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Consider a Bernoulli Process(p). What is the probability you will get 5 heads before 3 tails, flipping a coin.

Im having trouble answering this question because, from my understanding Bernoulli process has to have independent trials, however a failure of say tosses 1-8 flips, directly affects 2-9 (you know atleast 3 tails in 1-8)

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    $\begingroup$ I would do it recursively. Let $P(n,m)$ be the probability that you get $n$ Heads before you get $m$ tails. Then $P(n,m)=\frac 12\times (P(n-1,m)+P(n,m-1))$ and it is easy to sort out boundary values. $\endgroup$ – lulu Sep 20 '18 at 23:42
  • $\begingroup$ aaah @lulu yes perhaps interpreting question wrong! $\endgroup$ – Mehness Sep 20 '18 at 23:45
  • $\begingroup$ @Mehness the way I read it, the game takes at most $7$ tosses. After $7$ you must have either five heads or three tails. So, it is perfectly possible to just sort out the various cases over $7$ trials. My recursion, though, will handle much larger numbers. $\endgroup$ – lulu Sep 20 '18 at 23:48
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    $\begingroup$ On reflection, you can just assume that $7$ tosses are made (even if the game is over before then). Your problem is the same as getting $≥5$ heads over seven tosses. $\endgroup$ – lulu Sep 20 '18 at 23:51
  • $\begingroup$ yeah sure on reflection eminently readable as actually terminating if 5 heads occur first rather than BC2 type thing (it's late)! $\endgroup$ – Mehness Sep 21 '18 at 0:00
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Hint: In order to obtain 5 heads before 3 tails you must obtain at least five heads among the first seven tosses.

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