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I was reading this problem (Page 6, THE BASKETBALL PROBLEM, MEMORYLESS PROCESSES AND THE GEOMETRIC SERIES) and stumbled upon the solution using the memoryless property.

I cannot understand the intuition and logic behind writing $x_{B}=p_{B}+(1-p_{B})(1-p_{M})x_{B}$

why is $(1-p_{B})(1-p_{M})$ multiplied by $x_{B}$ ?

I can follow the geometric series solution and I can go backwards from $x_{B}=p_{B}\sum_{0}^{\infty }((1-p_{B})(1-p_{M}))^n $

to derive $x_{B}=p_{B}+(1-p_{B})(1-p_{M})x_{B}$

but I cannot understand how one can derive to $x_{B}=p_{B}+(1-p_{B})(1-p_{M})x_{B}$ from only the formulation of the problem. Is there any intuition that I am missing?

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Hi: The equation results from a recursive argument. If Bird and Magic both miss their first throw, then the game is essentially starting over right because Bird is now shooting ( and he shot first according to the rules of the game ).

So, if $x_b$ is the probability that Bird wins the game, then

$x_b = p_b + (1-p_{m})(1-p_{b}) x_b$ in words is just saying that

The probability of Bird winning the game is equal to

the probability of Bird making his first foul shot + [probability of them both missing their first shot] * the probability of Bird winning the game.

The argument ( heuristically speaking ) is that, if they both miss their first first free throw and and then a new spectator comes in the gym to watch them shoot, then from the perspective of the new spectator, the game is starting off from the beginning as if neither of them has taken any free throws.

This recursive argument is used often when dealing with Markov Chains.

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  • $\begingroup$ The equation shows that P(Bird winning the game) = P(Bird making the shot) + P(both missing their first shot) * P(Bird winning the game) . If the shots were independent i would expect something like P(Bird winning the game) = P(Bird making the shot) + P(both missing their first shot) * P(Bird making the shot*). Why is P(both missing) being multiplied by P(winning the **game)?? $\endgroup$
    – ECII
    Nov 12, 2020 at 15:17
  • $\begingroup$ but what you wrote, is the prob that bird misses THE FIRST TIME, and magic misses THE FIRST TIME and then bird makes. but that's the proba of bird winning in less than or equal to two rounds. $x_b$ is the prob bird wins eventually. Keep in mind that, if they both miss, the game starts over. In fact, think of what I wrote as prob( of bird winning once he's standing at the line ready to take A foul shot) = prob(of him making THAT shot ) + prob(bird misses THAT shot and magic misses (THAT + 1) shot) * prob(bird winning once he's standing on line ready to take THAT+2 shot ). $\endgroup$
    – mlofton
    Nov 12, 2020 at 20:44
  • $\begingroup$ But, in the last equation above, the last term on the right hand side has to equal the term on the left hand side because the process is memoryless. Therefore, you can replace both of them with the left hand side term which is prob(bird winning once he's standing at the line line ready to take A foul shot ). $\endgroup$
    – mlofton
    Nov 12, 2020 at 20:47
  • $\begingroup$ Finally, note that prob(bird winning once he's standing at the line ready to take A foul shot ) is the probability of him winning the game so it's $x_b$. $\endgroup$
    – mlofton
    Nov 12, 2020 at 20:54
  • $\begingroup$ Thank you for your comments. I get it now. Explaining prob( of bird winning once he's standing at the line ready to take A foul shot) = prob(of him making THAT shot ) + prob(bird misses THAT shot and magic misses (THAT + 1) shot) * prob(bird winning once he's standing on line ready to take THAT+2 shot ) helped me. Please include it in your answer. Thanks a lot $\endgroup$
    – ECII
    Nov 13, 2020 at 15:44

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