A and B play until one has 2 more points than the other. Assuming that each point is independently won by A with probability p, what is the probability they will play a total of 2n points? What is the probability that A wins?
I suggest the answer to the first question is 100% of the time. The lowest number of points that can be played with a winner is 2 then any other victory must be 2n greater than that number.
The second question has me stumped.
From the question:
"Assuming that each point is independently won by A with probability p"
A = \(\displaystyle \mathscr{Pr}(p)\)
B = \(\displaystyle \mathscr{Pr}(p^c)\)
The problem I have is that \(\displaystyle A^3B\) can be arrived at in 2 ways. But as n gets larger and \(\displaystyle A^{n+2}B^n\) the number of ways you can arrive at each result seemingly get endless.
I think I am definitely missing something.
I suggest the answer to the first question is 100% of the time. The lowest number of points that can be played with a winner is 2 then any other victory must be 2n greater than that number.
The second question has me stumped.
From the question:
"Assuming that each point is independently won by A with probability p"
A = \(\displaystyle \mathscr{Pr}(p)\)
B = \(\displaystyle \mathscr{Pr}(p^c)\)
The problem I have is that \(\displaystyle A^3B\) can be arrived at in 2 ways. But as n gets larger and \(\displaystyle A^{n+2}B^n\) the number of ways you can arrive at each result seemingly get endless.
I think I am definitely missing something.