## Poll

 1/4 1 vote (1.72%) 1/3 No votes (0%) 1/2 14 votes (24.13%) 2/3 39 votes (67.24%) 3/4 No votes (0%) Not information information given. 1 vote (1.72%) I believe in the "other math." No votes (0%) I'm a bigot 1 vote (1.72%) I'm a bully No votes (0%) I miss Vermenti 2 votes (3.44%)

58 members have voted

pmpk
Joined: May 6, 2014
• Posts: 2
May 6th, 2014 at 11:20:11 PM permalink

It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.

It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.
sodawater
Joined: May 14, 2012
• Posts: 3321
May 6th, 2014 at 11:33:26 PM permalink
Quote: pmpk

It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.

It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.

Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.

Here's how. There are 4 ways you could pull a coin out and look at one side.

1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
4. Observe top is black, bottom is white.

You pull out a coin and look at the top and see it is white. This leaves 3 options:

1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.

Two of these three are the all-white coin.

Think of it this way. If you reach in and pull out a coin without looking at it, it's 1/2. But now that you look at it and see that the top is not black, that hurts the chances you have the black and white coin. Sometimes when you have the black and white coin, the top will be black. Since it is not, you have eliminated those times.

If you're still not getting it, think of this:

Say it's the same set up, but you pull out one coin and the top is black. Now the chances of picking the all-white coin have dropped from 50 percent (unseen) all the way down to 0 percent (seen.) Seeing extra information changes probability
.

One final example. You buy a Mega Millions ticket and wait till the drawing to look at your ticket. Before you look at your ticket, the chance you have hit the jackpot is 1 in 259 million. Now you pick up the ticket with your thumb covering the Mega ball and you see with excitement that you've matched the first 5 numbers. With your thumb over the Mega Ball, your probability has increased from 1 in 259 million to 1 in 15 for the jackpot. Note that you have either won the jackpot or you haven't, but it's not 50-50.
pmpk
Joined: May 6, 2014
• Posts: 2
May 6th, 2014 at 11:49:17 PM permalink
Quote: sodawater

Quote: pmpk

It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.

It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.

Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.

Thanks for the detailed explanation! I stand corrected.
AxiomOfChoice
Joined: Sep 12, 2012
• Posts: 5761
May 7th, 2014 at 1:33:31 AM permalink
Quote: sodawater

I haven't read this thread at all, but are there really seven pages of debate over the most well known conditional probability problem of all time?

Yes
RS
Joined: Feb 11, 2014
• Posts: 8623
May 7th, 2014 at 1:51:21 AM permalink
Looks like part of it has to do with interpretation. I didn't think that "observing black" was even an option, as nothing about observing black was in the OP, as if when the black/white coin was chosen, the white side had to be the observed side.
BleedingChipsSlowly
Joined: Jul 9, 2010
• Posts: 968
May 7th, 2014 at 4:11:11 AM permalink
I think the crux of the incorrect thinking (mine included) is that if you are looking at the white side of one particular coin, of which there are only two, there is a 1/2 chance the other side will be white. However, when you consider multiple samples you will be working with the white/white coin twice as often because the question is predicated on observing a white side first. That is, the possibility of selecting the white/black coin and observing the black side first, a very real possibility, is not included in the sample population. As I said in an earlier post, it took me a while to see that 2/3 was the correct probability and even longer to realize that holds true both before and after a particular coin is drawn.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
JoePloppy
Joined: May 2, 2014
• Posts: 82
May 7th, 2014 at 8:04:31 AM permalink
Quote: BleedingChipsSlowly

... In that case I'm still willing to go head-to-head with any takers - provided I get to wager the \$5 the flip side being black against a \$3 wager it is white.

Edit : What about \$4000 winner on black , \$3000 winner on white (me)? 100 random draws?
2/3
BleedingChipsSlowly
Joined: Jul 9, 2010
• Posts: 968
May 7th, 2014 at 4:20:13 PM permalink
Did you note the "Slowly" part of my handle? Your proposed bet would be a few notches past slowly, at least for my bankroll.
“You don’t bring a bone saw to a negotiation.” - Robert Jordan, former U.S. ambassador to Saudi Arabia
Sonuvabish
Joined: Feb 5, 2014
• Posts: 1342
May 12th, 2014 at 11:48:43 AM permalink
Quote: sodawater

Quote: pmpk

It has to be 1/2. There are two coins. The color configurations are just the identifying information for each coin, and the question is asking for the probability that you chose one coin over the other.

It might as well be two otherwise identical coins of two different years (year A and year B for example). If you select one at random and happen to see its tails side first, there's still a 50% chance that the heads side will show that the coin is year A.

Sorry, this is totally wrong. Looking at one of the coins provides information you can't ignore. The answer is 2/3.

Here's how. There are 4 ways you could pull a coin out and look at one side.

1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.
4. Observe top is black, bottom is white.

You pull out a coin and look at the top and see it is white. This leaves 3 options:

1. Observe top is white, bottom is white
2. Observe top is white, bottom is white
3. Observe top is white, bottom is black.

Two of these three are the all-white coin.

Think of it this way. If you reach in and pull out a coin without looking at it, it's 1/2. But now that you look at it and see that the top is not black, that hurts the chances you have the black and white coin. Sometimes when you have the black and white coin, the top will be black. Since it is not, you have eliminated those times.

If you're still not getting it, think of this:

Say it's the same set up, but you pull out one coin and the top is black. Now the chances of picking the all-white coin have dropped from 50 percent (unseen) all the way down to 0 percent (seen.) Seeing extra information changes probability
.

One final example. You buy a Mega Millions ticket and wait till the drawing to look at your ticket. Before you look at your ticket, the chance you have hit the jackpot is 1 in 259 million. Now you pick up the ticket with your thumb covering the Mega ball and you see with excitement that you've matched the first 5 numbers. With your thumb over the Mega Ball, your probability has increased from 1 in 259 million to 1 in 15 for the jackpot. Note that you have either won the jackpot or you haven't, but it's not 50-50.

The coin is always white when you pull it out, 100% of the time. It is immaterial which white side comes out on the all-white coin because you are not observing 2 coins. That means there is a 50% the other side is white. Re-read the question. 2/3 is the probability that one of the remaining 3 faces is white; you either have the all-white coin, or you don't.

2/3 is simply the ratio of white faces to total unseen faces. This is the initial thought that everyone should have; everyone knows the chance of randomly selecting a white face, excluding the given, is 2/3--that was learned in elementary. If this is actually the correct answer, it is a trick question with poor wording. I would not call that controversial.
JoePloppy
Joined: May 2, 2014
• Posts: 82
May 15th, 2014 at 8:01:29 AM permalink
Quote: Sonuvabish

The coin is always white when you pull it out, 100% of the time. It is immaterial which white side comes out on the all-white coin because you are not observing 2 coins. That means there is a 50% the other side is white. Re-read the question. 2/3 is the probability that one of the remaining 3 faces is white; you either have the all-white coin, or you don't.

2/3 is simply the ratio of white faces to total unseen faces. This is the initial thought that everyone should have; everyone knows the chance of randomly selecting a white face, excluding the given, is 2/3--that was learned in elementary. If this is actually the correct answer, it is a trick question with poor wording. I would not call that controversial.

False. Don't blame the question.

Quote: Wizard

You randomly draw one coin, and observe one side only, which is white.
What is the probability the other side of that coin is white?

2/3

I was trying out some of the math questions and found this today. The 2 coin problem and solution on the Wizards Math Problem Site. http://mathproblems.info/?page_id=188#s16
2/3