Quote:
Originally Posted by Poogs
First thing I would do is come up with the chances of a 35+ ball on the first pull. Since there’s 5 winning balls out of 39, I get 12.82 percent chance. Is this right so far?
Right. The trouble starts with the 2nd ball because 2(5/39) would double-count the chance of two highballs, as shown by Siegmund in the weighted average formula.
Both balls have the same 5/39 chance of being high, but 5/39 is the chance of this ball being high and the other ball being anything (including high). That is, 5/39 = P(HL) + P(HH).
Adding 5/39 + 5/39 is the same as adding [P(HL) + P(HH)] + [P(LH) + P(HH)]
So you see that P(HH) is counted twice. And then with 3+ draws, multiplying by 5/39 counts some possibilities
more than twice. No good. This overcounting is why your formula especially breaks down if you draw say 10 balls like nickthegeek pointed out. With 10 balls your chance is not 128%!
Instead of that, we can add only what wasn't already counted: P(HL or HH) + P(LH) = 5/39 + (34/39)(5/38)
Or we can take 5/39 + 5/39 - P(HH) since we've established that 10/39 double-counts P(HH).
(Or as statmanhal showed, from the start we can just calculate P(LL) and subtract it from 100%.)
If you were only interested in the 39, that would be different. You could multiply 1/39 by the # of balls picked because there wouldn't be a possibility of more than one ball being the 39.
Overcounting multi-success possibilities is also why you can't say there's a 100% chance of flipping a Tails if you flip a coin twice.
This concept is known as the principle of inclusion-exclusion: P(A or B) = P(A) + P(B) - P(A and B)
Whenever P(A and B)>0, you can't just add P(A)+P(B)