I guess that's enough of that. MikeM's "bag" sentence is tighter than anything I could hope to come up with. All you learn when you draw it, is that before you drew it you were more likely to draw it by an amount equal to the amount which drawing it reduces it. In other words, you only learn about the tick you withdraw, which is no longer relevant. If you're still not satisfied, consider that the long-run realized volatility of the bag has to be the same as the long-run realized volatility of the coin flips. If the bag was counter-trendy at the 1-tick frequency, it would have to be trendy somewhere else to make up for it. But any logic which predicts it is counter-trendy, intuitively prohibits an inflection point where this counterbalancing trendiness might begin to take hold. It should only get more counter-trendy, right?

HAHAHA I can't believe that I am arguing with the creator of the question. I thought that you were going to step in and save the day and you just provided an incorrect solution also. OK pretend you are flipping 2 coins, do you agree that there is a 50% chance they are the same and a 50% chance that they are different?Let's pretend you agree to that.there is a 25% chance you have 2 heads, 25% chance you have 2 tails and a 50% chance that you have one of each.(so you flip the coins and hide them)So now in your mind there is a1) 1 in 4 chance they are both HEADS2) 1 in 4 chance they are both TAILS3) 2 in 4 chance they are DIFFERENTif I then tell you that one of them is HEADS you can eliminate 2) and one of the situation so we are left with1) 1 in 3 chance they are both HEADS3) 2 in 3 chance they are DIFFERENTgiving us a 2 in 3 chance that the other hidden coin is TAILS.Greg

Greg, Wrong.Assume there are 4 bags containing1)hh2)th3)ht4)ttYou reach into a bag (at random) and randomly pull. It turns out you pull H.What are the odds that you pulled that H from bag 1.(this is similar to the 3 drawers and gold/silver coins teaser).There are 4 possible ways to pull H.2 of those ways are in bag 1.Therefore, if you pull H, the probability that you pulled from bag 1 is 0.5, from bag 2 is 0.25 and bag 3 is 0.25or either bag 2 or 3 is 0.5So now we can combine bag 2 and 3 and call it bag 2now we have:1)H2)Tnow we have to add a H or T to both bags.We end up with our original distribution:1)hh2)ht3)th4)ttthe odds of drawing a head on the 2nd pull is 0.5.Wilbur

Greg,I disagree with you. You aren't being told that one of the coins was heads... you're being told that the one you pulled out of the bag is heads.Think about the following two games. Both start we me flipping two coins and hiding them from you. You are allowed to ask me question about the coins that I must truthfully answer.In game 1, you ask me "Is coin #1 a head?". I reply yes.In game 2, you ask me "Is one of the coins a head?". I reply yes.In game 1, the probabilty of both coins being heads is still 0.5. I agree (mirroring your reasoning) that in game #2 the probability is 1/3.In the original question, we are in the situation of game 1 (at least as I interpret the question).

I considered titling the question "the glass half full." I expected to create an argument between A) people who thought that drawing an uptick made it more likely to come from a bag with 4 upticks, and B) people who thought that drawing an uptick left the probably originally balanced bag with a downtick bias.Then Tripitaka came out with some unstoppable "transition matrix" and put me in my place. I also thought maybe there might be some interesting interplay between a finite bag and an infinite coin. But I guess nobody else is unemployed enough to be bothered with meditating on such a deep concept.

Wilbur, I will admit that I was wrong to you. You are absolutely right I forgot one part of the bayesian equation. so it is still 50/50.I digress

farmer,Can you please explain to a guy who is too employed to have the time to meditate on the subject, what an "infinite coin" is? Also, it sounds like a nice thing to have, where can I get one for cheap?Best regards,Matt

I assume when you toss an "infinite coin" you get a one side up out of infty. Then here is a "paradox": what is the probability to get a particular side of an "infinite coin" when you toss it? Zero. How come then when tossing it you choose one side with a probability 1?

Measure theory.More to the point, how can you ever really observe which side is up? You can only decide that it (the up side) lies in, or does not lie in, certain subsets of sides that can be determined using a finite amount of information.

This reminds me of the old 3 doors question. I would post it as a separate question, but I'm sure it's been a brainteaser before, it's just too standard a question. But for those who haven't heard it...3 doors. Behind one of them is a pot of gold. You get one guess, then a chance to change your guess.For example:You pick door #1. The moderator opens door #3 to reveal nothing behind it.Should you stay with #1 or change to #2? Does it matter?The answer is that you should change. I argued with my stats prof in college about this, then felt like an ass when I realized my mistake.It's a different problem, of course. It would be more similar if Farmer told you that the # of ticks in his bag happened to have exactly a normal distribution before you picked any.-t.

Tristain,You are missing the most important part of the Monty Hall teaser.The moderator (monty) knows where the gold is.He does not open a door at random.He opens a door that he knows contains no gold.If he opens a door at random, and reveals no gold, the problem is reversed.Wilbur.

That's what I was implying with: "a chance to change your guess". You always get a chance to change because the moderator opens a door that doesn't have the gold behind it.Sorry that I wasn't more clear.-t.