Imagine someone has a bag in it with 10 balls in it. Just one of them is marked with your name. They flip a coin. If heads they pull nine balls from the bag, if tails they pull one ball.
Sounds familiar.
Now, suppose you are told that your ball was pulled; you are called to be a judge. P(heads|called) = P(heads) × P(called|heads) / ( P(called|heads)×P(heads) + P(called|tails)×P(tails) ), applying Bayes’ theorem and completeness. Is this updated probability different than the prior of 1/2? Well, we need to know that P(called|heads) and P(called|tails) are. Since the problem is non-perverse, they’re the nice round 9⁄10 and 1⁄10.
So P(called|heads) / ( P(called|heads)×P(heads) + P(called|tails)×P(tails) ), the “likelihood ratio” that’s basically how much evidence you get (there’s a log in there somewhere) evaluates to 9⁄10 / (9/20 + 1⁄20 ) = 9⁄5. Therefore being called as judge gives you some evidence about whether the coin landed heads or tails.
Ok. You’re offered a bet that is only valid if your ball is selected. That is enough to bring updating into the situation- you don’t even need to know whether or not your ball will be selected! You say “Ok, there’s a half chance my ball is not selected, and the bet is off. The other half of the time, the bet is on, and there’s a 9/10ths chance that the coin came up heads, since I know my ball has been selected.”
As suggested earlier, this strategy only works if you throw away half of the outcome space under the assumption that you can’t impact what happens there, despite the formulation of the problem being such that you do impact what happens there.
Imagine someone has a bag in it with 10 balls in it. Just one of them is marked with your name. They flip a coin. If heads they pull nine balls from the bag, if tails they pull one ball.
Sounds familiar.
Now, suppose you are told that your ball was pulled; you are called to be a judge. P(heads|called) = P(heads) × P(called|heads) / ( P(called|heads)×P(heads) + P(called|tails)×P(tails) ), applying Bayes’ theorem and completeness. Is this updated probability different than the prior of 1/2? Well, we need to know that P(called|heads) and P(called|tails) are. Since the problem is non-perverse, they’re the nice round 9⁄10 and 1⁄10.
So P(called|heads) / ( P(called|heads)×P(heads) + P(called|tails)×P(tails) ), the “likelihood ratio” that’s basically how much evidence you get (there’s a log in there somewhere) evaluates to 9⁄10 / (9/20 + 1⁄20 ) = 9⁄5. Therefore being called as judge gives you some evidence about whether the coin landed heads or tails.
Ok. You’re offered a bet that is only valid if your ball is selected. That is enough to bring updating into the situation- you don’t even need to know whether or not your ball will be selected! You say “Ok, there’s a half chance my ball is not selected, and the bet is off. The other half of the time, the bet is on, and there’s a 9/10ths chance that the coin came up heads, since I know my ball has been selected.”
As suggested earlier, this strategy only works if you throw away half of the outcome space under the assumption that you can’t impact what happens there, despite the formulation of the problem being such that you do impact what happens there.