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Suppose financial analyst is predicting market raise/drop, and they are of 2/3 probability correct. It means if market raise, they have 2/3 probability predict market is raise, and if market drop, they have 2/3 probability predict market is drop.

If I am asking 3 analyst (suppose they are independent), what is the probability two of them are predicting the right results? Could it be simply calculated as 2/3 * 2/3 * (1-2/3)?

Another way of thinking is, P(two predict raise, one predict drop | market raise) * P (market raise)+ P(two predict drop, one predict raise | market drop) * P (market drop).

But the latter solution is more complex and require know prior probability (P (market raise) and P (market drop)). Just wondering if the first one is correct? Thanks.

Edit 1, BTW, in my scenario, analyst are independent.

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If I am asking 3 analyst (suppose they are independent), what is the probability two of them are predicting the right results? Could it be simply calculated as 2/3 * 2/3 * (1-2/3)?

Nope, you can't do this. There is more than one way that two analysts could be correct. For instance, if we label our analysts 1, 2, and 3, then both scenarios would give us two analysts being correct:

  1. 1 and 2 are correct, but 3 isn't
  2. 1 and 3 are correct, but 2 isn't

So you need to multiply this probability by the number of ways you can get 2 being correct. This is a combinatoric, which is 3 choose 2 and is given by the formula $\frac{3!}{2!(3-2)!}$ which gives you "the number of ways of choosing 2 objects from a set of 3, such that the order they're chosen doesn't matter."

Assuming independence, and also that the probability of being correct is the same for all 3, this number of correct analysts is a Binomial random variable with $p=\frac{2}{3}$ and $n=3$

So the probability is $P(X=2) = \frac{3!}{2!(3-2)!} \frac{2}{3}^2 \frac{1}{3}^{(3-2)}$

In case you're unaware what "$x!$" means, it is this: $x! = x \times (x-1)\times (x-2) ... 2 \times 1$

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    $\begingroup$ I see that this is what the question was actually looking for, but keep in mind that it is only true if the analysts are talking about different investments $\endgroup$ – shadowtalker Sep 3 '16 at 11:35
  • $\begingroup$ Thanks ssdecontrol, actually in my case, I am talking about the same stock/ETF, I am collecting feedback on independent finance analyst, and try to predict the probability. Wondering what will be wrong if for different investments? $\endgroup$ – Lin Ma Sep 4 '16 at 6:44
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    $\begingroup$ @LinMa this is the correct answer for different investments. $\endgroup$ – shadowtalker Sep 5 '16 at 18:58
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The first one has to be wrong because you can't suppose they are independent.

If there are only two events ("drop" and "rise") then if one analyst predicts "drop" and has a objective 2/3 probability of being correct, then every other analyst with a 2/3 probability of being correct must also have predicted "drop." That's because they only other thing they could have predicted, "raise," would have been correct with probability 1/3 based on our assumptions about the first analyst. You can shift the numbers around here, but at some point assuming that the analysts' forecasts are independent leads to a contradiction with the setup of the problem.

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  • $\begingroup$ Thanks ssdecontrol, if they are independent, then the first method is correct? $\endgroup$ – Lin Ma Sep 3 '16 at 4:08
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    $\begingroup$ @LinMa that's like saying "I calculate humans will be populating the galaxy in the next 10 years, assuming faster than light travel becomes possible." $\endgroup$ – shadowtalker Sep 3 '16 at 11:33
  • $\begingroup$ Thanks ssdecontrol, do not quite get your points. If the analyst are independent, then the first method is correct? $\endgroup$ – Lin Ma Sep 4 '16 at 6:43
  • $\begingroup$ @LinMa they can't be independent unless you're talkin about different underlying markets. I've said it literally three times and presented a short proof. Moreover, the other answer (which you accepted) states that no, the first method is not correct even if you manage to assume independence $\endgroup$ – shadowtalker Sep 4 '16 at 11:24

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