Nail-Tinted Glasses

From outcomes back to probabilities

We continue with our bags of biased coins, this time considering two different questions. Having flipped the coins, and collected 82 that all landed tails each time, we ask

What is the probability that a randomly chosen coin in this group is biased against tails?
What is the average probability of tossing tails again with each of these coins?

The first of these is pretty straightforward: we had 82 coins, one of which was actually biased against tails, so the probability is 1/82 = .0122.

Flipping 81 of the coins will produce tails again 75% of the time, so we should expect 60.75 more tails from that group. The other coin lands tails 25% of the time, so in total we expect 61 tails. We are flipping 82 coins to achieve that, so the probability of flipping tails on these coins is 74.39%

As it happens, there is a more direct way to calculate these, thanks to Bayes Theorem. Bayes theorem allows you to work a probability problem backwards - instead of asking "given a biased coin, what is the probability that it lands tails" you ask "given that a coin landed tails, what is the probability that it is biased".

We begin by defining the following symbol P(A|B) to represent the probability of event A, given that B occurred. Mathmatically, this is defined by the expression
P( A&B ) = P(A|B)P(B)

In our problem A might represent the event "flip tails four times in a row" and B the event "pick a coin biased toward tails". Of course, there's no particular reason to take the events in this order. One might instead write
P( A&B ) = P(B|A)P(A)

If we set these two equal to each other, we discover
P(B|A) = P(A|B)P(B)/P(A)

Reasonably, B or !B must be true, so we can write
P(A)
= P(A&B) + P(A& !B)
= P(A|B)P(B) + P(A|!B)P(!B)

So what is the probability that we flipped a biased coin (B), given that the coin we did flip landed tails four times in a row (A)? P(A|B) we determined
to be 81/256, P(A|!B) is 1 / 256, P(B) = P(!B) = 1/2.

P(B|A)
= P(A|B)P(B) / P(A)
= P(A|B)P(B) / [ P(A|B)P(B) + P(A|!B)P(!B) ]
= 81/256 * 1/2 / [ 81/256 * 1/2 + 1/256 * 1/2 ]
= 81/82

Hooray, it works.

What about the average probability? The expectation of a function f is
= ∑ f(n)p(n)
and the expectation of a function f, given A is
= ∑ f(n)p(n|A)
So the average probability of the coin landing tails is
= .75 * 81/82 + .25 * 1/82 = 61/82 = .7439....

May 7, 2005 8:50 PM | TrackBack