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I have problem where you are given the number of times you must win, and the chance of winning a game. I need to find the probability that I win all 4 times.

Example: I am only given two inputs, 4 (the number of times I must win) and 0.4 (the chance of winning each time).

I would expect the answer to just be 4 * 0.4, but the answer that is correct for this scenario is 0.290 (rounded off). I've tried multiple methods to achieve this answer, but I cannot for the the life of me get it.

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  • $\begingroup$ This question doesn't seem to have much to do with java. I'll replace the tag with a relevant one (probability). Your question is very unclear. Please edit to explain what you mean more clearly. How many attempts are there. Is the 0.4 the chance to win at least once over several trials? The chance to win per trial? What is the precise condition being satisfied? What is the unknown quantity you want to find? Please remove the several ambiguities that are in it presently. $\endgroup$ – Glen_b Oct 17 '14 at 1:10
  • $\begingroup$ @Glen_b I fixed it. $\endgroup$ – Devin Oct 17 '14 at 1:14
  • $\begingroup$ "I would expect the answer to just be" -- the answer to what? Up to that point there's no question to answer, no probability to find. What is the quantity you want to find? You improved your question slightly but you didn't really deal with the problem with the question. $\endgroup$ – Glen_b Oct 17 '14 at 1:19
  • $\begingroup$ @Glen_b I gave you all of the information, I don't see what you're confused about. 4 trys with a 0.4 (40%) chance of winning each time. I need to find the probability that I win all 4 times. $\endgroup$ – Devin Oct 17 '14 at 1:21
  • $\begingroup$ " I need to find the probability that I win all 4 times. " --- where in your question does it say that? $\endgroup$ – Glen_b Oct 17 '14 at 1:23
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You said you want "P(Win on all four trials)".

You use the fact that the trials are independent to compute the joint probability.

That is, you use the rule $P(A\cap B)=P(A)P(B)$ (for independent events A and B). You'll need to apply that rule recursively to get the equivalent for four events.

See here


Interpreting the question to be

"In a series consisting of a maximum f 7 games, the only possible outcomes on a game are win (W) or lose (L) -- the series is won as soon as one side wins a majority of - games (i.e. 4 wins, making them the winner). Our side has a 0.4 chance to win each game, and the outcomes of each game are independent of all the other games. What is the chance our team wins the series?"

Then you need to consider the mutually exclusive cases

(i) The series is won in exactly 4 games (4W, 0L)
(ii) The series is won in exactly 5 games (4W, 1L)
(iii) The series is won in exactly 6 games (4W, 2L)
(iv) The series is won in exactly 7 games (4W, 3L)

and solve each separately. Within each you have a simpl binomial problem.

-- but take care! note that in each case the last game must be a win for our side, so that's not part of the possible arrangements of the W/L ordering within each case.

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  • $\begingroup$ How would I compute events A and B if I only have one event? $\endgroup$ – Devin Oct 17 '14 at 1:38
  • $\begingroup$ If the answer given to me in the textbook is 0.290 (rounded off), and if I'm understanding correctly, P(A)P(B)P(C)P(D) != 0.29 $\endgroup$ – Devin Oct 17 '14 at 1:41
  • $\begingroup$ I can only answer the question I have. I don't know why they think it's 0.29 $\endgroup$ – Glen_b Oct 17 '14 at 1:42
  • $\begingroup$ Did you see this picture? $\endgroup$ – Devin Oct 17 '14 at 1:45
  • $\begingroup$ Your question must stand on its own; I relied on your interpretation of the question (you said "4 trys with a 0.4 (40%) chance of winning each time. I need to find the probability that I win all 4 times."). I expect that instead, "win the series" only requires you win a majority of games. How many possible games in the series? It's not in your question nor in the image. $\endgroup$ – Glen_b Oct 17 '14 at 1:50
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Looking at the question you posted, it looks like you need the probability of winning 4 games before the opposing team wins 4 games. Then there are four possibilities: your team wins the first four, your team wins 3 of the first 4 and the last game, your team wins 3 of the first 5 and the last game, your team wins 3 of the first 6 and the last game. If you add up those probabilities, you should get about .2898.

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  • $\begingroup$ What do you mean "and the last game $\endgroup$ – Devin Oct 17 '14 at 2:03
  • $\begingroup$ Meaning you can break up the series of games into two parts. For example, suppose your team wins in 6 games. Then this means you can have your team win 3 of the first 5 games in any order, but then they would have to have won the 6th (last) game in order to have won in 6 games. $\endgroup$ – hard2fathom Oct 17 '14 at 2:04
  • $\begingroup$ I still don't really follow :/ Is there any way you could reword what you said? $\endgroup$ – Devin Oct 17 '14 at 2:06
  • $\begingroup$ So the probability that team A wins 3 of the first 5 games can be calculated as $P(Y=3)$ where $Y \sim Binomial(5,.4)$. To get the probability of 4 in 6 games, it would then be .4*P(Y=3). You need to repeat that idea for each possible case.The key idea is that your team has to win the last game to win the series in k games. $\endgroup$ – hard2fathom Oct 17 '14 at 2:18

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