One of two independent flip sequences reaches two heads simultaneously. What is the distribution of others tosses.
In this answer a way to find the probabilities.
Let $p$ denote the probability at start position, let $q$ denote the probability under condition that in the first round a tail and a head are thrown and let $r$ denote the probability under condition that in the first round two heads are thrown.
If it concerns the probability on HH then:
- $p=\frac14p+\frac12q+\frac14r$
- $q=\frac14p+\frac14q$
- $r=\frac14+\frac14p$
Solving this we find $p_{HH}=p=0.12$.
If it concerns the probability on TH then:
- $p=\frac14p+\frac12q+\frac14r$
- $q=\frac14+\frac14p+\frac14q$
- $r=\frac14p$
Solving this we find $p_{TH}=p=0.32$.
If it concerns the probability on HT then:
- $p=\frac14p+\frac12q+\frac14r$
- $q=\frac14p+\frac14q$
- $r=\frac12+\frac14p$
Solving this we find $p_{HT}=p=0.24$.
If it concerns the probability on TT then:
- $p=\frac14p+\frac12q+\frac14r$
- $q=\frac14+\frac14p+\frac14q$
- $r=\frac14p$
Solving this we find $p_{TT}=p=0.32$.
Edit:
Let me expand on the first case. There $3$ states to be discerned: $S_0,S_1,S_2$ where the index denotes the number of heads that where thrown in the former round. At the start we are in position in $S_0$.
$p$ denotes the probability of the other guy ending with HH if we start in $S_0$, $q$ denotes the probability of the other guy ending with HH if we start in $S_1$ and $r$ denotes the probability of the other guy ending with HH if we start in $S_2$.
Starting from $S_0$ the probability to arrive in $S_0$ again (i.e. both throw tails) equals $\frac14$. The probalility to arrive in $S_1$ is $\frac12$ (one of them throws tails and the other throws heads). The probability to arrive in $S_2$ is $\frac14$ (both throw heads).
That gives us the equality:$$p=\frac14p+\frac12q+\frac14r$$
Starting from $S_1$ on a similar way there is probability $\frac14$ to return to $S_0$ and also probability $\frac14$ to stay in $S_1$. Next to that there is the probability of $\frac12$ that the game ends in such a way that not both end with HH.
That gives us the equality:$$q=\frac14p+\frac14q+\frac120=\frac14p+\frac14q$$
Starting from $S_2$ on a similar way there is probability $\frac14$ to return to $S_0$. Also there is a probability $\frac14$ that the game ends in such a way that both end with HH. Also there is a probability $\frac12$ that the game ends in such a way that not both end with HH.
That gives us the equality:$$r=\frac14p+\frac141+\frac120=\frac14p+\frac14$$
Short inexact answer for intuition.
Let's say A got HH on throws $n$ and $n+1$. We know that player B didn't get HH on throws $n-1$ and $n$. This makes it a bit less likely that player B got H on throw $n$. So this explains why HH and HT are lower than TH and TT.
But what about HH vs. HT? To gain intuition, consider just both players tossing their coin twice, without any extra conditions. What is the probability that (a) both get HH, or (b) one gets HH and one gets HT? The answer to (a) is $1/2^4$ because there are 4 tosses that have to give a specific result. But the answer to (b) is actually twice that because there are two possible outcomes that give this result, as we didn't specify who gets a HH!