You can find the results of a random switch quite easily without calculating anything, it’s the reason so many people intuitively think the odds are 50-50.
If you re-read my post your response is actually illustrating the phenomenon I was noting quite nicely.
Perhaps you should re-read the post youself.
You might want to consider the significance of the sections I have bolded.
There is no error.
I didn’t say that there was an error. Perhaps you might explain to our audience why the terms 1/3 and 2/3 appear where they do in that formula?
Ok I’ll explain it to you Spiritus Mundi.
My first post was aimed at people who accept there is a 2/3 chance of winning if you switch doors. The equation clearly establishes this belief is entirely compatible with the belief that the contestant has a 50% chance of winning. Fifty percent of the time you switch doors (and win 2/3 of the time) and fifty percent of the time you stick (and win 1/3 of the time).
Obviously this 50% winning outcome figure is entirely unaffected by whether you happen to know the switch/stick probabilities, and there are other ways of arriving at it.
If this simple fact was stressed more often when explaining the solution, I believe the solution would be a lot less hotly contested, and attention would be drawn to the fact that often when people debate this puzzle they’ve been hoodwinked into not realising they’re not actually trying to answer the same question.
I apologise that this wasn’t easy for you to understand.
I wouldn’t talk down to Spiritus if I were you, rosasaks. He’s one of the brighter mathematical minds on this board.
Personally, I find your explanation to be extremely confusing, and I’ve got a math degree. I don’t think it’s going to be very helpful to those who don’t already understand the problem.
Fair enough. I am one, and I neither dispute nor find interesting the fact that a coin flip to determine which of 2 doors hides a prize when exactly 1 is guaranteed to find a prize yields a 50% probability.
The issue in dispute is with those folks who do not agree that switching every time yields a greater probability to win. In your analysis, you begin from that point and then say that you are adding something new to ther discussion. I disagree.
True, and if you could use the compound probability to argue backward from this point to the probability of switching or not switching that might be useful. Unfortunately, no such rigorous argument is possible. .5x + .5(1-x)=.5 has infinite solutions.
I don’t see why. This fact adds nothing to the analysis of whether ones odds improve by switching.
It is not difficult to understand. I simply disagree.
ultrafilter, you are too kind. There are far more competent mathematicians on this board than I, yourself included.
This speaks to the point I was trying to convey.
I suspect that more often than not the fault is not with the analysis of whether ones odds inprove by switching. The fault is that people get mesmerised by the fact the contestant has a 50-50 chance of winning irregardless of whether she sticks or switches and so fail to analyse the proper question entirely. This is compounded by the fact it would hold true irrespective of whether the contestant is even aware of the outcome of the initial choice of door. Furthermore I believe it’s surprisingly common for people who do know the correct answer to fail to identify this mistake as the real obstruction preventing others from comprehending the solution. Maybe you feel this is a worthless observation, but if you read some of the protestations against 2/3 upthread you’ll see it really is close to the heart of why this topic gets rehashed again and again. C’mon, have you never wondered why this relatively easy probability puzzle always causes so much consternation when people who really should know better see it for the first time? It’s such an old chestnut shouldn’t the discussion really be about why the Monty Hall problem confuses people?
I’m just suggesting that if you correct the simple misconception I describe above the analysis of whether ones odds improve by switching ceases to be problematic because people will either pay attention to it for the first time, or will stop feeling resistant to it because they find the approach counter intuitive.
I was not providing a solution to the puzzle. I was outlining why I believe people either reject the solution or fail to look for the right one.
And you think that pointing out the 50-50 chance that would exist if it were a different problem will help people see why the current problem does not represent a 50-50 outcome?
shrug I see no reason to share your optimism, but if it works for you knock yourself out.
Not quite. The 50-50 chance is a legitimate part of the Monty Hall problem. The contestant with no strategy really does have those odds of winning or losing. In order to convince somebody who believes it doesn’t matter whether you stick or switch that one is beneficial and one is detrimental you need to introduce the concept that 50-50 and 66-33 are not contradictory in the way that she has assumed. Rather 66 can be the chance of winning if you pick the first “50” and 33 can be the chance of winning if you pick the second “50”. Add it all up and contestants do indeed win 50% of the time, it’s just more winners switched than stuck. This way you’re not throwing out somebody’s intuition that the contestant has a 50-50 chance of winning, you’re affirming it, but showing the need to add another layer.
Yes, but the whole point is that they cannot agree with the 67% and 33% outcomes. If someone refuses to be convinced by an exhaustive case listing, then I remain skeptical that they will convinced by the 50-50 chance of a 67%-33% split Still, as I say, if you find that illustration gets you better results in fighting ignorance, more power to you.
BTW – the Monty Hall problem explicitely asks which strategy is best. “No strategy” is not an option. “Random chance” is not the correct answer, since it yields a sub-optimal result.
I’ll add my disagreement that this observation really helps. To amplify a point that Spiritus Mundi makes in post #127…
If the player adopts a strategy of flipping a coin to determine whether he switches or stays, then no matter what the winning odds are for switching or staying, he will win 50% of the time. That is, if P is the probability of winning under Always Switch, then the Flip a Coin strategy (or pseudo-strategy) will win on a portion of 0.5P + 0.5(1-P) = 0.5. The value of P actually becomes irrelevant then. It could be 2/3, or 1/2, or anything from 0 to 1.
People who misunderstand the Monty Hall problem sometimes say that you “might as well” flip a coin. When they do, they’re effectively claiming that P = 1/2, and this is probably what they have in mind. But that right there is the whole crux of the dispute. We know that actually P = 2/3, and therefore the player shouldn’t flip a coin. (In fact the player shouldn’t randomize his decision at all; he should always switch.) I don’t think considering the effects of a random decision helps to see this though.
Well now, I’m on your side, but surely you’ve encountered math problems before that ask trick questions? (Man I know I have.) The Monty Hall problem isn’t one of them, but there certainly are puzzles that ask for a solution that doesn’t really exist, or a best strategy where there is none, or a number satisfying certain conditions when there is, provably, no such number. In principle Monty Hall could have been another one of those puzzles, as in fact people must believe when they’re giving their “50-50” answers.
Sure, but the point that I guess I did not make clear is that “random choice” is not the same as “no strategy”. Random choice is actually a quite accetable strategy in Game theory (though some folks prefer to call it a random selection between strategies.) Some games might conceivably have a “no strategy” means of playing (can’t think of one off-hand), but in the Monty Hall problem the only “no strategy” I can see would be to not play the game (make the choice) at all.
On review, I see that I could have phrased that more clearly.
I’m still reeling at the thought that after Monty reveals a zonk, the odds are not yet fixed. We have to ask him if he knew the door was a zonk or not; if he just randomly picked one, (which resulted in a zonk), then the odds are 50-50. If he intentionally opened a zonk he was aready aware of, then it’s 33-67.
Despite being physically identical situations, his intangible “motivation” actually determines the concrete probability.
That’s some fucked up shit right there.
You got that right! I spent two hours reading this thread, and it was somewhere around post #90-something before I grasped it. I think it was the card problem that made me say “oh yeah, duh!”.
If somebody messes around, knowingly intervenes with the probabilities somewhere in the process, it ain’t what it seems like it ought to be. If they don’t know, the last two doors or cards are 50/50. Fucked up shit indeed!
I’m glad there’s math guys out there to worry about this stuff, 'cause I damn sure don’t want to do it.
Well, I’ve been busy lately. One of the things I was busy with was cleaning out my mail box including the mail I sent my Dad.
I’ll see if he can send it back to me. If that happens it’s gonna take a bit because he is on vacation and now going somewhere without phone service so obviously he won’t have email.
I probably stated the problem unclearly but don’t remember what I wrote. I dpn’t know if I meantioned that Monty always picks a door without the prize.
Slee
This seems rather obvious to me. Suppose you and I are betting on the “up” side of a coin: heads => I win; tails => you win. Consider these two situations:
Unless you have very unusual skill in coin tossing, #1 will give me a 50% chance of winning. But in #2, the odds are whatever you choose. IOW, these are physically identical situations, in one of which your motivation determines the outcome.
How is that physically identical?