No, now you are guaranteeing from the outset that there is a door that never wins, which completely changes the rules of the game.
I think those are precisely the rules of the game. The only difference is that monty doesn’t eliminate one of the choices until you have made yours. Once he does, 2 choices remain - one winner and one loser.
In the original game, you pick first, and Monty is forced to open on of the doors you haven’t picked. If Monty picks first and forces you to choose between the two left-overs, that’s a different game.
I think the difference would be intuitively easier to grasp if you increased the number of doors to 1,000,000, of which Monty opens 999,998.
Odds can change when you receive information. For example, you’re playing poker with someone, and you can see his hand reflected in a mirror that’s on the wall right behind him. The new information changes the odds of winning at poker.
Information doesn’t have to be blatant to change the odds either. Martin Gardner wrote a whole article about the Monty Hall problem for Skeptical Inquirer. In it, he gave a scenario of an alien being landing his spacecraft as soon as Monty opens a door to show you the goat behind it. The alien has no knowledge of your first guess or which door Monty opened. What’s the odds of the alien picking the right door? Only 1:2. He has less information than you do because he didn’t see your first guess.
There’s the classic one: A man has two kids, and says “one of them is a boy”. What’s the possibility that they’re both boys? Answer is 1:3.
A man has two kids, and says “the older one is a boy”. What’s the possibility that they’re both boys? Answer is 1:2.
By merely stating that one particular child is a boy, the odds changed. The same thing could have happened this way too:
A man has two kids, and says “one of them is a boy”. Suddenly a little boy comes down stairs to give his dad a good night kiss. What’s the possibility that they’re both boys? Answer is 1:2 even though you don’t know if the boy who ran down the stairs was the older one or younger one.
The question is whether Monty’s knowledge changed the odds. Let’s say Monty doesn’t know which door has the car, you pick a door, and he randomly opens another.
You pick Door #1, Monty says "Let’s see what’s behind Door #2. Oh no! It’s the car. Would you like to change your answer from Door #1, which contains a goat, to Door #3, which also contains a goat? Won’t do you a lot of good.
However, sometime Monty doesn’t pick the door with the car. You pick Door #1, Monty picks Door #2, and it’s a goat. What are the chances of winning if you switched? The answer is 1:2 – Monty doesn’t know the answer, and his opening a door gave you no new information.
Information can change the odds in subtle ways. Gary Forshee offered a twist on the two boy puzzle. He gave a small speech at the G4G conference a few years ago, here’s how one person described it:
You can read the rest here: geregistreerd via Argeweb. By the way, in this case the correct answer is not 1/2, but 13/27.
The second part of your sentence does not follow from the first. The fact that the number of possible choices is two does not imply they have an equal chance of winning.
Why wouldn’t it be closer to when there are a million doors and monty only opens one?
Regardless, the odds are still 50-50. It’s like the lottery ticket example in Cecil’s column. Out of a million tickets, 2 are left. You know one is worth $1M. Is it worth anything to buy out the other ticket holder? Of course it is. Your odds go from 1 in 2 to 2 in 2.
I completely agree. But where is the new information? You know in advance that you will see a goat at the end of the first selection process. That is true whether or not you guessed correctly. Therefore, Monty’s act of showing you a losing door tells you precisely nothing new.
Take yet another example.
I have a twin brother. He’s hiding backstage. I pick door #1. While Monty is opening a losing door with a goat behind it, my brother and I switch places without anyone knowing.
Further, my brother has no idea what has transpired. He doesn’t know what my initial selection was. All he knows is that he has 2 choices. Show me how his odds of winning are 2/3 if he picks, purely by chance, the door that I did NOT pick and 1/3 if he picks the door that I DID pick.
As you’ve described this, there’s no way for your brother to know which door to choose, but if he happens to pick the door you didn’t pick, his chance of winning will indeed be 2/3 - because unless the prize was behind the door you originally selected (probability = 1/3), it’s got to be behind that door.
A good way to understand this is to realize that the following is equivalent to the MH problem:
Of 3 doors, two conceal a goat and the other a valuable prize. You have two choices:
- Choose one door and keep what’s behind it.
- Choose two doors, agreeing to give back one goat.
I guess until someone can provide a rigorous mathematical proof that settles the issue one way or the other, there won’t be any resolution. But in that case, it’s unlikely I would understand it anyway, so I would have to trust that the proof was in fact validate - which I would do were it appropriately peer reviewed.
My position is that while your odds of picking the correct door are only 1 in 3, knowing that one door will always be excluded means that the relevant choice is the second one. If you flip a coin at that point, I don’t see how you can argue for any result other than the fact that half of the time you will win and half of the time you will lose.
To my mind, that makes the odds 50-50 - even if you consider the fact that you arrive at these odds by taking 1/2 of the 2/3 probability and adding it to the 1/2 of 1/3 probability (see post #77 ).
This is perfectly correct - basing the choice to switch on a (fair) coin flip indeed gives a 50-50 chance of winning.
But why do that when always choosing to switch gives a much better chance of winning the big prize?
I think naita basically did that in post #78, by considering all possible distributions of goats & prize, and the outcomes they produce.
I’m just not convinced that is the case when flipping a coin gets you the prize half of the time.
That’s not a complete truth tree. I actually did one several years ago and the result was a 50-50 chance. Of course at this point I don’t remember how I did it and what mistakes I may or may not have made in the process.
Okay, how about this:
Call the doors 1, 2 and 3. “1-G” means that door 1 conceals a goat; “1-P” means it has the big prize.
There are 3 possible setups, each offering three ways to play the “always switch” strategy - as follows:
Case 1: The setup is 1-P 2-G 3-G. If you choose 1, then switch, you lose; if you choose 2 or 3 then switch, you win.
Case 2: The setup is 1-G 2-P 3-G. If you choose 2, then switch, you lose; if you choose 1 or 3 then switch, you win.
Case 3: The setup is 1-G 2-G 3-P. If you choose 3, then switch, you lose; if you choose 1 or 2 then switch, you win.
Of 9 possible games, the “always switch” strategy thus yields 6 winners. Probability of winning = 6/9 = 2/3.
Okay, try this out.
Imagine that Monty has decided to make this game really easy.
You pick a door. Monty opens another door and shows you that it’s empty. There are now two doors left. Monty now opens both of these doors and shows you what’s behind them. It’s all there - you can see which door has the prize behind it. You can see if your picked door has the prize or if the other door has the prize. Monty now asks you if you want to stay with your first pick or switch to the other door. Remember you can see where the prize is.
At this point, you’re going to get the prize 100% of the time. If you picked the right door the first time, you’ll stay with it and win. If you picked the wrong door, you’ll switch and win. So if you picked the right door first, staying will always win. If you didn’t pick the right door, then switching will always win.
Now think, what are the chances you picked the right door on your first pick when all three doors were closed? One chance in three. What are the chances that you didn’t pick the right door on your first pick? Two chances in three. That means that one third of the time you’re going to win by staying and two thirds of the time you’re going to win by switching.
Now Monty decides the game has become too easy, so he’s stop showing you were the prize is. But the odds are still the same - one third of the time you’re going to win by staying and two thirds of the time you’re going to win by switching.
Here is how to construct a probability tree.
First draw - 1/3 P, 2/3 G
Second draw
for P there are 2 outcomes - 1/2 G and 1/2 P
for G there are 2 outcomes - 1/2 G and 1/2 P
The result is 4 outcomes
- 1/3 x 1/2 = 1/6 for P
- 2/3 x 1/2 = 2/6 for P
- 1/3 x 1/2 = 1/6 for G
- 2/3 x 1/2 = 2/6 for G
Total probability for P = 3/6
Total probability for G = 3/6
You have here accurately modeled what happens if the choice to stick or switch is based on a coin flip. But this isn’t terribly useful, since there has never been any question that this leads to a 50-50 chance of winning.
Here’s what happens when we model the strategy we’re actually trying to consider - the one I call “always switch”:
First draw - 1/3 P, 2/3 G
Second “draw”
for P there is one outcome - 1 x G (i.e. if your first choice was P, you will wind up with G)
for G there is one outcome - 1 x P (i.e. if your first choice was G, you will wind up with P)
The result is 2 outcomes
- 1/3 x 1 = 1/3 for G
- 2/3 x 1 = 2/3 for P
Total probability for P = 2/3
Total probability for G = 1/3
I think you matched up the wrong probabilities.
1/3 of the players will initially pick the right door
2/3 of the players will initially pick the wrong door
Let’s say half of each group switches their door and half of each group stays with their first door
People who are right and switch lose - 1/3 * 1/2 = 1/6
People who are right and stay win - 1/3 * 1/2 = 1/6
People who are wrong and switch win - 2/3 * 1/2 = 2/6
People who are wrong and stay lose - 2/3 * 1/2 = 2/6
So people who switched won 2/6 of the time and lost 1/6 of the time.
People who stayed won 1/6 of the time and lost 2/6 of the time.
That’s not how the tree works AFAIK. You ONLY look at the gross probabilities of each outcome. You don’t consider the rationale of the individual players. What they think is superfluous. All that matters is the total probability of each outcome at each drawing.
I see what you’re getting at and I’ve already pointed this out in previous posts. The first time was post #77. I said that the probability of winning on the second draw was 1/2 of 2/3 when you switched plus 1/2 of 1/3 when you didn’t.
So I see that switching contributes more to the win percentage than not switching. But the fact remains that regardless of the source of the probabilities at the time of the second draw, the odds for a win or loss are still even. Unless you can challenge how I constructed my probability tree, you are obliged to agree.
I’ll admit it seems somewhat magical to go from a 1/3 chance to a 1/2 chance simply by virtue of one choice being eliminated. Alternatively, in the case of the switching strategy, it seems magical that you would go from 2/3 chance to only 1/2.
Clearly, if you could pick 2 doors and none were eliminated, your odds would be 2/3.
I’m beginning to think that as long as it is guaranteed that one of the losing doors will be eliminated after the first draw, that perhaps your odds are 1/2 right from the beginning - that in fact, they were never 1/3.
What I mean to say is that although you get to pick one of 3 numbers, there are in fact, even at the beginning of the first draw, only 2 actual choices. You could regard the initial choice as a sort of buy one, get one situation. You may pick door number 1, but you know that you will also get door 2 or 3 - whichever is the loser.
This would be the same as allowing you to have 2 doors to start but where one of the 2 is chosen by Monty and the other is chosen by you.
We can account for this by adding an intermediate draw - Monty’s draw.
First draw - contestant
1/3 chance of prize
2/3 chance of goat
Second draw - Monty
if prize was chosen, then
1/2 chance of goat
1/2 chance of goat
if goat was chosen
2/2 chance of goat
0/2 chance of prize
Third draw - the switch
if prize then goat, 1/2 chance of prize
if prize then goat, 1/2 chance of goat
if goat then goat, 1/2 chance of prize
if goat then prize, 1/2 chance of goat (even though we know Monty never picks the prize)
We get 8 outcomes as follows
prize - goat - prize = 1/3 * 1/2 * 1/2 = 1/12 = prize
prize - goat - goat = 1/3 * 1/2 * 1/2 = 1/12 = goat
prize - goat - prize = 1/3 * 1/2 * 1/2 = 1/12 = prize
prize - goat - goat = 1/3 * 1/2 * 1/2 = 1/12 = goat
goat - goat - prize = 2/3 * 2/2 * 1/2 = 4/12 = prize
goat - goat - goat = 2/3 * 2/2 * 1/2 = 4/12 = goat
goat - prize - prize = 2/3 * 0/2 * 1/2 = 0/12 = n/a
goat - prize - goat = 2/3 * 0/2 * 1/2 = 0/12 = n/a
Then let’s make this decision tree really simple.
First draw - contestant
1/3 chance of prize
2/3 chance of goat
If you have the prize and you switch you will always lose
If you have the prize and you stay you will always win
If you have the goat and you switch you will always win
If you have the goat and you stay you will always lose
So you will lose by switching 1/3 of the time and you will win by switching 2/3 of the time.
You will lose by staying 2/3 of the time and you will win by staying 1/3 of the time.