Lotto Probability

[Molinaro]

I may well have missed the change in problem that he was refering to, thanks.

 

[ynot]

Molinaro,

I think ynot could be right. He's arguing a game with two contestants that are not choosing the same door. Therefore, Monty could get stuck having to pick the prize door.

This is different than the scenario you describe with only one contestant where Monty can always pick an empty door.

I don't think that ynot is arguing against the odds being 2/3 for switching in the one contestant scenario?

If he is then, as many have already proven, he's wrong.

Yes, my post (#478) assumes that the contestants cannot choose the same door. The host must always therefore reveal the remaining third door. 2/3 of the time this door would be revealed to be empty and the contestants are effectively being told that one of them has made the correct choice (one can win). 1/3 of the time however, the host would have to reveal that the third door was not empty, and that both contestants had made the wrong choice (neither can win). This effectively negates I/3 of the games being played and reduces the overall games in which a contestant can win from 1/3 to 1/2. Neither contestant would improve their chance of winning by changing their choice.

If the contestants could both choose the same door the host would always be able to reveal one of the remaining doors as being empty and effectively two “normal” Monty Hall games would be being played concurrently. Both contestants would improve their chances by changing their choice.

Hmmmm . . . Consider this scenario: There are three contestants, two have to choose the same door, the other a different door . . .

 

[Just thinking]

If the contestants could both choose the same door the host would always be able to reveal one of the remaining doors as being empty and effectively two “normal” Monty Hall games would be being played concurrently. Both contestants would improve their chances by changing their choice.

Maybe not ... just because the two contestants could choose the same door, doesn't mean they must do so -- they might still choose different doors, in which they would be as in your first scenario. With that option open to them (different or same doors allowed), there are 9 possible outcomes -- two of which are immediately invalid as neither chose the winning door and it must be opened revealing the prize. So switching (if a non-prize door can be opened in this variation) yields a 4/7 chance of winning, for either player.

 

[Just thinking]

Here is a simple probabilities question that seems to always lead to a similarly long set of explanations as to why the correct answer is in fact correct:

- There are 2 cards. 1 is blue on both sides. 1 is blue on 1 side, red on the other.

- You pick 1 card at random, and look at the side that is face up. If it is blue, what is the probability that the other side of that card is also blue?


Naive answer: 1/2 since that can only happen with 1 of the cards, and you have a 50-50 chance of picking that card.

Correct answer: 2/3

That's a good one. Here one must not consider the number of cards but instead the number of sides of cards. Three sides have blue on them, and one red -- given that a blue side shows up, the red side is now out of the equation. There are now only 3 possibilities (or sides of cards) left -- and two of them are blue-blue.

Nice.

 

[ynot]

Maybe not ... just because the two contestants could choose the same door, doesn't mean they must do so -- they might still choose different doors, in which they would be as in your first scenario. With that option open to them (different or same doors allowed), there are 9 possible outcomes -- two of which are immediately invalid as neither chose the winning door and it must be opened revealing the prize. So switching (if a non-prize door can be opened in this variation) yields a 4/7 chance of winning, for either player.

Sorry - I should have wiritten "If the contestants could, and did, both choose the same door"

 

[BillyJoe]

Molinaro,

Regarding the highlighted bit below....

There are 2 cards. 1 is blue on both sides. 1 is blue on 1 side, red on the other.

- You pick 1 card at random, and look at the side that is face up. If it is blue, what is the probability that the other side of that card is also blue?

Correct answer: 2/3

The reason why 2/3 is correct is the same as with the monty hall discussion. Namely, that you have an extra piece of information given to you after you make the 1st random selection.

Maybe you need to point this out, because it is not obvious to me what this new information is.
Before you even draw a card, you know that, if you draw a blue card, the probability that the other side is blue is 2/3. This does not change when you actually draw the card. Therefore, no new information.
This puzzle would seem to me to be quite different from the Monty Hall Puzzle where you are shown an empty door after your selection.

BillyJoe

 

[BillyJoe]

....oops, you probably mean that you make a selection first and then are given the information about the cards.


edit:
Actually, looking back, you didn't really make that clear from your original post and, in my experience, as much time is often spent backpedalling about how the puzzle should have been phrased as there is about the solution.

BJ

 

[illuminatedwax]

Molinaro,

Regarding the highlighted bit below....

Maybe you need to point this out, because it is not obvious to me what this new information is.
Before you even draw a card, you know that, if you draw a blue card, the probability that the other side is blue is 2/3. This does not change when you actually draw the card. Therefore, no new information.
This puzzle would seem to me to be quite different from the Monty Hall Puzzle where you are shown an empty door after your selection.

BillyJoe

You're right; the reason the odds are 2/3 has nothing to do with the presence or lack of new information. This puzzle is easiest to understand if instead of cards you have balls: You have 4 balls, 3 blue and 1 red, two of them are numbered #1 and two numbered #2. You pick a blue ball which is numbered #1. What are the odds that the other ball numbered #1 in the bag is also blue? It's obviously 2/3, because any of the other balls in the bag could be labled #1, and there are two blue ones and one red.

 

[Just thinking]

Molinaro, ...
This puzzle would seem to me to be quite different from the Monty Hall Puzzle where you are shown an empty door after your selection.

BillyJoe

You're right; the reason the odds are 2/3 has nothing to do with the presence or lack of new information.

Hmmm ... I'm wondering if the Monty Hall problem actually gives you new information when an empty door is opened -- after all, picking just one door guarantees that at least one of the other non-chosen doors must have nothing behind it. When the host reveals the empty door, just exactly what is new to the contestant? Yes, it shows him exactly which of the two non-chosen doors is empty, but does that really matter? When offered to switch, the contestant really is getting both of the other two doors, along with being shown that at least one is a blank, which is nothing new -- and everyone knew that the chances are 2/3 that the prize is among the two non-chosen doors from the get-go.

 

[Rasmus]

Hmmm ... I'm wondering if the Monty Hall problem actually gives you new information when an empty door is opened -- after all, picking just one door guarantees that at least one of the other non-chosen doors must have nothing behind it. When the host reveals the empty door, just exactly what is new to the contestant? Yes, it shows him exactly which of the two non-chosen doors is empty, but does that really matter? When offered to switch, the contestant really is getting both of the other two doors, along with being shown that at least one is a blank, which is nothing new -- and everyone knew that the chances are 2/3 that the prize is among the two non-chosen doors from the get-go.

Well, he is not really getting two doors, he's getting one that's worth two. And that only works because one door has been opened. If no door was being opened, he could switch doors all day long and would always end up with 1/3 of a chance of winning.

 

[ynot]

Hmmm ... I'm wondering if the Monty Hall problem actually gives you new information when an empty door is opened -- after all, picking just one door guarantees that at least one of the other non-chosen doors must have nothing behind it. When the host reveals the empty door, just exactly what is new to the contestant? Yes, it shows him exactly which of the two non-chosen doors is empty, but does that really matter? When offered to switch, the contestant really is getting both of the other two doors, along with being shown that at least one is a blank, which is nothing new -- and everyone knew that the chances are 2/3 that the prize is among the two non-chosen doors from the get-go.

So instead of revealing which of the non-chosen doors is empty, the host could say “Do you want to keep the choice you have made, or do you want to switch to both of the other choices”. If nothing else, this is a very clear way of explaining to those that still don’t “get it“, that the final choice is between 1/3 or 2/3.

 

[Just thinking]

Well, he is not really getting two doors, he's getting one that's worth two. And that only works because one door has been opened. If no door was being opened, he could switch doors all day long and would always end up with 1/3 of a chance of winning.

Oh, but he (the contestant) is getting both doors -- it just looks like he's getting only one (the unopened non-chosen one) when he switches. (Remember -- this is how the Monty Hall scenario is presented -- one non-chosen door is opened.) There is absolutely no difference in outcomes or probability if the host gives the contestant both non-chosen doors for the switch and the contestant later opens them both up to show at least one was empty. He still gets exactly what was behind both doors -- be they one empty and one prized, or both empty.

So instead of revealing which of the non-chosen doors is empty, the host could say “Do you want to keep the choice you have made, or do you want to switch to both of the other choices”

Yes Yes Yes ... The rain in Spain stays mainly on the plain!

You've got it!

 

[Just thinking]

If nothing else, this is a very clear way of explaining to those that still don’t “get it“, that the final choice is between 1/3 or 2/3.

Yes ... it does sort of clear it up -- at least to me it did.

Thank you.

 

[BillyJoe]

Hmmm ... I'm wondering if the Monty Hall problem actually gives you new information when an empty door is opened.

Before the door is opened: You know that one of two doors is empty.
After the door is opened: You know which one of the two doors is empty,
This is a new bit of information.

 

[Just thinking]

Before the door is opened: You know that one of two doors is empty.
After the door is opened: You know which one of the two doors is empty,
This is a new bit of information.

OK, I'll rephrase that ...

"I'm wondering if the Monty Hall problem actually gives you any new useful information when an empty door is opened."

(Actually, I did mention this ... "Yes, it shows him exactly which of the two non-chosen doors is empty, but does that really matter?")

 

[drkitten]

"I'm wondering if the Monty Hall problem actually gives you any new useful information when an empty door is opened."

Yes, it does. The usefulness is demonstrated by the fact that you have a better chance of getting the prize after the door is opened than you had before.

(Actually, I did mention this ... "Yes, it shows him exactly which of the two non-chosen doors is empty, but does that really matter?")

Yes, it matters.

 

[Just thinking]

Yes, it does. The usefulness is demonstrated by the fact that you have a better chance of getting the prize after the door is opened than you had before.

Perhaps I'm not being clear -- which is nothing new.

If the host offers a switch to one of the non-chosen doors, and neither is opened, then the contestant does not increase his chances by switching. But once the host reveals one door as being empty, he is actually giving the contestant both that door (the empty one) and the other non-chosen one -- if he chooses to switch. I have argued early on that this new choice of only one door (the still unopened one) is actually the same as getting both doors. So opening one empty door is no different than getting both in the offer. In this regard (as opposed to saying 'You can have both doors if you switch'), it offers the contestant no new information.

 

[drkitten]

If the host offers a switch to one of the non-chosen doors, and neither is opened, then the contestant does not increase his chances by switching. But once the host reveals one door as being empty, he is actually giving the contestant both that door (the empty one) and the other non-chosen one -- if he chooses to switch. I have argued early on that this new choice of only one door (the still unopened one) is actually the same as getting both doors.

I think you're confusing mathematical equivalence with identity.

The actual structure of the Let's Make a Deal show disproved this. When Monty opened a door, he didn't open a door to reveal an empty chamber -- he opened it to reveal a worthless prize, the classic example being "a goat," which would cost more to feed and take care of than it would be worth.

Monty didn't offer the contestants' "both doors." If you switched, you didn't get the goat as well as the car. (More's the pity.)

But we can extend this analysis a little further to show the difference. I have three envelopes, one with a dollar bill, one with a hundred, and one with ten C-notes ($1000), and I get to pick one. My "expected value" from that pick is $1101/3, or 367 My expected value is different if Monty shows the hundred than if he shows the single. Depending upon Monty's strategy ( if he always shows the hundred if it's available, for example), then I might actually be better off not switching.

 

[Just thinking]

I think you're confusing mathematical equivalence with identity.

The actual structure of the Let's Make a Deal show disproved this. When Monty opened a door, he didn't open a door to reveal an empty chamber -- he opened it to reveal a worthless prize, the classic example being "a goat," which would cost more to feed and take care of than it would be worth.

Monty didn't offer the contestants' "both doors." If you switched, you didn't get the goat as well as the car. (More's the pity.)

Yes, yes ... I know (I used to watch the show). But I'm just referring to the Monty Hall brain teaser problem. In that scenario it's one prize and two empty doors; and with that, getting an empty door and the other door or getting just the other door are equivalent.

 

[drkitten]

Yes, yes ... I know (I used to watch the show). But I'm just referring to the Monty Hall brain teaser problem. In that scenario it's one prize and two empty doors; and with that, getting an empty door or getting nothing are equivalent.

That's the point. They are mathematically equivalent in that situation, but not in the general case (they're not identical -- the numbers just happen to balance.)

You can see that in the Monty Hall problem variation I proposed, where the three numbers are different. You still benefit (in the general case, assuming only that Monty will never show the grand prize, but with no information about which goat he will show) from switching because of the new information.