Mageth

I'm not sure I understand this analogy - it doesn't appear to equate to to the original problem. If you revealed the 999,998 tickets that lost, then the analogy would be better - and it would still be better to switch. (apologies if this is what you meant!).

I think Bill indeed means that you would reveal 999,998 loser tickets and retain one. It's either the winning ticket or the one you're holding is the winning ticket.

This equates to the original problem because Monty opens/reveals a "loser" and leaves exactly one choice for you to switch with, if desired.

Think of it with four doors, and Monty reveals two empty ones, leaving your original choice and one option.

Or with ten doors, and Monty reveals eight empty ones, leaving your original choice and one option.

Etc. etc.

Bill Snedden

That is indeed what I meant.

I've just found that increasing the difference between the odds of winning from 1:3 to 1:1,000,000 seems to help with perspective. It's easier to see that the original choice had less chance of being the winning one than any of the tickets left on the table. If you think of it as two sample spaces, the chosen and non-chosen, you can see that the chosen has a 1:1,000,000 chance and the non-chosen has a 999,999:1,000,000 chance. Revealing 999,998 tickets of the non-chosen doesn't change the initial probabilities of the chosen/non-chosen spaces. The chosen still has a 1:1,000,000 chance and the non-chosen, which now consists of only one ticket, has a 999,999:1,000,000 chance.

I like the way Armchair dissident put it: "If I hold the winning card, then you have discarded 999,998 cards entirely at random knowing that I have the right card, and can then present me with one random wrong card. But this will only happen once in every million times the game is played. Of the remaining 999,999 times the game is played, you will have the card, and I loose."

Why don't lotteries work like this in RL?

Regards,

Bill Snedden

Quantum Ninja

No, because rows 1 and 2 below are "equivalent" outcomes.

1 2 3
-------
X O -
X - O

If I chose door number 1, and the other two doors are empty, opening door 2 to leave with me the option of door 3 is really no different than opening door 3 to leave me with the option of door 2.

Think about it. My initial choice is door 1 and the prize is behind that one. The host knows that doors 2 and 3 are empty. It doesn't make a difference which one of the two doors he decides to show me. Whichever of the two empty doors he decides to show, it does not effect the probability of which door the prize is behind.

If he randomly chooses which of the two empty doors to reveal, then the two outcomes above ("X O -" and "X - O") are equally like to occur. BUT, these two outcomes can only happen in the event that the prize is behind door 1. The prize is only behind door 1 one third of the time. Thus, this probability of 1/3 is split between the two outcomes, giving each of them the probability of 1/6 (as others have already pointed out).

xorbie

I loved this problem when I first saw it 4 years ago in 9th grade and I still love it now. Trust me, there is no denying that you will win far more often if you change doors. This is a simple fact. It is also very interesting and mind boggling, but that does not mean you can deny it.

tronvillain

Here, let us try it with Bayes' Theorem:

P(h/e)=P(h)P(e/h)/[P(h)P(e/h)+P(~h)P(e/~h)]

h: the prize is behind the door you picked
e: Monty knows which door the prize is behind, and opened one of the other to doors to show that there was no prize behind it.

P(h)=1/3
P(e/h)=1 (since neither door has a prize behind it)
P(~h)=2/3
P(e/~h)=1 (since Monty will never reveal the prize)

P(h/e)=1/3*1/[1/3*1+2/3*1]=1/3

So, the new information does not change the probabilities of h (1/3) or ~h (2/3), but since ~h is now only one door rather than two, you should switch doors.

Tenpudo

Okay, I see how the 2/3rds comes about, thanks to Bill's lottery analogy (specifically, Armchair's analysis of it). I guess what I'm trying to do is come to a "gut" appreciation of the odds (I am strongly reminded of a similar situation regarding the "roll two dice, and one comes up 6: what are the odds the other is a 6?" thread).
Unfortunately, even though I loved math as a kid, probability was always relegated to the back of the book in school, and I don't think a single elementary, middle, or high school class of mine (in any subject) ever made it through a full textbook. Thus, some of the underlying principles of probability remain counter-intuitive for me.
I always seem to look at it from a symmetric perspective, that is, say two people are playing the same set of doors, and pick doors 1 and 3. Monty reveals door 2-- should they both switch to each other's doors? If not (which seems "obviously" the case), why does somebody else picking the other unopened door change things? If it's a matter of knowing it was picked, what if the players have no knowledge of each other's actions, and it's up to us in the audience (who see both picks) to encourage, say, player 1 to stay or switch, as logic dictates.

tronvillain

I will try that scenario with Bayes' Theorem again.

P(h/e)=P(h)P(e/h)/[P(h)P(e/h)+P(~h)P(e/~h)]

h: the prize is behind the door you picked.
e: Monty opened a door without a prize behind it.

P(h)=1/3
P(e/h)=1 (since the one door Monty can open does not have a prize).
P(~h)=2/3
P(e/~h)=1/2 (since the one door Monty can open is one of two door which could have the prize behind it).

P(h/e)=1/3*1/[1/3*1+2/3*1/2]=1/3*3/1=1/2

When two people each pick a different door, that leaves Monty with no choice about the door to open, so him opening the door does not provide the same information as when only one person picks a door. In this scenario, Monty might be forced to reveal the prize when he opens the door (that assumes he has to open a door, but anything else needlessly complicates things).

Now, if the players do not know that there is another player, then they "should" switch. If they had the information the audience has then they would realize switching is pointless, but without that information they will estimate the probability of winning if they switch to be 2/3 rather than the 1/2 the audience will estimate.

Artemus

Deleted post. The question I posted was essentially the same as the "one die is a six" question that was refered to above.