The problem is that you don’t know if the $2 bill you pulled out is the original bill, of heretofore unknown denomination, or the one you know for certain is a $2 bill. It’s that uncertainty that leads to the intersting problem.
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bnorton**s first post, with jcgmois caveat takes care of this pretty well.
Actually your answer was perfectly comprehensible, and, now that I’ve reread the OP, it was also right. For some reason I was under the impression that the bill he pulled from his wallet was randomly selected (which is true) and unknown (which is false). I was clearly stated that it was a $2, so my case #3 will never happen just like you said.
The question I was answering was actually more interesting than the OP, so I guess I can claim that I gave the right answer to the wrong question. Alas, that still makes it wrong, doesn’t it?
Now, IzzyR:
I’ve got to ask - do you really see your post as being clear and lucid? So much so that you don’t even know where to start to make it more comprehensible? Surely not.
Hey, this place is for fighting ignorance, right? Well I’m either Horrendously ignorant or a Genius (and I’m betting it’s the 1st option…)
Why would it matter what $2 bill it is? The OP asked what the probability of picking a $2 was, not the probability of picking any particular $2.
There are 2 possabilities here.
- The bill you start with is a $20. you put a $2 in, and you take a $2 out. You’re obviously left with a $20.
2)the bill you start out with is a $2. You put a $2 in, and you take one of the $2s out. Yes, you might take the original $2 out, but why would it matter? No matter which bill you take out, you’re still left with $2 in your wallet.
So, if you start out with $20, you will not have a $2 left in your wallet at the end. However, if you start with a $2, you will have a $2 left in your wallet at the end, thus the .5 probability of either.
The reason, Mikahw, is that your pulling out a random bill and it turns out to be a $2 is empirical evidence of the current distribution of the bills in your wallet. Before you pull a random bill out, you have a 50/50 chance that there are either two $2 bills or one $2 and one $20 bill in the wallet. After you pull out a $2, since that is a more likely result (100% compared to 50%) of the random selection in the first case, the a posteriori probabilities have changed.
You may be able to see it more easily in a more extreme case. Suppose the two equal possibilities of the contents of your wallet were one $20 and 100 $2 bills, and one $2 and 100 $20 bills. When you reach in and pull out one of the 101 bills at random, and it’s a two, what do you conclude? That you almost certainly have a billfold full of $2’s with one $20-the odds of pulling out the one $2 when you had a wallet full of $20’s were very low. In the OP’s example, you have exactly the same situation, except that the odds of pulling out a $2 bill are 50% vs. 100% rather than 1% vs. 99.1%. The pulling out of the $2 bill is statistical evidence that the distribution in which that was the more likely result is the case.
And if you still don’t get it, I can only recommend finding a textbook that discusses Bayesian analysis.
The reason the odds aren’t 50/50 after removing the first bill is because you don’t actually know which bill was removed, the two different $2 bills are not interchangeable.
Before pulling the first bill out of the wallet we wave these two possible conditions:
$2[sub]Starting bill[/sub] and $2[sub]Added later[/sub]
$20[sub]Starting bill[/sub] and $2[sub]Added later[/sub]
We are then told a $2 bill was pulled from the wallet but not WHICH bill it was.
If that $2 bill was the starting bill then the next bill will be a $2 bill.
If that $2 bill was the added bill then you have either a $20 bill or a $2 bill.
After the first bill was drawn we only have enough information to reduce the possible outcome to three, two of which are that the second bill will be $2 bill. Now, if we had been told the exact identity of the drawn bill then the probably of another $2 bill would either be either 100% or 50% depending on that identification.
The flaw in your thinking is that you didn’t have as much information as you thought you did. (Unless I’m wrong in this explanation, in which case I don’t have as much information as I think I do.)
Hmm, so according to protocol, do I, in my reply to this question, snidely impugn your reading comprehension, your math skills, or your overall intelligence? Or, to put it this way: it seems to me that my presentation of Bayes Theorem is pretty straightforward, and should be comprehensible to someone who is capable of understanding the theorem itself. I could have certainly made it simpler had I geared my post towards someone of extremely limited mathematical ability, but I did not think such would be necessary is a thread such as this, and it is more effort than I am willing to put forth. So if there is merely some aspect of the post that you did not understand I would be glad to explain it, but if the entire thing went over your head I’ll give it a pass. Your approach of
is a strange one, but whatever works for you…
Without getting into specific theorems and whatnot, simply looking at this threads third and forth posts should clear any confusion.
Simply stated:
You have in your wallet a bill that is either A or B, but you do not know which. At this point, the odds are 100% that it is one or the other (it has to be something), but for the sake of argument, we say it’s 50/50 that it’s one or the other.
Okay, now you put a bill you know for certain is a B bill in your wallet.
At this point, you have one of two scenarios were you to remove both bills: both bills are B bills or you have one B bill, which you just put in, and one A bill, which was the original.
Continuing on: if you, by chance, remove a bill from your wallet that happens to be a B bill, you are now back in you original delema, except that you do not know if it was the second bill or the first bill. This last bit is where the added complexity comes to bear. You are no longer asking simply if the bill is an A or a B, the question of which bill you removed is in play.
Given that the bill removed was a B, what is the probability that the remaining bill is a B? Your possibilities: [list=1][li]you removed the second bill you put in and the remaining bill is an A[]you removed the second bill you put in and the remaining bill is a B[]you removed the first bill you put in and the remaining bill is a B. At this point, it does not matter what the original bill was, only that it was 50/50 of it being a B bill.[/list=1]Of these outcomes, there is a 66.6% chance that the remaining bill is, in fact, a B bill.[/li]
Okay, not so simply stated, but hopefully reworded enough to help clear things up?