Lotto Probability

[69dodge]

I have read the last several pages, which is why I made the post. Somewhere back there, everyone agreed that the odds remain 1 in 3 for the original door after the host opens the empty door. This is wrong because the host opening an empty door added information to the puzzle, thereby changing the odds. Probablilities are used only when there is less than perfect knowledge, and when additional information becomes available, the odds are changed.

Odds can change as a result of new information. In this case, they don't, because the new information happens not to say anything about what's behind the contestant's door. (Of course, the new information does say something about what's behind the door that the host opened, and so its probability does change, from 1/3 down to 0.)

Suppose the contestant picks door A. At this moment in time, the probability is 1/3 that door A contains the prize. Now suppose the host opens door B. Here is the reason why the host's opening of door B tells us nothing new about whether door A contains the prize: prior to his opening a door, the probability that he would open B, rather than C, is the same on the assumption that A contains the prize as it is on the assumption that A doesn't contain the prize. (It's 1/2 in either case. If A contains the prize, the host chooses randomly between B and C, so B has probability 1/2 of being opened. If A is empty, the host opens whichever other door also is empty, and B has probability 1/2 of being that door, because B and C are equally likely to contain the prize and one of them is guaranteed to contain it if A doesn't.)

 

[Just thinking]

No. Both contestants have the same information about everything. How can their probabilities differ?

Because contestant A picked his choice among 3 doors, knowing 2 of them were empty. Contestant B picked his choice among 2 doors, knowing only 1 was empty. When contestant A is offered a choice of switching, he is getting both of the non-chosen doors -- one is then opened (the empty one, of course). The odds for him are still 2/3 that the prize is among those 2 doors, even though one is opened. Contestant B is simply given 2 doors to choose from with a prize behind only one -- that's 50/50. Contestant B never gets two doors if he switches -- contestant A does, and I think this is a key point. When offered the switch in the normal scenario, it helps to see that the contestant is actually getting both non-chosen doors if he switches -- after all, what's the difference between getting nothing or an empty door?

 

[69dodge]

Because contestant A picked his choice among 3 doors, knowing 2 of them were empty. Contestant B picked his choice among 2 doors, knowing only 1 was empty. When contestant A is offered a choice of switching, he is getting both of the non-chosen doors -- one is then opened (the empty one, of course). The odds for him are still 2/3 that the prize is among those 2 doors, even though one is opened. Contestant B is simply given 2 doors to choose from with a prize behind only one -- that's 50/50. Contestant B never gets two doors if he switches -- contestant A does, and I think this is a key point. When offered the switch in the normal scenario, it helps to see that the contestant is actually getting both non-chosen doors if he switches -- after all, what's the difference between getting nothing or an empty door?

Contestant B is offered the choice between the same two doors as contestant A, correct? And B knows everything about the doors that A knows, correct? (Namely, that the door which is now out of the game was opened by a host who was avoiding A's original door and also avoiding the door with the prize.) So, how can the probabilities for A and B possibly be different? They're playing precisely the same game.

I mean, suppose you're A. And you're about to switch to the other door, because that's the best thing to do. Suddenly, you're called away on important business. But a friend is allowed to substitute for you and to pick a door on your behalf. Wouldn't you instruct him, as you hurriedly leave, to pick the same door that you would have picked? Who cares which person actually picks the door? It's the same door.

 

[Just thinking]

Contestant B is offered the choice between the same two doors as contestant A, correct?

Incorrect. When contestant A is offered the option to switch, he is essentially given both non-chosen doors. The host opening one of them to reveal it's empty is nothing the contestant already knew. Yes, it does specify which of the two doors is empty -- but so what? Contestant A knew at least one of them had to be empty. It would make absolutley no difference if the contestant (after switching) took both doors and opened one and found it empty. He still gets the other unchosen door. So getting two doors for the price of one doubles (in this case) the chances for contestant A to win.

Now, contestant B only gets to choose switching with one other door -- a one for one choice. Hence, his chances remain the same ... 50/50.

And B knows everything about the doors that A knows, correct? (Namely, that the door which is now out of the game was opened by a host who was avoiding A's original door and also avoiding the door with the prize.) So, how can the probabilities for A and B possibly be different? They're playing precisely the same game.

Perhaps now you see the difference? It's not the same game for A is it is for B.

I mean, suppose you're A. And you're about to switch to the other door, because that's the best thing to do. Suddenly, you're called away on important business. But a friend is allowed to substitute for you and to pick a door on your behalf. Wouldn't you instruct him, as you hurriedly leave, to pick the same door that you would have picked? Who cares which person actually picks the door? It's the same door.

But arrived at differently. Try -- please try and see that contestant A is actually given both of his non-chosen doors if he switches -- and an empty one (which must be one of them, at least) is opened.

 

[69dodge]

Perhaps now you see the difference? It's not the same game for A [as] it is for B.

I totally do not see.

If A's initial choice was correct, then he'd win by sticking with it, and so would B. If A's initial choice was incorrect, then he'd win by switching, and so would B. They are both in exactly the same situation. If it's better for A to open a particular door, how can it not be better for B to open that same door? "A" and "B" are just names. The car doesn't care about the name of the person who opens the door in front of it. It is wherever it is.

 

[Just thinking]

I totally do not see.

If A's initial choice was correct, then he'd win by sticking with it, and so would B. If A's initial choice was incorrect, then he'd win by switching, and so would B. They are both in exactly the same situation. If it's better for A to open a particular door, how can it not be better for B to open that same door? "A" and "B" are just names. The car doesn't care about the name of the person who opens the door in front of it. It is wherever it is.

Doesn't contestant B (according to this from post 447) only enter the game after the host reveals one of A's non-chosen doors to be empty?

 

[69dodge]

Doesn't contestant B (according to this from post 447) only enter the game after the host reveals one of A's non-chosen doors to be empty?

Yes, that's my understanding. But why does that matter? B knows, just as A does, that the host's plan from the beginning was to open an empty door other than the door that A would initially choose. Since B knows everything that A does, he ought to make the same decision as A, namely, to switch from A's initial choice to the remaining unopened door.

 

[Just thinking]

Yes, that's my understanding. But why does that matter?

Well, according to the post, Contestant B is free to initially pick one of the two remaining unopened doors (after the host opens one), not the same door that contestant A chose. Perhaps this is the cause of your disagreement?

B knows, just as A does, that the host's plan from the beginning was to open an empty door other than the door that A would initially choose. Since B knows everything that A does, he ought to make the same decision as A, namely, to switch from A's initial choice to the remaining unopened door.

See? You're assuming contestant B chose as did A -- not true. And where exactly in the original post (447) does it mention contestant B knowing all that contestant A does? All contestant B sees are two unopened doors ... to him it's 50/50.

 

[69dodge]

Well, according to the post, Contestant B is free to initially pick one of the two remaining unopened doors (after the host opens one), not the same door that contestant A chose. Perhaps this is the cause of your disagreement?

Not sure what you mean here. B can pick whichever door he wants to: one is the door that A initially picked, the other is ... the other one.

See? You're assuming contestant B chose as did A -- not true.

He [edit: B] hasn't chosen anything yet. The question is, which door should he choose?

And where exactly in the original post (447) does it mention contestant B knowing all that contestant A does? All contestant B sees are two unopened doors ... to him it's 50/50.

Oh. I just figured that B knew the whole story. Otherwise, sure, it's different.

 

[Just thinking]

Not sure what you mean here. B can pick whichever door he wants to: one is the door that A initially picked, the other is ... the other one.

He [edit: B] hasn't chosen anything yet. The question is, which door should he choose?

Oh. I just figured that B knew the whole story. Otherwise, sure, it's different.

OK -- we're on the same page here, then. (I guess now you know why I asked those questions.) Yes, he should choose the same as A -- but if he just arrived on the spot ...

 

[BillyJoe]

There is a big difference between your two hypothetical contestants. The first one picks his winning prize from among 3 possible doors, the second only 2. The first has odds of 1/3 -- the second 1/2. If the host never exposed anything, and let each contestant choose from among all 3, then both have a 1/3 chance of winning.

To clarify this a little:

Assumption:
Contestant B enters the scene only after the host has opened one of the doors and he has no knowledge about what has happened up to that point. All he sees is two closed doors and one open door. He is simply told that there is a prize behind one of the closed doors.

With contestant A's extra knowledge of the game, the odds for him are 1/3 and 2/3. Therefore, he would switch to the other door.
However, with contestant B's limited knowledge of the game, the odds for him are 1/2 and 1/2 (ie same for both doors). Therefore, he would just choose randomly between the two.
In a series of such games contestant A would always switch and, therefore, win 2/3 of the time, just as he expected. Contestant B would choose randomly between the two and win (1/3 * 1/2) + (2/3 * 1/2) = 1/2 of the time, just as he expected.


BJ

 

[pdw709]

OK, you guys have convinced me, the contestant should always switch

Yes and no.......Everyone so far has talked about repeated trials. In the long run then the odds are clearly stacked 2/3 in your favour to switch.

However each contestant is effectivly given ONE shot. even if the odds are in your favour to switch you may still lose. This is what makes gambling so much fun!

Phil

 

[BillyJoe]

However each contestant is effectivly given ONE shot. even if the odds are in your favour to switch you may still lose. This is what makes gambling so much fun!

I have never understood gambling. And I have read Dostoyevsky's "The Gambler".

 

[Just thinking]

To clarify this a little:

Assumption:
Contestant B enters the scene only after the host has opened one of the doors and he has no knowledge about what has happened up to that point. All he sees is two closed doors and one open door. He is simply told that there is a prize behind one of the closed doors.

With contestant A's extra knowledge of the game, the odds for him are 1/3 and 2/3. Therefore, he would switch to the other door.
However, with contestant B's limited knowledge of the game, the odds for him are 1/2 and 1/2 (ie same for both doors). Therefore, he would just choose randomly between the two.
In a series of such games contestant A would always switch and, therefore, win 2/3 of the time, just as he expected. Contestant B would choose randomly between the two and win (1/3 * 1/2) + (2/3 * 1/2) = 1/2 of the time, just as he expected.


BJ

Yes ... that's how I took it also. I even tried it with a different number of doors ... all the way up to my lottery ticket example where one ticket is essentially a sure win and the other a sure loss. That's pretty much a 50/50 chance. One can also look at the Monty Hall scenario with contestant B this way -- instead of having 3 doors to choose from with one prize, he gets only 2. Can't be more 50/50 than that.

Nice job, BJ.

 

[BillyJoe]

Well, thank you, JT.

 

[macgyver]

Holy Crap! I LOVE lobbing this stuff into here and watching what happens...I'm not getting my update emails nearly as often as there's activity on this thread though....

However, I was right along side joe87 when I first asked the question, but Rasmus's Evil Twin, and JT's "prize is behind door B" posts finally drove it home for me!

Thanks to everyone for ironing out those wrinkles.

I'm cereal.

 

[BPScooter]

I love it too, because it helps me talk to people that are otherwise intelligent understand the no-win nature of lotteries or keno. Sure take a chance bet, it's fun, but don't try to use a system of lucky numbers on a regular basis.

 

[ynot]

Consider this scenario:

We will label the three doors, 1, 2, and 3.
There are two contestants, A and B.
Contestant A chooses door 1.
The odds that the prize is behind door 1 is 1/3.
Contestant B chooses door 2.
The odds that the prize is behind door 2 is 1/3.
The host opens door 3 and reveals it to be empty.
The host allows contestant A and B to switch.
They both switch to increase their odds to 2/3.



BillyJoe

I have been travelling all week and haven‘t had any time to keep up with events. Haven’t had time to read the other posts yet but here is my response to your scenario.

A and B are in the same draw and share the same odds of winning or losing, regardless of whether or not they swap their choices. The odds that either will win or lose are always therefore 50/50 and no advantage is gained or lost by either player switching their choice. If you and I were to each purchased a lotto ticket in the same draw, would you want to swap tickets with me if you were told that one of our tickets had won? Half the time you would be very happy if you did, and half the time you would be very sad. It doesn’t matter how many people are in the draw, if everyone shares the same odds the odds don’t change by any swapping.

 

[Molinaro]

ynot, post #451 should convince you otherwise


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

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.

 

[macgyver]

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.