The "Deal or no Deal" dillemma

[Jay-qu]

 

The key to this is something you alluded to, only you didn't give it enough emphasis. Your own eliminations always change the probability of what your card is. The host's eliminations never do.

 

Thanks Craig, I like the maths better - for some reason I cant get my head around the above, it just doesnt seem logical to me. I think it would be more correct worded in another manner, random eliminations change probability while targeted ones do not?

[Simon]

I think it would be more correct worded in another manner, random eliminations change probability while targeted ones do not?

 

For "targeted", we could say "non-random", but otherwise yes. The puzzle made clear that the host's eliminations were indeed "targeted" - i.e blacks only.

 

However it would be more accurate to say:

 

"Random eliminations uniformly change the current probability of every item remaining, including those previously selected. Targeted eliminations uniformly change the current probability of every item remaining, except those previously selected".

 

You selected a card, then randomly revealed 9 red cards. As a result, every remaining card uniformly increased it's chance of being black to 4/5, including your own.

 

The host then used his knowledge to target 11 black cards. As a result, every remaining card uniformly decreased it's chance of being black to 1/15, except your own. Your card's probability was preserved at 4/5.

 

If the above sequence of eliminations had occured, but you had not made a prior selection, then it wouldn't matter whether they were random or targeted. All 4 remaining cards would have uniformly changed their odds of being black - up to 4/5, then down to 1/4.

 

Simon

[Qfwfq]

Thanks Craig, I like the maths better - for some reason I cant get my head around the above, it just doesnt seem logical to me.

Actually, Craig's math could be simplified and doesn't give a reason for the mysterious distinction.

 

A: After having removed 9 cards, all of these red, there are 15 remaining cards, 3 red and 12 black. Dividing these three numbers by 3, we get 1/5 for red and 4/5 for black.

 

Now the real mystery is:

 

B: After the host reveals 11 more cards, all of them black, there are 4 remaining cards, 3 red and 1 black. This would give 3/4 for red and 1/4 for black.

 

Why then is this not the best way for the probability to be judged? Why is it OK in case A and not in case B?

 

I think it would be more correct worded in another manner, random eliminations change probability while targeted ones do not?

There are different ways to put it but:

Your own eliminations always change the probability of what your card is. The host's eliminations never do.

 

Once you beleive this, everything else follows.

What isn't obvious is why one should believe that. The important thing is to give the reason.

 

First of all, I was about to reply "But it's the other way around!!!!!!!" and then I saw what Simon meant, it depends on what "change" means. He meant compared to before removing the cards and I meant it as compared to case B above, a purely frequentist judgement that doesn't partly exploit the information known to the host. Actually it's a good way of seeing it and makes it easier.

 

Revealing a card randomly changes the probability by changing what's left on the table. If you were sure the host had revealed cards just as randomly as you, it just happened that altogether 9 red and 11 black cards were revealed, you should consider the above point B just as valid. Why then is it different if you're sure the host knew which cards were which? As I mentioned, you can partially exploit information known to the host. One way to see it is by applying Craig's math (conditional probability) to when the host reveals 11 cards, but with a difference: For each of these, the probability of it being black is exactly 1 because the host knows which ones are black. That's what Simon puts as "doesn't change".

[Simon]

To really confound the mystery and put it in the spotlight, consider the following example.

 

Imagine two 24 card games, side by side - A and B.

 

You have randomly chosen a card from each.

 

Suppose the following:

 

a) The host does not know the layout of Game A

:xparty: The host does know the layout of Game B.

c) Whatever colour is randomly revealed in Game A is knowingly revealed by the host in Game B.

 

So at every stage - where the same colour card is removed - the number of reds and blacks remaining in both games exactly mirror each other. Whatever the host reveals in Game B depends on what you randomly reveal in Game A.

 

Therefore, before any round, the next colour to be revealed in Game B is not pre-determined. It has a varying probability - identical to that of Game A. In turn, this probability is based on the how many of each colour remain - which is always the same in both games.

 

One by one, 9 red cards were randomly revealed in Game A.

One by one, 9 red cards were knowingly revealed in Game B to match.

 

The host has done his job and leaves the room.

 

In both games, there are now 3 red cards and 12 black cards left.

 

In Game A, the probability that your card is black progressed from 50% to 80%. The same is true of each the remaining cards.

 

Under these circumstances, is Game B different?

 

Simon

[Qfwfq]

Yes. They're not the same.

[Jay-qu]

 

There are different ways to put it but:What isn't obvious is why one should believe that. The important thing is to give the reason.

 

First of all, I was about to reply "But it's the other way around!!!!!!!" and then I saw what Simon meant, it depends on what "change" means. He meant compared to before removing the cards and I meant it as compared to case B above, a purely frequentist judgement that doesn't partly exploit the information known to the host. Actually it's a good way of seeing it and makes it easier.

 

Revealing a card randomly changes the probability by changing what's left on the table. If you were sure the host had revealed cards just as randomly as you, it just happened that altogether 9 red and 11 black cards were revealed, you should consider the above point B just as valid. Why then is it different if you're sure the host knew which cards were which? As I mentioned, you can partially exploit information known to the host. One way to see it is by applying Craig's math (conditional probability) to when the host reveals 11 cards, but with a difference: For each of these, the probability of it being black is exactly 1 because the host knows which ones are black. That's what Simon puts as "doesn't change".

 

You put into words what I suspected but didnt fully understand the subtleties of