Imagine you can flip a coin up to a maximum number of X times. You stop if during this sequence, the coin lands on tails 3 times in a row. Otherwise you keep flipping.

You repeat this game some N number of times.

The "real world" application of this might be a shooter in a basketball game. Say he can shoot up to X=25 times (i.e. 25 potential FGA), and N=82 (games in a season). His coach tells him to stop shooting if he misses 3 in a row.

I simulated this scenario and found that the stopping condition does not affect the overall probability of the coin landing heads. This seems counterintuitive to me, as it "feels" (my intuition tells me) that the stopping condition would somehow bias the results.

Can someone explain to me why the stopping condition here doesn't affect the overall distribution of heads and tails in this scenario?

Imagine you can flip a coin up to a maximum number of X times. You stop if during this sequence, the coin lands on tails 3 times in a row. Otherwise you keep flipping.

You repeat this game some N number of times.

The "real world" application of this might be a shooter in a basketball game. Say he can shoot up to X=25 times (i.e. 25 potential FGA), and N=82 (games in a season). His coach tells him to stop shooting if he misses 3 in a row.

I simulated this scenario and found that the stopping condition does not affect the overall probability of the coin landing heads. This seems counterintuitive to me, as it "feels" (my intuition tells me) that the stopping condition would somehow bias the results.

Can someone explain to me why the stopping condition here doesn't affect the overall distribution of heads and tails in this scenario?

Edit: This appears to be different from the "birth problem" in that the value of p does not matter (i.e. it does not have to be 1/2). The stopping condition in this case seems to have no effect on the overall probability, regardless of the value of p. It's not clear to me from the answers to the birth problem why this has to be the case.