Another way of looking at it is if Monty doesn't open either of "his" doors but asks "Do you want to change from your door to the best option of my two doors?" If you swap, there is a 2/3 chance that one of his doors has the prize. As at least one of the doors must be non-optimal, Monty opening the non-optimal door doesn't make any difference to the game.
Likewise if Monty opens both of his doors and asks "do you want to swap to the best of these doors" you would want to change 2 out of three times.
So calling all STEM teachers, we need help! So this is a "word problem", I was taught the whole trick to "word problems" (real life stuff) is to construct the equation correctly. The proof is the equation itself, seem like a simple 2 stage, one line equation is all that is needed. The problem I have is how to express the subtraction of the choice that that is the less of the two remaining choices, I think what He really does is subtract the most valuable choice that is less than the grand prize. (So Bob Bubank could describe all the things America could just go out and buy). The odds will work out the same regardless which of the two algorithms are used as long as the grand prize is not subtracted from the remaining choices.
I can construct the whole thing in python with one line, but that really a computer command not a equation. The equation should express the odds of one or the other remaining choices as one can get the other by subtracting the one you have from 100.
Wow…. 20 pages.
I’ve been out for the last half dozen pages, so I’ll take a swing.
No one reads my sh!t anyways.
Nick,
If Monty was selecting from his two doors at random and happened to select a goat, then you would be correct, the odds of either door containing the car would be 50/50.
But he’s not. Monty knows the locations of the car and goats from the outset. The draw is only random for the contestant.
Once Monty makes the informed choice to remove a goat from his two, the remaining door now has the odds are locked in a 1/3 and 2/3.
what is the relationship that affects the outcome ?!??
don't you see it by now ?
by choosing the first door we have removed that one door from the pool of three !
that is exactly it !! that is the exact event that affects the outcome of the final choice
MONTY NOW ONLY HAS TWO DOORS FROM WHICH TO PICK THE PRIZE WHEREAS WE HAD THREE
that is why he has better odds than the first player
Last edited by coelcanth; 06-15-2022 at 07:20 PM.
no that's not it
as long as the contestant has to pick first, their door will always have worse odds than Monty's two doors
Monty is not cheating at the game and he's not 'removing' the goat from play
even with a 1 in 3 chance and without switching doors, the player will still win the prize sometimes, there's just a smaller chance
it's never 50/50
Flatus has the distinct odor of our last Pratus and Flotus
I have a separate question:
The Monty Hall fans are legendary. The participants are rabid.
If there was a double probability of winning the car if you switch, and since this problem has been published on the internet for a long time, and since the fans watch the show religiously, wouldn't every participant always switch to the point that the show would stop running this game? I don't think they have because I don't think the viewers have seen the pattern that some claim is so clear.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
This simply has to be leg pulling. No-one could be so intentionality blind……could they?
Last edited by Decourcy; 06-15-2022 at 08:29 PM.
Correct.
Nick keeps saying it is between randomised events of equal probability and it isn't.
Monty opening one door showing a goat doesn't make any difference, but that could only be random if the the player picked the car to start with. Otherwise which door he opens is determined by the position of the car.
monty hall is dead.
long live the monty hall problem!
no, you and Nick are both stuck on the same point..
revealing a goat in the second round does nothing to affect the probability of a pick that was made in the past
thinking about the reveal as 'removing' that goat from play is probably the wrong way to go about it and that's what's confusing you.
revealing the specific location of the goat really makes no difference (we already knew Monty had to have AT LEAST ONE goat, right?)
the particular door that it is in has no effect on whether Monty has the prize or not
if we think of Monty's two doors as his entire hand we can say that his entire hand has a 2 in 3 chance of containing the prize
opening the one door means that entire 2 in 3 chance now rests on the only unopened door in Monty's hand
that is why his last door has a 2/3 chance and your one earlier pick has only a 1/3 chance of winning
The show has known about this for most of the show, I watched a show years ago where Monty Hall himself talked about it. And the internet was decades away! The idea the show has a problem with this is silly, they have loved it. The show is for advertisers to highlight their products while the audience is watching a show. This is the stuff dreams are made of for those guys!
I do think the equation is to subtract the highest priced prize that is less than the grand prize from the two remaining choices.
33%, best offer!
There's a lot of things they didn't tell me when I signed on with this outfit....
Also bear in mind that in the real Let's Make a Deal!, there's a whole lot more going on.
Monty sometimes offers cash up front to entice the contestant to switch (or not switch).
The lovely Carol Merrill pops up out of a trap at random with a cart carrying a box that might or might not contain something of value, and that gets figured into the equation, too: "Stick with the door you selected? Or switch to the box that the lovely Carol Merrill has for you?"
Sometimes, it's even, "You switch doors? Great! I've got another deal for ya -- the lovely Carol Merrill has just brought this box on stage. Would you be interested in switching to that?"
The "Monty Hall Problem" is the stripped-down minimal example for analysis.
You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)
How can the probabilities of choosing between 2 different doors be 30% and 50%?
I apologise for not having read everything since I originally posted, but am somewhat short of time.
However, respondents seem to have missed my point. Perhaps I was not clear enough.
Contestant chooses a door. Monty opens one of the two remaining doors to reveal a goat.
Statistics say that Contestant can double his chances of winning the car by switching.
But it doesn't seem to matter which door the contestant originally chose.
He chose door 1; his chances double by switching to door 2.
He chose door 2; his chances double by switching to door 1.
This is the peculiarity i would like to see explained.
Because all doors are equal. They all have an equal chance to win.
Once you have chosen a door, you have a 1 in three chance of winning. Monty has a 2 in 3 chance. You know one of his is wrong from the start. Your choice of door doesn’t affect the odds. And he can always show you one door with a goat because he HAS to have one, and he knows which it is.
Tom
Simple explanation
Comprehensive lecture by a PhD at Harvard, first 25 minutes of his talk.
Edited to add, the lecturer is Joe Blitzstein, Professor of the Practice in Statistics, Harvard University, Department of Statistics
Last edited by oldcodger; 06-16-2022 at 04:42 AM.
What seems to be the consensus amongst some here is that the set of probabilities of first choice (1/3) is carried over somehow to the second set of probabilities in the second choice (1/2).
Nick (and I) can not see that they are connected in anyway. How are the 2 sets of probabilities connected?
To my mind it doesn't matter a damn which door I chose on first choice, or which door is the goat reveal, the second choice of 2 doors is completely unrelated to the first choice of 3 doors, yet some here say they are entangled some how?
So what's the mechanism?
The are connected because the “game” was never reset. Monty didn’t open a door randomly leaving you with new choices. You new choice is really do I stick with my one door, or do I choose both doors that Monty had. 796 is a good explanation.
What information did you really gain by him showing you what was behind one of his doors? If, before opening a door to reveal a goat Monty asked if you wanted his two doors or stick with your 1 door would you stay or switch? You would switch, because you now have two chances. You know one of them has a goat but the odds of a win are better.
Tom
I want an equation, I know it has to exists. else the answer is wrong!
And 1 car 1 medium prize and 1 goat, not 2 goats!
OK. Go here: https://medium.com/@judaikawa/the-mo...m-4415e50e233f. The article to which I've linked shows a different statistical equation solving the solution than that in the Harvard video (using the Law of Total Probability). The Harvard video is very good. It is also clear (I could follow it).
After viewing the video and then reading the linked article and working through the Bayesian equation, actually play the game yourself. Do what JimmyW suggested, take an Ace and two Jokers out of a deck of playing cards. Enlist a friend to take the Monty Hall role. Play over and over for an hour. Bet a dollar to win on each play. Never switch. Be sure to return and tell us if your friend thanked you.
.
Last edited by Tom Montgomery; 06-16-2022 at 05:53 AM.
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
You accept that the initial choice is a 1/3 change of getting the car.
If you choose to switch your first choice for Monty's other door, then you reverse your choice - if you chose the car and switch, you get a goat, if you chose a goat and switch you get a car.
Therefore, unless you can come up with a scenario where switching to Monty's unrevealed door does not reverse the outcome of your initial choice, switching doors must have a 2/3 chance of winning the car.
'When I leave I don't know what I'm hoping to find. When I leave I don't know what I'm leaving behind...'
.
The offer is to switch your one door (with a 1/3 chance of having the car) for the aggregate of the remaining doors (which have a 2/3 chance of having the car). The odds the aggregate has a 2/3 chance of having the car do not change if one of the doors of the aggregate is revealed to not contain the car. You knew that already.
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly