Eight people are given a fair coin and each tosses the coin until they land a tails. What's the probability that nth toss is the final toss? I am trying to understand how the sample space and probabilities for a problem like this would look. Starting with the basics, each person having a probability of 0.5 for tossing a tails for the first toss. For the second toss, each person having a probability of 0.75 (HH, HT, TH, TT) of tossing a tails. I believe this would continue through n tosses as $1-p^n$ where $p=0.5$.
Now looking at the probabilities across all 8 people, the probability of each person tossing a tails after the first toss would be equal to tossing 8 consecutive tails, $0.5^8$.
Then what would be the probability that the second toss would be the last toss (i.e., all 8 people tossed a tails within the first two tosses), then the probability that the third toss would be the last toss, all the way through n tosses.
I've looked at the probability of tossing a tails after each toss, but this is always increasing and I suspect that eventually the probability of toss n being the last toss will start to decrease (say after 7 tosses) because of the high probability to land a tails in 6 tosses. Any insights into how I should be thinking about this would be appreciated.

Number the persons with $i=1,2,\dots,8$ and let $X_i$ denote the number of tosses that are needed for person $i$ to arrive at tails.
Then at stake is the distribution of:$$X:=\max\{X_1,\dots,X_8\}$$where the $X_i$ are iid random variables with geometric distribution equipped with parameter $p=0.5$.
Here for nonnegative integer $k$: $$
\begin{align} 
P(X\leq k)&=P(X_1\leq k,\dots, X_8\leq k)\\&=\prod_{i=1}^8P(X_i\leq k)\\&=P(X_1\leq k)^8\\&=[1-P(X_1>k)]^8\\&=(1-2^{-k})^8
\end{align} 
$$
enabling you to find for positive integer $k$: $$
\begin{align} 
P(X=k)&=P(X\leq k)-P(X\leq k-1)\\&=(1-2^{-k})^8-(1-2^{1-k})^8
\end{align} $$
