0

Here is a problem I thought of:

lets say unfair coin that has Probability to get heads $P(h)=0.4$ for winning I have to get at least once heads of 7 flips.

so $P(loose) = (1-p(h))^7$ and $P(win)=(1-p(loose))=0.972$

now lets say that for full win I have to win 10 times (if I loose once I loose). $P(fullWin) = (p(win))^{10} =0.753$

can you please tell me if my calculation is good?

Ori
  • 11
  • 1
  • 1
    Yes, this all looks good. Assuming, of course, that the coin tosses are independent of each other, but that's a standard assumption. – lulu Apr 13 '22 at 18:39
  • Thanks! but I saw this Q and I'm not sure why not solve it like I did:https://math.stackexchange.com/questions/59738/probability-for-the-length-of-the-longest-run-in-n-bernoulli-trials/59749#59749 – Ori Apr 13 '22 at 18:47
  • That is a radically different problem. – lulu Apr 13 '22 at 18:49
  • why? in my game n is 7 flips X 10 games and m= 7 – Ori Apr 13 '22 at 18:54
  • @OriSpokoini because in your game, having a run of at least 7 does not guarantee a win. Suppose the first seven flips, you flip heads every time, but tails every time after that. You lose your game. So, setting $n=70$ and $m=7$ does not help you calculate anything about the question you asked. If you were to play $10$ games, it could tell you the probability that you wind up with a streak of at least $m$ winning games among them, but that is a different question. – SlipEternal Apr 13 '22 at 19:08

0 Answers0