Bertrand's_box_paradox

This has probably been done to death, but I just re-encountered this paradox and I'm stuck on the answer 1/2, where the stated solution is 2/3
Can someone persuade me that it is indeed 2/3 and not 1/2

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

"There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

I'm stuck thinking the answer is 1/2
Can you provide me with an excel workbook demonstrating that it's 2/3

Quote: OnceDear

This has probably been done to death, but I just re-encountered this paradox and I'm stuck on the answer 1/2, where the stated solution is 2/3
Can someone persuade me that it is indeed 2/3 and not 1/2

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

"There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

I'm stuck thinking the answer is 1/2
Can you provide me with an excel workbook demonstrating that it's 2/3
link to original post

Sounds like the same math from the Monty Hall problem.

Try putting it a little differently for a better mental picture: Each box has a million coins. One box has a million gold coins, another box has a million silver coins, and the third box has 1 gold coin and 999999 silver coins.

If you pick a coin at random from a box at random, and it's gold, what are the chances you picked from the box with just one gold coin out of a million and got the gold coin? One in a million, right? Could be, but there's a 99.9999% chance you picked from the box that's all gold.

Quote: OnceDear

This has probably been done to death, but I just re-encountered this paradox and I'm stuck on the answer 1/2, where the stated solution is 2/3
Can someone persuade me that it is indeed 2/3 and not 1/2

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

"There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

I'm stuck thinking the answer is 1/2
Can you provide me with an excel workbook demonstrating that it's 2/3
link to original post

Well. I've created said workbook myself and demonstrated that it is indeed approaching 2/3

Quote: OnceDear

This has probably been done to death, but I just re-encountered this paradox and I'm stuck on the answer 1/2, where the stated solution is 2/3
Can someone persuade me that it is indeed 2/3 and not 1/2

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

"There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

I'm stuck thinking the answer is 1/2
Can you provide me with an excel workbook demonstrating that it's 2/3
link to original post

I’m no excel workbook guy. But this is an easy one. If I just tell you given those boxes he blindly picked a coin, and it was gold, you would know there was a 2/3 chance he picked from the box with two gold coins, right? Since 2 out of the 3 gold coins were in that box. Sooooo….. 2 out of 3 times the other coin is gold.

Similar type question. (Assume boys and girls are 50% of the population, and ignore identical twins). A couple has two children. The only thing you know for certain is that one of them is a boy. What are the chances the other is a boy?

The quick (incorrect) response is 1/2.
4 families
BB
BG
GB
GG
Since the only fact we have is GG is eliminated, there are 2 chances for a G as the second sibling is a G, while only 1 where it’s a B. Hence the correct answer is 1/3.

You (instinctively?) interpret the question as ‘the first child is a boy, while it specifically just says ‘one of them is a boy’.

Quote: OnceDear

Can you provide me with an excel workbook demonstrating that it's 2/3
link to original post

Here you go.

The coin you observe could be the gold coin in B2, B3 or C2. For each, look at the other coin in the same box. In 2 out of 3 cases it will also be gold.

There are three boxes, with one box containing only silver coins. That box is eliminated, leaving two boxes with four coins. (Three gold and one silver. )You remove one gold coin and you are left with three coins, two of which are gold. Choose a coin at random and the odds of it being gold are two out of three.

There was a time in this forum when a topic like this would result in someone coming in and contradicting the correct answer. Then the stubborn SOB would argue about it for months. You could count on it. I never knew if they were really that thick or just enjoyed keeping it going. I apologize if I have insulted someone who is still around.

Here’s another way to look at the Bertrand’s Box problem:

Have a deck of only three cards. One of the cards is yellow on both sides; another is black on both sides; and the third is yellow on one side and black on the other.

Without looking at the deck, shuffle the cards. Besides shuffling the usual way that cards are shuffled, be sure to shuffle the backs of the cards with the fronts.

Now look at the top of the deck and ask yourself, “What is the probability that both sides of the top card are the same color?”

Quote: JimRockford

There was a time in this forum when a topic like this would result in someone coming in and contradicting the correct answer. Then the stubborn SOB would argue about it for months. You could count on it. I never knew if they were really that thick or just enjoyed keeping it going. I apologize if I have insulted someone who is still around.
link to original post

I imported a question about the blind man and a pair of sixes to my comic forum. The debate there turned nasty, with two guys threatening to meet up in a comic book store parking lot.

Quote: billryan

Quote: JimRockford

There was a time in this forum when a topic like this would result in someone coming in and contradicting the correct answer. Then the stubborn SOB would argue about it for months. You could count on it. I never knew if they were really that thick or just enjoyed keeping it going. I apologize if I have insulted someone who is still around.
link to original post

I imported a question about the blind man and a pair of sixes to my comic forum. The debate there turned nasty, with two guys threatening to meet up in a comic book store parking lot.
link to original post

If I owned the shop I would have invited them to come and do that, on the condition that they wear superhero costumes and sign actor releases.

Quote: billryan

There are three boxes, with one box containing only silver coins. That box is eliminated, leaving two boxes with four coins. (Three gold and one silver. )You remove one gold coin and you are left with three coins, two of which are gold. Choose a coin at random and the odds of it being gold are two out of three.
link to original post

I modified my workbook to eliminate the box with two silver coins.
That box was a red herring.
Picking randomly from only two boxes still led to 2/3 probability.

All these names, Bertrand, Monty Hall, and etc., are very hard to memorize. I’m a gambler, so I tend to summarize all these brain teasers into one. Here is a problem in terms of blackjack card counting:

There are three single-decks on a table for a card counter to play two hands of blackjack. If the counter wins the first hand, should he continue playing the same deck to win the second hand? Or, switch to a new deck?

Quote: aceside

All these names, Bertrand, Monty Hall, and etc., are very hard to memorize. I’m a gambler, so I tend to summarize all these brain teasers into one. Here is a problem in terms of blackjack card counting:

There are three single-decks on a table for a card counter to play two hands of blackjack. If the counter wins the first hand, should he continue playing the same deck to win the second hand? Or, switch to a new deck?
link to original post

Without any other information- switch decks.

But not for reasons having anything to do with Bertrand's Box or Monty Hall.

Quote: OnceDear

This has probably been done to death, but I just re-encountered this paradox and I'm stuck on the answer 1/2, where the stated solution is 2/3
Can someone persuade me that it is indeed 2/3 and not 1/2

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

"There are three boxes:

a box containing two gold coins,
a box containing two silver coins,
a box containing one gold coin and one silver coin.

A coin withdrawn at random from one of the three boxes happens to be a gold. What is the probability the other coin from the same box will also be a gold coin?

I'm stuck thinking the answer is 1/2
Can you provide me with an excel workbook demonstrating that it's 2/3
link to original post

I don't have an "Excel Workbook" handy, especially as I use LibreCalc, but here's an explanation:

Number the gold coins 1, 2, 3, and the silver coins 4, 5, 6.
One box has coins 1 and 2 (both gold)
One box has coins 5 and 6 (both silver)
One box has coins 3 and 4 (one of each)

You selected a coin at random, and it was gold, so you selected one of 1, 2, or 3
There was a 1/3 chance of selecting 1; the other coin is 2, which is also gold
There was a 1/3 chance of selecting 2; the other coin is 1, which is also gold
There was a 1/3 chance of selecting 3; the other coin is 4, which is silver
The probability that the other coin was gold = 1/3 + 1/3 = 2/3.