You net odds of winning in either scenario you just presented are (1 in 3) + (1 in2). Sorry I don’t know exactly how to do that math, but it does not equal 2 in 3.
This is absolutely true which is why if instead of 3 doors, there were 100 and Monty opened 98 of them to reveal 98 goats, it’s just as likely that I’ve chosen the door with the car as the one with the goat.
I think the assumption is that if Monty opens a door and reveals a car, you are not allowed to select that car. But it’s really an assumption, so yeah, if it works the way you described then it works in your favor.
The point is that since Monty knows where the car is* and is never going to reveal the car* (because if he does, the game is over and you’ve lost), and he always offers one chance to change doors, then the odds remain set at 1/3 for the door you intially picked and 2/3 for the two doors you didn’t initially pick.
If Monty doesn’t know where the prize is, then he’s guessing 50/50 (because he can’t pick the door you’ve picked), which means 1/3 of the time you picked right first, 1/3 of the time Monty picks the car and you lose, and 1/3 of the time you picked wrong and Monty picked wrong. If you have the option to pick the door that Monty opens, then the odds remain at 2/3 in favor of switching, with the bonus that 1/3 of the time you see the car before you choose the second time.
Ultimately, your problem is thinking about it in terms of the actual show, rather than in terms of the probability calculation demonstration that the problem represents. It has already been acknowledged that the actual show did not run like the ideal problem the demonstration presents. Monty was not constrained by rules to always offer a chance to change doors, nor was he constrained by rules that he was only allowed to open a non-winning door.
This is where you are wrong. Wrong wrong wrong. The terms of the probability demonstration set up the constraints. Given those constraints, you have an initial choice of 1/3. Monty then gives you the option of staying with your original door or taking both other doors. He helpfully opens one of the other doors and shows you there’s no car there, but there’s the remaining door.
The key feature is that Monty was constrained on which door he could open by where the prize was. In the 1/3 situation that you guessed right initially, he can open either other door. But in the 2/3 situation that you guessed wrong initially, he can only open one of the other doors, so he doesn’t change the probability assigned to your door choice. You had 1/3 of being right and 2/3 of being wrong, and you have the option to swap sides. The door he reveals doesn’t change that proportion distribution.
I don’t believe that is accurate. Given your constraints, you will always end with a choice between 2 doors. 1 of those 2 doors will always be a winner and 1 will always be a loser so the odds will always be 50/50. Anything that happens before that final choice is irrelevant.
Stating something does not necessarily make it true. Mathematically, the fact that when you chose the initial door, the odds were 1 in 3 is still relevant when asked if you want to switch from the door you’ve already chosen to a new door. If Monty opened the first door you picked to reveal a goat and then asked if you wanted to switch to one of the other two doors, then you’d have a 50/50 chance.
It doesn’t matter if Monty opens one of the doors to show you it’s empty or just tells you one of the doors will be empty. That doesn’t change the odds.
You start with three choices so you have a 1/3 choice of guessing right and a 2/3 choice of guessing wrong. If you guessed right, then you obviously shouldn’t switch. And if you guessed wrong, then you should switch and you’ll always win if you switch (you can’t guess wrong, switch, and then lose anyway). So 1/3 of the time you win by staying and 2/3 of the time you win by switching. Your odds are better if you switch.
The odds can change if Monty sometimes decides to not offer you a chance to switch. But that’s not the way the game is set up here. Monty always offers you a chance.
Try it.
Seriously, get a friend and try it. Or watch the Mythbusters episode where they proved it experimentally.
If you stick with your original choice, you will lose 1/3 of the time. Guaranteed. Don’t believe it? Prove otherwise, experimentally.
How could it be otherwise? Think about it: You had a 1/3 chance of choosing the car, right? Now Monty shows you that one of the other two doors doesn’t have a car behind it. YOU ALREADY KNEW THAT. You already knew that one of those two doors didn’t have a car behind it. So by Monty telling you something you already knew, all of a sudden your odds of having chosen correctly in the first place go up from 33% to 50%? That doesn’t make sense. Your odds of having originally chosen correctly are still 33%. If you stick with your original choice, you still lose 67% of the time.
Powers &8^]
Just like it doesn’t matter how many times a coin flip has come up heads, the next time, the chances of heads is 50/50, it doesn’t matter what has occurred leading up to that last decision of 1 door or the other. 1 is a winner and 1 is a loser. It is always 50/50 if one door is eliminated as an option.
You could also look at it like one door is a throw away. You pick your door and have a 50/50 chance that it is a winner (since you know that one loser will be eliminated regardless of what you choose). Then you are asked if you want to switch. There is no way of determining if you are switching from a winner to loser or vice versa. The eliminated door has no bearing on whether or not you’ve picked the winner. It is irrelevant.
jacolly, how do you explain the math, then. Because there is ample evidence that when it comes to switching, you’re better off if you do. By your reasoning, they should be equal.
You’re not flipping the coin again. You’ve already flipped the coin, it’s on your arm under your hand. Do you call heads or tails? Do you swap? It’s 50/50, but the point is you aren’t changing the coin, only what you call for your win.
In this case, take three cards, shuffle, lie them on the table. Pick one. Now you aren’t changing the cards on the table, you’re just deciding whether to keep the first card or swap to the other two cards.
Get a friend and 3 cards and play the game. Let the friend be Monty who looks at the cards, and reveals a non-winning card every time. Track the results.
For fun and profit, bet money against your friend. If you pick correctly, you get a dollar. If you don’t pick correctly, you pay Monty a dollar. Play your strategy of keeping your intial guess, since the odds are supposedly even. If you are correct, then over a large number of games (say 100 games), you will break even. You may have to pay a couple bucks or get a couple bucks, but it will be essentially even. If you are wrong, then you will owe Monty ~$33. You were right ~33 times, he won ~66 times. Or you can play the swap strategy and make Monty pay you the $33.
I’m betting you will come back and argue the point rather than actually try it yourself.
Agreed. Therefore, as Powers neatly states: when there were three doors, your chance of being correct was 1/3. It doesn’t matter what has occurred leading up to that last decision to switch (or not). It is always 1/3 when you’re choosing among three doors.
Now, if Monty opened a door BEFORE you chose one, then you’d have 1/2 of the remaining doors. But that’s not what happened. When you chose it was 1/3, and whatever Monty may say (“Your shoe’s untied” or “You need to put on a blindfold”), it’s still 1/3. And the chance of the other doors (whether 2 still closed or 1 open) is still 2/3.
The door question is simple and obvious when you think about it, pick the right door at first (1/3) switch and lose. pick one of the 2 wrong doors at first (2/3) switch and win. basic.
A more interesting and less trodden ground is the male/female problem mentioned in the article which I’ve analysed earlier. Anybody want to leave the 3 door question behind and defend the wise old master’s probablity reasoning?
But it’s not the “next time” in this case. You’re still in the same game. You picked one door out of three so you had a one out of three chance of picking the right door and a two out of three chance of picking the wrong door.
Here’s a spreadsheet of the probabilities I created one time. Sorry about the format, I made it in XL but it doesn’t paste into the board.
You pick Car Location Host Opens Total Prob. [FONT=Calibri]Stay Switch
(prob) [/FONT](prob) (prob)
----------------------------------------------------------------------------------------------
Door 1 Door 1 Door 2 1/18 car goat
1/2
---------------------------------------------------------
1/3 Door 3 1/18 car goat
1/2
------------------------------------------------------------------------------
1/3 Door 2 Door 3 1/9 goat car
1/3 1
[FONT=Calibri] ------------------------------------------------------------------------------
[/FONT] Door 3 Door2 1/9 goat car
1/3 1
----------------------------------------------------------------------------
Door 2 Door 1 Door 3 1/9 goat car
1/3 1
[FONT=Calibri] ------------------------------------------------------------------------------
Door 2 Door 1 1/18 car goat [/FONT]
1/3 1/2
[FONT=Calibri] ---------------------------------------------------------
1/3 Door 3 1/18 car goat [/FONT]
1/2
[FONT=Calibri] ------------------------------------------------------------------------------
Door 3 Door 1 1/9 goat car [/FONT]
1/3 1
----------------------------------------------------------------------------
Door 3 Door 1 Door 2 1/9 goat car
1/3 1
[FONT=Calibri] ------------------------------------------------------------------------------
Door 2 Door 1 1/9 goat car [/FONT]
1/3 1
[FONT=Calibri] ------------------------------------------------------------------------------
1/3 Door 3 Door 1 1/18 car goat [/FONT]
1/3 1/2
[FONT=Calibri] -------------------------------------------------------
Door 2 1/18 car goat [/FONT]
1/2
-----------------------------------------------------------------------------------------------
Sum of Prob =1
[FONT=Calibri][FONT=Calibri] -------------------------------------------------------
[/FONT] Prob win= 1/3 2/3 [/FONT]
You can see the first column shows 3 door options for you to pick, and your probability for picking each assigned at 1/3.
Second column is the car location, assigned evenly 1/3 of the time for each door, for each door you pick. Should be all 1/3 in that column.
Third column is the Host Opens, with his options constrained by the door without the car, unless you picked the car initially, in which he has a 50/50 option on which door to open. Numbers are either 1 or 1/2 to reflect this.
Total probability is the calculated probability for that option from all available options.
Stay shows what you get if you stay, switch what you get if you switch. Goats are red for lose, car is blue for win.
Note that there are an equal number of goats and cars in the columns. The difference is to compare the probability number with the outcome. If you look, you should see that the 1/18 prob aligns with car in the Stay column, while 1/9 prob aligns with car in the Switch column.
6 x 1/18 = 1/3
6 x 1/9 = 2/3
QED, if you stay, you will win 1/3 of the time, switch you will win 2/3 of the time.
Actually, the probability of the door you pick strictly isn’t important. You can choose a strategy of always picking the middle door or the right door or whatever. That would affect the numbers, but not the distribution. I used a random selection to make the math clean, and not constrain your choice of door.
Thanks Powers. Something about the way you explained it finally made it click. I’m still having issues with the 2/3 number if the contestant switches, but I think I’m finally good with the less than 50% chance if the original door is kept.
Cmoore, I’m trying to make sense of what you’ve done and how it’s different. What strikes me is that biologically, the options include 4 outcomes, including two male children. That is how the probabilities for the options are originally presented. Because one of 4 options is eliminated, that cuts out that percentage of options, but leaves the remaining biologically determined distributions unaffected.
Something about what you are doing is reassigning those distributions.
This isn’t really my forte.
…and now I’m getting the 2/3 number. Thanks all!
The tricky part with the male-female problem is that it’s extremely difficult to set it up in a real-world-plausible way. The counterintuitive answer depends on you having exactly the information that there are two children, and that at least one of them is a boy, without knowing anything about any specifiable child. But in the real world, almost all of the ways you would know that there was at least one boy involve information about a specific child, and that’s what informs our intuition. See one kid walk into the room? The other kid is 50-50, just like you’d expect. Hear a mention of “my oldest child, who is a boy”? 50-50. Favorite child is a boy, or first child to get married, or child who lives closest to mom? All 50-50. It’s really hard to construct a plausible scenario where the 2/3 answer is correct.
The simplest is a dialog:
A: “I have two children.”
B: “Is at least one of them a boy?”
A: “Yes.”
Short of trivial cases like that, the only one I’ve come up with is as follows:
Suppose my high school (which is all-boys) throws an appreciation dinner for parents of alumni. My mother goes, and sits with another parent she doesn’t know. Over the course of dinner, the other parent mentions that he has two children. Mom now knows that he has at least one boy (since he couldn’t be a parent of an alumnus otherwise), but that’s not a specific identifier (since, for all she knows, both of the kids could have been alumni).
Cecil, while the Monty Hall problem has appeared on the board several times and your original answer matches the Parade Magazine event that Monty Hall participated in way back in the early 1990s, I don’t know why you haven’t fixed your two child answer. Martin Gardner addressed this puzzle back in 1959. This must be the third time this puzzle has appeared on the board.
“Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says “at least one is a boy.” If both are girls, he says “at least one is a girl.” And if both sexes are represented, he picks a child at random and says “at least one is a …” naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases – BB, BG, GB, GG – and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.”
Scientific American, October, 1959
Martin’s description is very important to solve this problem. If the original selection of a child is completely random, the answer is 1/2. The 1/3 solution is only obtained if we select a specific sex for the original “clue”. By doing this we have eliminated one family group, because one type of family will always say they don’t have a girl or boy and they are gone from the puzzle before the question is addressed. That is the same requirement that Scientific American stated in 1959. In the families that have both a girl and a boy, we have to assume that they will always answer one way. All the families with girls (if we are looking for girls) will always say they have a daughter even if they also have a son. As pointed out in the 1959 article, if the families with mixed children answered randomly, then the answer for original question would be 50/50.
So was Martin Gardner wrong?
No, the coin tosses have nothing to do with one another, but your door choices in a single instance of the “Monty Hall” problem do not have nothing to do with one another.