Count heads across the stopping rule before setting a fair fee

The source calculations assume fair, independent coin outcomes. H and T denote a head and tail. The first two outcomes belong to the initial round; any following three belong to the second round. They are not five successive rounds. Initial HH contributes two heads to the total even when the next round gives no heads.

The paths are HH followed by TTT, or TT followed by two heads and one tail in any of $3!/2!=3$ orders. Thus $P(X=2)=(1/2)^5+(3!/2!)(1/2)^5=4/32=1/8$.

For zero heads, TT must be followed by TTT: $P(X=0)=(1/2)^5=1/32$.

For one head, the initial round can stop with TH or HT, or TT can be followed by one head and two tails: $P(X=1)=(1/2)^2+(1/2)^2+(3!/2!)(1/2)^5=19/32$.

For three heads, TT can be followed by HHH, or HH by one head and two tails: $P(X=3)=(1/2)^5+(3!/2!)(1/2)^5=1/8$.

For four heads, HH must be followed by two heads and one tail: $P(X=4)=(3!/2!)(1/2)^5=3/32$.

For five heads, HH must be followed by HHH: $P(X=5)=(1/2)^5=1/32$.

Let Y be the payout in dollars. Since each head earns one dollar, its distribution is

$E(Y)=0(1/32)+1(19/32)+2(1/8)+3(1/8)+4(3/32)+5(1/32)=56/32$. A fair game requires $E(Y-k)=0$, so $k=E(Y)=56/32=7/4$. The fee is 1.75 dollars.