DFCarlson
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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
OK, so this is similar to the original Monty Hall game in that there is only one host and that host shows one card, except there are now four cards, two Jacks and two Queens for you to choose among. And is it the case that you as contestant will win if you identify one of the Queens? And is the question whether or not to swap the card you originally chose for one of the two unrevealed cards?

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
Thanks, Simon! I believe I understand what you're saying.
The difference is in recognizing that:
1) The odds that you have a Queen and the game survives + the odds that you have a Jack and the game survives + the odds that the game does not survive = 1
compared to:
2) the odds that your card is the Queen + the odds that your card is a Jack = 1
This is why in my analysis the chance of a cancelled game appeared to increase as the number of hosts got larger, whereas in reality the odds of a cancelled game get smaller.
I'll do the other analyses on my own and I'll look for your new Monty Hall problem, although I've already been turned topsy turvy more than once!

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
Here is my attempt at the proof applied to one game with a series of knowing hosts. I

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
I believe that you have convinced me. It looks like the number of ignorant hosts is immaterial, and it's the number of hosts who know where the Queen is that determine the final odds, as long as the game is not cancelled.
Each host builds on the odds existing at the time that that particular host appears, and since the resultant odds from an ignorant host are 50  50, the net odds remain unchanged by his actions, as long as the game is not cancelled.
However, let me try to work out the odds for a few other combinations of ignorant and knowing hosts to see if my calculations make sense.
For instance,
1) a simple succession of knowing hosts, one after the other;
2) two ignorant hosts followed by a knowing host, followed by two ignorant hosts, and so on;
3) two knowing hosts followed by an ignorant host, ..., and so on.
It may take me a bit of time, let's see what I can come up with. Give me a couple of days.
Very interesting!

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
OK, I can accept, if I cannot prove, that as long as one host knows where the Queen is, your original 2/3 advantage is restored to 2/3 after a succession of ignorant hosts who do not cancel the game. It's like the knowing host's choice removes the possibility of a cancelled game and your odds revert to the odds in the original game where the knowing host was the only player.
But it seems to me that if a game has not been cancelled by two knowing hosts one after the other (regardless of how many random ones there were) that the odds cannot rise above the original odds set by your original choice of a card. Any knowing host wll make the same choice by definition. It seems to me that every knowing host's action simply negates (or affirms) the actions preceding him and the odds that you have a Jack remain at 2/3.
The idea that If there were 10 knowing hosts in succession, the odds would be 99.9% that you have a Jack, seems fantastic since your original choice of any card set the odds that you have a Jack at 2/3. Repitition of the same event cannot raise the odds!?
What if it was the same knowing host? What if there was an alternating sequence of ignorant and knowing hosts, none of whom cancel the game, ending up with a knowing host? If the final host is a knowing host, no matter how many there are, the odds of your having a Jack still must be the original odds of 2/3 set by your original choice of card!

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
So what happens to the probability of the cancelled game(s)? It doesn't seem reasonable that they could simply be scrapped!

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
OK. In both instances there is the possibility of a cancelled gane so the same analysis applies as in the case of the ignorant host and the odds are 50/50.

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
Firstly, you said I changed the rules.
I said, either:
1) You have the Jack, I have the Queen.
2) You have the Queen, I have the Jack.
You said, either:
1) The remaining card not turned face up is the Queen; OR
2) I have the Queen
Sorry, I didn't mean to imply that you changed the rules with regard to these two statements. I restated the situation in a different way, but I agree the two sets of statement say the same thing. I was trying to emphasize the location of the Queen. My statement about not playing by the same assumptions was related to playing 3 games simultaneously as compared to playing only a single game (but you address this later).
Correct me if I'm wrong but I think your postion is this: when an event happens, all that matters is the event itself. You believe that whether the card was removed by design or chance is of no consequence.
My position is: this is not true. The host's state of knowledge in the Monty Hall game is an indespensable factor that determines the odds and cannot be ignored.
A clear statement of our conundrum!
Whatever odds apply to all random host games must also apply to one. The first time you get an uncancelled game, whether it occurs immediately or not, the odds must be identical. If you wanted to test it statistically, you would then play until you got the next one. And the next. If you took a sample of 1000 uncancelled random host games, the pattern will be clear. You should find that in around half of them you chose a Jack and in the other half you chose the Queen. This bears out the 50/50 odds. In the cancelled games, of course, you will have chosen a Jack. Those are the missing games that would have given you your swapping advantage, if the host had not revealed the Queen.
Now that statement makes it a lot clearer!
Now if you concede that this is what would be born out statistically, you can't then say the odds don't apply to each individual game. There can be no exceptions.
Let's confine ourselves to the original setup. The key here is distinguishing the difference between the same event occuring with both the knowing host and random host. In each case the same event is an uncancelled game. But the probability of having achieved this is different in both cases.
When either host turns over a card, you seem to be faced with an identical situation. After all, an uncancelled game has definitely occured in both cases. However, before the random host removed the card, the probabilty of getting an uncancelled game was less than 100%. This must be taken into account. Basically you're assessing the probabilty of an event, after the fact, of how we got to where we are.
This time I'll offer a more simple proof along these lines and keep to the original format, using a Queen and two Jacks.
You have chosen a card. The host knows where the Queen is and has eliminated a Jack.
Because the host

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
I'll start by relabelling the cards  one Jack, one Queen, one King.
OK
The rules are exactly the same. Whoever ends up with the Queen wins. You randomly choose a card. I then randomly choose a card. The other card is eliminated. If the eliminated card turns out to be the Queen, the game will never be played again.
I don't think that's the same rulebook I started with. The rules that I was going by were that the person who turns up one of the cards of the remaining two (the game host) has no stake in the game. His function, whether he knows where the Queen is or not, is simply to flip over one of the remaining two cards so that I can see what it is. After he turns one of the cards face up I have the choice of switching from my original choice of card to the remaining card which is face down. The question is, do I improve my chances of finding the Queen if I switch?
If he is ignorant of the location of the Queen and turns it over then the game is nullified and we start over again.
If I am the contestant, and I see that the card turned face up is the King, it makes no difference who turned that card face up or what he knows or doesn't know. What I see is the same display of cards whether the person who turned the card over knew where the Queen was or not.
As it happens, the card I eliminated (turned face up) was the King. There are now two possibilites:
1) You have the Jack, I have the Queen.
2) You have the Queen, I have the Jack.
I would describe the situation at this point in the game as:
1) The remaining card not turned face up is the Queen; OR
2) I have the Queen
You might say that it was always more likely that the Queen would be among the two cards you didn't select. True enough, but this can also be said of the Jack.
It always was and is still more likely. Flipping a card over does not change this, which was set when I originally chose one of the three cards. It is also true of the Jack, but the Jack is not the winner and I don't care what the probability is that the Jack is one of the two cards.
When I chose one of the original 3 cards, the chance that the card I chose was the Queen was 1 in 3 and the chance that the Queen was in the group of two cards that I did not choose was 2 in 3. When the breeze blowing through the room flips one of the other 2 cards face up and it is not the Queen, then the chance that the other card is the Queen is still 2 in 3.
Since both statements carry equal weight, the odds must be 50/50.
But IMHO we're not talking about the same thing. The game where the Queen is turned face up does not enter into the calculations, since if this happens the game is null and void, it is as if it was never played and the game is restarted.
Fascinating discussion! 8)
(BTW I had a devil of a time getting on this website earlier today. I'm located in Thailand so I'm what? 7 hours ahead of you (in London?)? Do you know if there was a problem with the site? I had no trouble accessing my other frequently visited sites.)

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
I understand what you're saying.
But I am talking about playing only one game, not three.
And if a Jack is turned up in that game the odds are still 2 in 3 that the remaining card is the Queen. It doesn't make any difference how the Jack ends up face up.
For any single game, do you not agree with:
2b2) He turns over the Jack, in which case the existing situation is identical to 2a).

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
Very interesting! Wait a minute. I thought about this all last night and I have become confused. (Easy for me!).
IMO we have to be careful what scenario we're talking about. I want to simplify the discussion. (I know, I'm the one who made it more complicated!) Let's go back to the original 3card game, with the proviso that if the location of the Queen is revealed by the person who does not know where the Queen is located, then that game is null and void and the game is restarted.
First, you pick a card. What are the possibilities? Clearly,
1) you have the Queen, in which case the other two cards are Jacks, OR
2) you do not have the Queen, in which case one of the other two cards is the Queen and the other is a Jack.
The chance that you will pick the Queen is 1 in 3 and the chance that the Queen will be one of the other two cards is 2 in 3. OK so far...?
In case 1) it doesn't matter which card of the remaining two cards gets turned over, whether the person knows where the Queen is or is ignorant of the Queen's location. It will be a Jack.
In case 2a) if the person knows where the Queen is he will turn over the Jack, thereby not revealing the location of the Queen;
In case 2b) if the person is ignorant of the location of the Queen, there are two possibilities:
2b1) He turns over the Queen and the game is nullified and restarted. It's as if this game never existed.
2b2) He turns over the Jack, in which case the existing situation is identical to 2a).
In no way do the actions of either person change the original odds. So, in my opinion, in this scenario, if you see a Jack turned over you switch.
Now  is this the scenario that you had in mind for your analysis? If so, how can you argue that the odds in 2a) are 2 in 3 that the remaining card is the Queen and in 2b2) the odds have changed to 1 in 2?
If not, how does your scenario differ from this one? Let's establish clearly that we're talking about the same game and the same rules.

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
Your explanation is very clear.
Actually, I think you have got through this thick skull  to a point! Bear with me...
(Rhetorical) What happens if the person who knows where the Queen is turns up 98 cards of 99 and all are Jacks? You switch, because his choice of the 98 cards to turn face up is not random. This is the standard Monty Hall conundrum with 100 cards.
Now, what happens if the person who does NOT know where the Queen is turns up 98 Jacks of 99 cards on his first try? Your observation of events is identical in both cases. The ignorant person gets the same result as the knowledgeable person on the first try.
If you do not know which person knows where the Queen is located, the two situations appear the same. So... your knowledge of which person knows where the Queen is and which person is ignorant of the Queen's location will cause the odds to change for you, even though ON THE FIRST TRY for both men the result is the same.
It actually makes some sense to me even in this case because the choice of 98 Jacks by the ignorant chooser is a random event while the same choice by the knowledgeable chooser is not a random event.
So even if you do not play hundreds of games with the ignorant chooser, where on the average in 98 of 99 of which the Queen will turn up "prematurely", the random nature of the choice of 98 Jacks by the ignorant chooser will change the odds even if he turns up 98 Jacks on his first try.
Interesting. You have to know who knows where the Queen is and who doesn't. Definitely not an intuitive conclusion.
What happens if you don't know who knows and who doesn't, and the 98 Jacks are chosen on the first try? In one case the odds are 50 in 100 and in the other case the odds are 99 in 100 that you'll improve your choice by switching. Hmmmm...
Thanks for bearing with me!
I'll still have to solidify my thoughts over the next few days but I think you've convinced me.
It might be interesting to run some actual experiements in both cases to see what the results are. (You understand I'm a retired engineer...!)

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DFCarlson added a post in a topic VARIATION ON THE MONTY HALL CONUNDRUM  for the initiated
You suspicions are right!
Your original choice of one card assigns probabilities that the Queen is the card you chose (1 in 3) and that the Queen is in the remaining group of cards (2 in 3). Now turn one of the cards in the group of two face up. It's a Jack! What is the probability that the remaining card in that group is the Queen? It's 2 in 3 whether the person who turned the card face up knows that it is a Jack or doesn't know that it is a Jack. His awareness or lack of it does not change the fact that the card turned up was a Jack and that the other card now has a 2 in 3 chance of being the Queen.
To make it clearer,
Suppose you start with 100 cards, of which one is a Queen and the others are not. You pick one. The chance that you have the Queen is 1 in 100. Now someone (whether he knows where the Queen is or not) turns over 98 of the remaining 99 cards and each card that he turns over is not the Queen.
No matter what the person knows, either the card you picked is the Queen or the remaining card in the group of 99 is the Queen. And since the probability that the Queen is in the group of 99 is 99 in 100, and there is only one card left, then the probability that that card is the Queen is 99 in 100. So you switch, and you have a .99 probability of success.
Would you not switch in this case?

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DFCarlson added a post in a topic How did you get here?
I found this site by chance when I was Googling for information on classic logic and probability problems.

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