A Fistful of Brain Teasers

Some entrenchment beginning here! Nothing has swayed me from my original assessment!

hungryhalibut posted:

fatcat posted:

Don Atkinson posted:

hungryhalibut posted:

The probability of the second coin being also gold is 2/3, therefore it being silver must be 1/3. It’s the opposite of the standard solution to the Bertrand’s box paradox, which is rather like that ‘do I switch to the other cup’ Monty Hall problem. My annoyingly good at maths son tries to explain it to us, invariably after we have had too much wine. 

HH,

What you have stated in the first two sentences above is absolutely correct.

I will take in good faith that what you have said in your last sentence is also absolutely correct.

Once the first gold coin is revealed. And it has been revealed. The chance of revealing a second gold is 1/2.

You might think that, but it’s not. It’s 2/3. Therefore the probability of it being silver is 1/3, as the probabilities sum to 1. 

But the question is not whether the other lid on the same box will reveal silver, so fatcat is right in my view! 

Just google Bertand’s Box paradox. 

Mulberry posted:

Innocent Bystander posted:

[...] the question is not what is the probability of the second lid of the same box revealing silver [...]  but what is the probability that this half opened box is the one with the other half containing a silver coin.

I think these are actually the same question. Only if the second lid reveals silver, the half-opened box is the one with the other half containing silver (to reassamble your post).

No, they are not the same question. You are considering the remaining three lids and the chance of one being silver, whereas the lids are tied in pairs in the boxes, and as the part opened box is not the silver-silver one then that part opened one under consideration has to be either the silver-gold or the gold-gold, = one of two.   

hungryhalibut posted:

Just google Bertand’s Box paradox. 

As my response above - Bertrand’s box is considering what the next one out of the same box will be (or in the adjoining compartment). But that is not the question!

There are six lids, yes ?

TOBYJUG posted:

There are six lids, yes ?

Yes, but the one opened is one of only three having a gold coin underneath.

Innocent Bystander posted:

[...] as the part opened box is not the silver-silver one then that part opened one under consideration has to be either the silver-gold or the gold-gold, = one of two.   

Lets stay with these two remaining boxes. We see an open lid with a gold coin underneath. Isn't that coin twice as likely to be in the g/g box, as there are two of them in that box?

hungryhalibut posted:

fatcat posted:

Don Atkinson posted:

hungryhalibut posted:

The probability of the second coin being also gold is 2/3, therefore it being silver must be 1/3. It’s the opposite of the standard solution to the Bertrand’s box paradox, which is rather like that ‘do I switch to the other cup’ Monty Hall problem. My annoyingly good at maths son tries to explain it to us, invariably after we have had too much wine. 

HH,

What you have stated in the first two sentences above is absolutely correct.

I will take in good faith that what you have said in your last sentence is also absolutely correct.

Once the first gold coin is revealed. And it has been revealed. The chance of revealing a second gold is 1/2.

You might think that, but it’s not. It’s 2/3. Therefore the probability of it being silver is 1/3, as the probabilities sum to 1. 

Are you sure it's 2/3.

why.

Mulberry posted:

Innocent Bystander posted:

[...] as the part opened box is not the silver-silver one then that part opened one under consideration has to be either the silver-gold or the gold-gold, = one of two.   

Lets stay with these two remaining boxes. We see an open lid with a gold coin underneath. Isn't that coin twice as likely to be in the g/g box, as there are two of them in that box?

Let’s say yes for this moment: a gold coin is twice as likely to be in the g-g box as the g-s. So the second coin being gold is twice as likely to be gold as silver. So the probability of the second coin being gold is 2:1. So 2:1 against the box being g-s?

Suppose there are three cards:

• A black card that is black on both sides,
• A white card that is white on both sides, and
• A mixed card that is black on one side and white on the other.

All the cards are placed into a hat and one is pulled at random and placed on a table. The side facing up is black. What are the odds that the other side is also black?

N.B be very, very careful when reading this teaser. And it might be a good idea to re-read the Gold Coin teaser as well, before coming to any conclusions.

IMHO these two teasers are about as carefully worded as any statement by Rees-Mogg, Boris, the PM or virtually any politician who has spoken with regard to Brexit in the past two weeks !!!!

2/3. It’s the opposite of the way you phrased the box question.

Noel Edmunds is joining "I'm a celebrity, get me out of here" and has vowed that if he wins he will leave presenting on the telly.

What are the odds of him leaving the jungle ? Or leaving for good ?

That depends on whether he is sharing a tent with Mr Blobby. 

1/2 surely? We can discard the white card, so you either have the black card or the mixed card. It’s 50/50.

best

David

hungryhalibut posted:

That depends on whether he is sharing a tent with Mr Blobby. 

Why can’t all the “celebrities” be left wherever it is they are, and make the world better for everyone else? Starting with the presenters.  (Who celebrates them, anyway?)

hungryhalibut posted:

2/3. It’s the opposite of the way you phrased the box question.

Where does your 2/3 come from in the coin problem.

Surely the card problem is also 1/2.

It’s the solution to the Bertrand’s Box Paradox, as I said previously. If you look it up, you’ll get a better explanation than I can give. It’s counterintuitive, hence it’s a paradox. 

The question about odds is asked after the card is revealed as having one black side facing up. So it’s the odds of having two black sides after the white card has not been picked. So 1/2.

If the question was what is the chance of picking a card of two black sides, then it would be 1/3.

But I don’t see how it could be 2/3....

best

David

hungryhalibut posted:

It’s the solution to the Bertrand’s Box Paradox, as I said previously. If you look it up, you’ll get a better explanation than I can give. It’s counterintuitive, hence it’s a paradox. 

Hmm. Interesting.

I'd say that a very convoluted method of getting the answer wrong.

Innocent Bystander posted:

Mulberry posted:

Innocent Bystander posted:

[...] as the part opened box is not the silver-silver one then that part opened one under consideration has to be either the silver-gold or the gold-gold, = one of two.   

Lets stay with these two remaining boxes. We see an open lid with a gold coin underneath. Isn't that coin twice as likely to be in the g/g box, as there are two of them in that box?

Let’s say yes for this moment: a gold coin is twice as likely to be in the g-g box as the g-s. So the second coin being gold is twice as likely to be gold as silver. So the probability of the second coin being gold is 2:1. So 2:1 against the box being g-s?

Hi IB,

unless something is lost in translation (non-native speaker here): yes 

Mulberry posted:

Innocent Bystander posted:

Mulberry posted:

Innocent Bystander posted:

[...] as the part opened box is not the silver-silver one then that part opened one under consideration has to be either the silver-gold or the gold-gold, = one of two.   

Lets stay with these two remaining boxes. We see an open lid with a gold coin underneath. Isn't that coin twice as likely to be in the g/g box, as there are two of them in that box?

Let’s say yes for this moment: a gold coin is twice as likely to be in the g-g box as the g-s. So the second coin being gold is twice as likely to be gold as silver. So the probability of the second coin being gold is 2:1. So 2:1 against the box being g-s?

Hi IB,

unless something is lost in translation (non-native speaker here): yes 

OK, I can see that argument. 2:1 is 2 in 3, so the converse for it being silver, 1 in 3. However that starts with consideration before choosing the first one (and I thought that was the point of the Bertrand argument, the starting point being before the first choice). Inthis case the starting point effectively does not involve that initial chance of choosing gold, but starts from bering told there is a g-g, a g-s and an s-s and that you have one that is g-?. In this case it simply has to be either g-g or g-s, as it can’t be the s-s, hence 1:1 (or 1/2).

hungryhalibut posted:

The probability of the second coin being also gold is 2/3, therefore it being silver must be 1/3. It’s the opposite of the standard solution to the Bertrand’s box paradox, which is rather like that ‘do I switch to the other cup’ Monty Hall problem. My annoyingly good at maths son tries to explain it to us, invariably after we have had too much wine. 

Ah ! Monty Hall. I know it's been done a few times on the Naim Forum but...........

............Just to remind you of the "paradox" ......

You are the finalist in a quiz show hosted by Monty Hall

Your Prize is a brand new car but...... it is hidden behind one of three doors Labeled A, B and C

A goat is hidden behind each of the other two doors

(To keep it simple, and in line with Monty's show, he knows which door the car lies behind)

You are asked to choose a door, in the hope of winning the car. (who really would prefer a goat ?)

You choose (say) door B

Monty opens one of the other doors ie A or C and, as always, reveals a goat.

He then invites you to either stick with Door B or change your mind and choose the other, as yet, unopened door.

Is the probability of winning the car greater if you stick, or greater if you change your mind, or is it the same either way ?

Now, I know that you all know the correct answer, (including IB this time around ) so the problem isn't so much "what's the best strategy" but...........

..........who can explain it most clearly and concisely ?

I'll invite HH to be the judge, but only after he's had too much wine

The explanations out there for the Monty Hall conundrum suggest that one's chance of winning increases from one in three to two in three if one switches from the original choice to the other door remaining.

This seems to be based on the premise that by sticking with one's original choice, when the odds were one in three, the odds have not improved, and that by changing doors, one's odds shorten to two in three.

Seems to me that this overlooks the fact that there are now only two doors from which to choose, so that the often dismissed view that the odds are now 50 / 50, is, in fact correct.

The popular explanation overlooks this, and puts forward a theory based on the original choice of three doors ............. completely overlooking the fact that the choice has started all over again, now that the first door has been removed, i.e. 50 /50?

Exactly Dave, which is why the answer to the three card problem isn’t as Don explains, because he hasn’t read carefully enough the question he actually asked!

best

David

Oh dear.........

Let’s stick with Monty Hall for the moment.

The odds on winning the car are most definitely 2/3 if you change your mind, ie you accept Monty’s offer to open the other door. If you stick with your original door, the odds on winning are 1/3.

........yes, I know it isn’t blatantly obvious, and I also know that many eminent mathematicians disputed these probabilities when first presented, but..........