I heard this today but couldn’t figure out the answer:
You have one bill in your wallet. It’s either a $2 bill or a $20 bill, but you don’t know which. Later, you put a $2 bill into your wallet. Later still, you go to pay for something and pull a $2 bill out of your wallet. What is the probability that the last remaining bill in your wallet is a $2 bill? (Note: it’s not 50-50).
So what is the probability? Why? Thanks much.
– Jer
At first glance, since there is one bill to start with, it has only one denomination, and the chance of it being a $2 is either 100% or 0%.
But if you started with a $20 and a $2, one fell out unbeknownst to you, you set up the unknown. But that is semantics. Before I go any further, smarter people? Please correct me.
When you had one bill it was 50/50.
Add the $2, now your chances of pulling EITHER bill is 50/50. Your chance of pulling the known $2 is 50%, and your chance of pulling the UNKNOWN bill is 50% and within that bill is an overall 25% chance of it being a $2 and 25% chance of it being a $20. Overall, your probability of pulling the $2 is 75%
So, you pull a bill and PRESTO it is a $2. It was EITHER the 50/50 $2, or the unknown one. If it was the true $2 (50% chance), you have a 50/50 for the next one being a two. If it was the “unknown” (25% chance), you have a 100% chance of the next being the true $2. I’ll take a SWAG and say there is a 75% probability the last bill is a $2. Additive.
Now please correct me. I am not at all confident in this.
You start out with 2 possibilities: (1) $2 or (2) $20.
When you add the $2 you still only have 2 possibilities (1) $2a + $2b or (2) $20 + $2c. (I’m adding the a, b, and c to the $2 to distinguish each bill.)
When you pull out a bill at random (I assume) you have these possibilities:
(1) $2a + $2b - $2a
(2) $2a + $2b - $2b
(3) $20 + $2c - $20
(4) $20 + $2c - $2c
In cases 1, 2, and 3 you are left with a $2 bill, and only in case 4 are you left with a $20. So it’s 3 to 1 that it’s a $2.
bnorton–case 3, pulling the $20 bill, never happens.
So only cases 1, 2, and 4 have to be considered. They are all equally likely, and in 2 cases of the 3 you have a $2 bill still stashed away.
That means the chances are 2/3.
jcgmoi is correct.
Umm, folks. VarlosZ failed to provide a complete specification of the problem. People are assuming an initial 50-50 chance of there being a $2 or $20 in the wallet.
The correct response is to always point out the lack of initial information. One never fills in assumptions and works from there.
After all, this isn’t “Ask Marilyn” here.
FtG aka GLP
Well, that depends on how you read the question. Since it was unclear I went with the asumption that one of the two bills was pulled at random.
Sure you do. Why not?
ftg:
Exactly. Around here, we assume that people aren’t idiots until it’s proven otherwise. If the initial odds were other than 50-50, that would have been a rather important piece of information to leave out, yes? It’s been my experience that, in problems such as these, proportionate (sp?) probability is assumed unless otherwise specified (or, at least, implied).
However, for the sake of clarity: yes, the initial odds are 50-50. Yes, we’re choosing bills from our wallet randomly.
Thanks for your consideration, guys. That makes sense.
But what are the odds that the last remaining bill spontaneously vaporizes before you can find out what it is?
If this were an episode of Star Trek, I think those odds would be pretty damn good!
2/3 is correct. This is the general rule for all problems such as this, as per Bayes Theorem (aka posterior probability):
The probability of Hypothesis A is equal to the probability of Outcome X (i.e. the actual result) given Hypothesis A, times the prior probability of Hypothesis A, divided by the total probability of Outcome X.
In this instance, we are assuming that the prior probability of the $2 was 50%. The probability of the result given the $2 is 100%. The probability given the $20 is 50%. Therefore, the posterior probability is (1 x .5) / (1 x .5 + .5 x .5), or .5/.75 = 2/3.
Nifty rule.
Huh??? Did anybody follow that?
Math majors, I assume. But if you stop and think for a second it’s not too tough to wrap your head around it. I actually ran across Bayes Theorem in an intro philosophy course. I memorized it for the test, then promptly forgot it afterwards. Sounds like a good thing to know, actually.
I was a math major. Until someone can come up with an explanation that’s actually comprehensible, I’m sticking with my original answer.
bnorton
If you could be a bit more specific about what you don’t understand, I can try explain it better.
[slight hijack] I use Bayesian Estimates frequently. They are the results of applying Bayes’ Theorem to a set of results and are used in determining the endpoint of iteritive procedures. Specifically, they are used to find the slope estimators in Hierarcical Linear Modeling (among other things).
[/slight hijack]
In all probability questions there are two ways to go about solving them.
Apply appropriate rules and/or theorems and calculate results,
give up and draw out all possible combinations and count occurences.
True statisticians do both. 
Statistically yours,
Spritle
bnorton said:
I thought my answer was reasonably comprehensible, especially since I followed your lead.
Here’s your error. One of the terms of the problem was “Later still, you go to pay for something and pull a $2 bill out of your wallet.” But your case 3 is “You have $20 and $2 in your pocket and pull out a $20.” This is impossible.
Modify your last statement to read “In cases 1 and 2 you are left with a $2 bill, and only in case 4 are you left with a $20. So it’s 2 to 1 that it’s a $2.”
VarlosZ spake:
“Exactly. Around here, we assume that people aren’t idiots until it’s proven otherwise. If the initial odds were other than 50-50, that would have been a rather important piece of information to leave out, yes? It’s been my experience that, in problems such as these, proportionate (sp?) probability is assumed unless otherwise specified (or, at least, implied).”
So, you assumed that Marilyn meant that Monty’s choice of which door to show you was 50-50?
There are two possibilities: either it’s $2 or $20. Unless modified some other factor, the probability of either is 1/2. Since I didn’t mention any other factors that would modify the probability, even odds were implied (had I said that you either had $1, $2, or $20 bill, you would assume a probability for each of 1/3, given a lack of any other information).
I’m not sure to which part of the Monty Hall problem you are referring. I can think of at least two parts, neither of which apply to my rationale. Please specify. Do remember, though, that I am not implying that “50-50” is some universally assumed probability. I am saying that proportianate probability is assumed as a premise (not a conclusion), all else equal.
Why wouldn’t the probability be 50-50?
You put a $2 bill in. Given that you automatically take a $2 bill out, you’re left with what you started with, be it $2 or $20 bill.
And, since you said that there is a 50-50 probability of you starting with a $2 or $20, why wouldn’t there be a 50-50 probability of you ending up with a $2?
Of course the probability if 50/50. It’s either a $2 or it’s not. 50/50. [/sarcasm]