * 0 + 2/3 *1).
If there are 3 heads (and you haven't already lost), then on the next flip the relative probabilities (excluding the 3 heads case) are:
1/7 -- 0 heads -- you lose
3/7 -- 1 head -- you lose
3/7 -- 2 heads --your probability of winning improves to 2/3
So, if there are 3 heads (and you haven't already lost), then your probability of winning is 2/7 (i.e, 4/7 * 0 + 3/7 *2/3).
If there are 4 heads (and you haven't already lost), then on the next flip the relative probabilities (excluding the 3 heads case) are:
1/15 -- 0 heads -- you lose
4/15 -- 1 head -- you lose
6/15 -- 2 heads --your probability of winning improves to 2/3
4/15 -- 3 heads -- your probability of winning improves to 2/7
So, if there are 4 heads (and you haven't already lost), then your probability of winning is 12/35 (i.e, 6/15 *2/3 + 4/15 *2/7), unless I just made a mistake.
Gotta go, but working backwards to 5 and then 6 will yield the required probability. I can already tell, though, that I don't like my chances of winning.