The probability that I am not lucky.

Factual Questions
21

[QUOTE=aptronym]
This seems close to what I am looking for in part (1) here.

Okay, so let’s say A is flipping 120 more heads than tails out of 4,000 flips, and B is having the heads-favoring coin.

P(A|B) is C(4000,2060)*0.5025[sup]2060[/sup]*0.4975[sup]1940[/sup].

P(B|A) should be 0.95, because we want to be 95% sure of B given A.

P(A) should be P(B)*P(A|B)+(1-P(B))*C(4000,1940)*0.5025[sup]1940[/sup]*0.4975[sup]2060[/sup].

P(B) could be solved from those three values.

Is everything I did kosher?
[/QUOTE]

That looks reasonable, but those exponents are going to give Excel fits. I don’t know of any free or inexpensive software that can handle them.

22

[QUOTE=ultrafilter]
That looks reasonable, but those exponents are going to give Excel fits.
[/QUOTE]
It’s actually not that bad when you simplify. You can divide through by C(4000,2060)0.5025^19400.4975^1940 to get:

P(B) = (0.5025^120+0.4975^120)/(0.5025^120/0.95+0.4975^120) = 0.961

Incidentally, my original idea that P(B)/P(not B) = P(A|B)/P(A|not B) would give:

P(A|B)/P(A|not B) = 0.5025^120/0.4975^120 = 3.32, or P(B) = 0.769.

Can someone tell me why these two give different answers, and which one is right?

23

[QUOTE=aptronym]
P(B)/P(not B) = P(A|B)/P(A|not B)
[/QUOTE]

This is only true if A and B are independent. Check with P(B) = 1/2 and A = B[sup]c[/sup] and you’ll see that it doesn’t hold.

24

Regarding: P(B)/P(not B) = P(A|B)/P(A|not B)

If A and B are independent, the right-hand side always equals 1, so the identity will not hold. If A and B are independent, then P(B)/P(not B) = P(A and B)/P(A and not B). You’re confusing the conditional probabilities with the joint probabilities.

25

Thanks a lot, guys, that was really helpful.