=probability=?

Question

say I have a wallet that contains either a $2 bill or a $20 bill ( with equal likelihood), but I don't know which one. I add a $2 bill. Later, I reach into my wallet( without looking) and remove a bill. It's a $2 bill. There's one bill remaining in the wallet. what are the chances that it's a $2 bill?

Answer 1

2/3, assuming $2 and $20 were equally likely initially.

This is an application of Bayes Law, which says that:

Prob[intial bill in wallet was $2 | first bill drawn was $2] =

(Prob[first bill drawn was $2 | intial bill in wallet was $2]*Prob[initial bill in wallet was $2]) divided by [(Prob[first bill drawn was $2 | intial bill in wallet was $2]*Prob[initial bill in wallet was $2]) + (Prob[first bill drawn was $2 | intial bill in wallet was $20]*Prob[initial bill in wallet was $20])]

= 1*.5/(1*.5 + .5*.5)

= .5/.75

= 2/3

Edit: Incidentally, if you think the answer should be 1/2, you can run this experiment a bunch of times to verify that the answer should be more. Even if you do not have $20 and three $2 bills handy, you can try the following experiment. Take two small idential pieces of paper, write $2 on the front of each and $2 on the back of one and $20 on the back of the other. Now put them both in a paper bag. Now a bunch of times (say 50 or 100) draw a paper out and look at one side. If it says $20 put it back and ignore the result (since the bill drawn was $2). If it says $2, then keep a running tally of how many times the other side says $2 and how many times the other side says $20. If you do a lot of trials, you will get nearly twice as many $2 as $20.

Answer 2

Interesting, but I don't think this is the same scenario as the Monte Hall problem.

There aren't three doors here, nor three wallets or three bills. The problem states one wallet contains either a $2 bill or a $20 bill (meaning it originally contains one bill of value being either $2 or $20). A $2 bill is added. A $2 bill is removed. "There's one bill remaining in the wallet" - further exemplifying that there was only ever two bills in the equation to begin with, the original unknown and the $2 added/removed.

Ex: Wallet Contents:
(1) $2 bill or $20 bill
+(1) $2 bill
-(1) $2 bill
= (1) $2 bill or $20 bill = A or B = Spoon or Fork = Apple or Orange = 50%

Answer 3

You have 2 bills in your wallet and add 1 to make 3 bills in your wallet. Then you remove 1 bill leaving 1 bill?

Answer 4

Super easy. Here are the possibilities for two draws:

Two possible wallet configuations are equally likely:
a) 2 , 20
b) 2 , 2

Two possible draw sequences from wallet a) are equally likely:
c) 2 then 20
or d) 20 then 2

Two possible draw sequences from wallet b) are equally likely (and they
are the same!!!!!)
e) 2 then 2
f) 2 then 2

You have eliminated choice d:
Only choices c, e and f satisfy criterion than 2 drawn first.
Only choice e and f satisfy 2 drawn second.

2 / 3 is the answer.

Answer 5

Let us take 2 cases
1. U hav a 2$ bill,
If u add another, and take out one u'll be left wid the other one.
So probability is 1
2. U hav a 20$bill
If u add a 2$ and take out the same u'll be left wid 20$ and so prob of 2$ bill is zero.

So probability of 2$ bill is (1 + 0)/2 = 0.5

Answer 6

1/2

Answer 7

Instinctively I would say 1/2 but I believe it's 2/3.

Answer 8

Is this the Monte Hall problem revisited? No, the odds are... for a 2 at 2/3.

I finally understand the Monte Hall problem, all I needed was an example with money and not goats.

Answer 9

the chance of being a 2-bill given that you drew a 2 bill is
still 0.5

Answer 10

can you give me a two dollar bill? and its a 66% chance