The next exercise is to generalize our previous result. We continue to maintain the overall average, and examine what happens when we vary the magnitude of the bias.
In other words, instead of bags of coins that land tails with probability .5 + .25 and .5 - .25, we work through the same exercise with bags of coins that land tails with probability .5 + x and .5 - x. For my own sanity, let us define the median value M = .5.
P(A)
= (M+x)^4 * .5 + (M-x)^4 * .5
= [ (M+x)^4 + (M-x)^4 ] /2
At this point, we expand the expression in the brackets. Notice that the odd powers of x vanish, and the even powers of x get a contribution from each term which will cancel out the denominator.
P(A)
= [ (M+x)^4 + (M-x)^4 ] /2
= M^4 + 6 M^2x^2 + x^4
It's a good idea at this point to do a spot check - what happens when x is 0? In this case we find p(A) = M^4 = .5^4 which we know is right from flipping true coins. This is a test will be generally useful for checking that we've done the our math correctly, so file the idea away.
So what is the average probability of those coins that landed tails every time?
= ∑ fP
= f(B)P(B|A) + f(!B)P(!B|A)
= 1/2 [ (M+x)(M+x)^4 +(M-x)(M-x)^4 ] / ( M^4 + 6 M^2x^2 + x^4 )
= 1/2 [ (M+x)^5 + (M-x)^5 ] / ( M^4 + 6 M^2x^2 + x^4 )
= ( M^5 + 10 M^3x^2 + 5 Mx^4 ) / ( M^4 + 6 M^2x^2 + x^4 )
= M ( M^4 + 10 M^2^2 + 5 x^4 ) / ( M^4 + 6 M^2x^2 + x^4 )
May 7, 2005 8:53 PM
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