Two-Headed Coin and Bayesian ProbabilityDate: 04/21/2003 at 17:12:44 From: Maggie Subject: Probability In a box there are nine fair coins and one two-headed coin. One coin is chosen at random and tossed twice. Given that heads show both times, what is the probability that the coin is the two-headed one? What if it comes up heads for three tosses in a row? I understand that there are 10 coins in total. My teammates tried it out also and they got 4/9 + 4 for the first part and 8/9 + 8 for the second part. I don't understand how they got this. Date: 04/21/2003 at 17:28:05 From: Doctor Mitteldorf Subject: Re: Probability Dear Maggie, Here's a way to think about it. Make a tree: flip two heads (1/4) / choose fair coin (9/10) / \flip anything else (3/4) 10 coins \ choose two-headed coin (1/10) -> flip 2 heads (1/1) Study this tree, and it becomes clear that there are 3 possibilities: 1 - the top one has probability (9/10)*(1/4) = 9/40 2 - the next one has probability (9/10)*(3/4) = 27/40 3 - the last one has probability (1/10)*1 = 1/10 Before you did the experiment, these were all the possibilities there were. Then you did the experiment. What did it tell you? It told you that the middle option is out. The coin did NOT show a tail, so we know it wasn't the second outcome. This narrows our universe to the 9/40 and the 1/10. The trick now is to re-normalize these probabilities so that they show a total probability of 1, but stay in the same ratio. Within that universe (all the possibilities that are left) lines (1) and (3) remain in the ratio 9:4. So the probability of the top one is 9/13 and the bottom is 4/13, where 13 is just the sum of 9 and 4. Can you extend this reasoning to come up with the corresponding result for three flips? (This kind of reasoning is called Bayesian probability, and it is one of the most confusing topics in probability at any level of study.) - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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