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Originally Posted by Ken G
Answers like this always depend on the "rules". I'm reminded of the great Monte Hall puzzle, in which you are shown three doors, one of which is a grand prize and the other two are goats. After you choose one door, Monte reveals another door has a goat behind it. Then he asks you if you want to change your choice to the remaining door. Is there a reason to do it?
Depending on the "rules" governing what Monte does, the chance of the new door being the grand prize will either be 1/2 or 2/3. It's pointless to argue about which is correct until you know Monte's rules. (And given the way Monte normally works, the rules indicate the answer will be 2/3).
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It's a similar sort of problem, yes, but not the same. In the Monty Hall problem, Monty has extra information that he indirectly gives you by ALWAYS opening a door without the grand prize. In the choosing balls from a tub problem, no such decision is made -- it's purely random. You don't get any extra information from having chosen the number 7. Your odds of having chosen 7 were small, sure... but once you've picked it, you have pruned your decision tree:
prior to selection:
Root (1.0)
---ten ball urn (0.5)
------1 (0.5 * 0.1)
------2 (0.5 * 0.1)
------3 (0.5 * 0.1)
------etc
---million ball urn (0.5)
------1 (0.5 * 10^-6)
------2 (0.5 * 10^-6)
------3 (0.5 * 10^-6)
------etc
post-selection:
Root (1.0)
---ten ball urn (0.5)
------7 (1.0)
---million ball urn (0.5)
------7 (1.0)
Actually, it might help to think about it this way. The problem is formulated in a deceptive way. They could pick any number from 1-10 and still make their point. What they could not do is pick 11-1 000 000. That means that the selection was not truly random, and given a random choice of numbers which would seem to make their point, there are ten options from urn #1 and ten options from urn #2. Even odds for each possibility, no matter which number is picked.
Or, this way: there is an urn with one ball, and an urn with 9 balls. You reach in and pull out a number... any number. The chance that it is a 1 is 20%. However, given that it is going to be a 1, the chance that it is a 1 is 100%, and you can flip a coin to determine which urn you're going to pull it from: it's equally likely that each urn is the one with the lone ball.
If you still doubt, try it! Put nine pennies and a dime on a table. Turn all the pennies but one to heads, and the dime and the other penny to tails. Now, mix them about with your eyes shut and poke a coin at random. If it's a heads coin, discard the measurement (you are not interested in selections that are not the 1 ball!), and if it's tails, record whether it's a dime or a penny. Except for the difference in coin sizes, you should end up with about 50% of each.