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I'm wondering if probability is influenced by order of actions. For instance, lets have a problem:

Suppose there are two holes close together in a golf course. The golfer can hit the ball into either hole with equal probability. He is allowed to hit 4 times. Assuming he can always get the ball into one of the two holes. What is the probability that he could get 2 balls in each hole?

My solution is 1/5. This is result of (accepted results)/(possible outcomes). This seems a little simple to me, is it possible that the probability is influenced by order in which the player hits the shots?. Could anyone tell me if the result is correct please? Thanks in advance.

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    Doesn't sound like real golf. Sounds more like an implausible variant of "what is the probability of exactly $2$ heads and exactly $2$ tails if you toss a fair coin $4$ times". And the answer to that is definitely not $1/5$.2011-05-17

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When the problem says "either hole with equal probability", they mean that every time he hits a ball, the probability it goes into a specific hole is 0.5. In other words, no, the problem does not allow for the probability to be influenced by what order shots are taken (this is completely irrelevant to whether or not it is the case in real life that the arrangement of grass or whatnot could influence the probability between shots).

Besides, each shot itself should be indistinguishable from the rest; otherwise, it's not "4 occurrences of one shot", but "1 occurrence each of 4 slightly different shots", making the problem impossible to analyze without knowing the exact probabilities related to each shot. Because the shots are indistinguishable, there is no real meaning to what "order" they are taken in.

By the way, this means that the correct answer is $\frac{\binom{4}{2}}{2^4}=\frac{3}{8}$.

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    Just pointing out the mistake in the OP is that he seems to have counted the total possibilities as "one way to get 0 balls in the left hole, plus one way to get 1 ball in the left hole, plus one way to get two balls in the left hole ...", the oversight being that there are different numbers of ways to satisfy the cases of "n balls in the left hole". As implied in Zev's answer, there's $4\choose{n}$ ways to do this. The $2^4$ in the denominator comes from the sum over $4\choose{k}$ for $k=0,1,2,3,4,5$.2011-05-17