Apparently, in order to have a better chance of winning the goat you have to choose a door then change your mind 
Ok. I'm convinced now and that without me writing a single line of code as previously threatened. Further, I have seen a strategy for achieving a 100% chance of a win at this game. Simply consider a 70's era Chevrolet and a goat to be of equal value.
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The logic should still work -- you stand pat until N-2 doors are left (your initial choice, and the other unopened door) and then switch.
I am having trouble accepting SLM (switch at the last moment) as the unique optimal strategy.
I claim (with no analysis or Monte Carlo), that in the progressive variation, it makes equal sense to switch into the set of remaining unopened doors.
With 100 doors, the first pick is locked in at 1/100.
Opening the first goat-door bumps the probability of 98 doors a small amount. So I would switch if I was not sure Monty would keep offering me a choice.
At every step, the unexamined doors are reduced in number, but share a fraction of the original probability, successively greater by increments.
Played out to the end, it is the same as SLM but for all stages in between, switching is the choice.
This may depend on subtleties of the problem expression.
TBH this is where my brain starts objecting, and other activities look like they would be more fun... please do not go to any trouble on my behalf with this, I am happy to retreat to my own understanding of just the three door thing.
Still not gonna do any bar bets around it.
a7
my gut feeling is that it does not matter if you get offered a switch until only 2 doors are left.
if you start with N doors, when 2 doors are left, switching has a probability of (N-1)/N.
if you had switched every time along the way you would have improved your probability during the game but it does not matter, the end result is the same probability of (N-1)/N.
Gut feelings are dangerous. Switching every time does indeed improve over the hold-with-the initial 1/N case forever, but switching every round n turns the game into into a fresh 1/(N-n) versus (N-n)-1)/(N-n)` game, and at n=N-3 it degenerates into the 1/3 chance of winning if you don't switch on the last round and 2/3 chance if you do switch on the last round. if you held your initial 1/Nth choice until the last round and then switched, you'd get the complimentary (N-1)/N result.
I did not get what you said
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I meant that repeated switching doesn't reach (N-1)/N odds of winning.
Maybe:
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interesting - thanks fo sharing.
Haha, more Jason Rosenhouse.
I'm not going to even try, and I repeat only a simulation will be what I can work with.
I would not be surprised to find many articles just as thorough and dense and plausible (to the regular person who never knew or has forgotten any kind of maths like they use) as the several you've shared that are totally wrong.
I'm not saying these guys are, it's just those 10000 letters that arrived after M von S published a correct analysis, many from people who should have known better and doubt supplied what they felt was rigorous proof of her error.
I will see if they clarify how a progressive game play is exactly meant to go.
But from the original idea, that nothing you do will ever change the locked-in nature of 1/N being the probability that the first pick is correct, it seems like all the switching amongst the others makes no difference as long as at the end you can pick the remaining unopened door.
Which opening is informed by MH knowing where the car is.
At that point, there are two doors, one has 1/3 probability, so the other must 2/3.
That's my gut feeling of an intuition with a bit of head and arm scratching. Which have no place in this matter. 
a7
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Yes, if you switch every time up until only two doors remain, the simulation would be like all the on-line simulators with the 1/2 to 2/3 probabilities for the last choice of "hold" or or "switch"
And if you don't do any switching up until the last two doors, you've got the locked in 1/N for your initial pick, and the 1-1/N for the other door.
Folks who program Monte Carlo simulations professionally use lots of tricks/cheats/equivalences/degeneracies/optimizations to speed the simulations up, like this The Monty Hall problem -- reorder the cases to always put the car behind door number 1 and trust the RNG not to cheat. Programming a simulation for the wait-until-the-last-chance-to-switch strategy optimizes quickly into some 1/N vs 1-1/N code.
It can too be true.
Do not just think it. Do the math, enumerate the possibilities, or do a simulation to show that "It's the same [50%?] question as if the goat was already revealed when you selected".
My favorite enumeration is to think of the doors having a female goat, a male goat, and a car: Gf, Gm, & C. And using the switch strategy, there's a an even 1/3 chance of:
- You pick Gf, (in which case Monty shows Gm, and you switch to C and win.)
1/3 * 1 - You pick Gm, (in which case Monty shows Gf, and you switch to C and win.)
1/3 * 1 - You pick C, (in which case there's a 50% chance Monty shows Gf, and you switch to Gm and lose, or a 50% chance Monty shows Gm, and you switch to Gf and lose.)
1/3 * (0.5 *0 + 0.5 *0)
p_win_switch = 1/3 *1 + 1/3 *1 + 1/3 *( 0.5* 0 + 0.5*0) = 2/3
And if you stand pat, you are stuck with your first choice:
p_win_stand = 1/3 *0 + 1/3*0 + 1/3*1 = 1/3
Both strategies are different than the obvious 50:50 chance of making the winning choice in a 2-door problem of goat vs car.
Yes.
Making the choice after 1 goat door has been picked is this bit:
p_win = 1/2 *1 + 1/2*0 = 1/2
or maybe if you keep the Gf and Gm names, and the enumeration before Monty randomly picks a goat, there are two cases:
- Monty picks Gf, (in which case there's a 50% chance you pick C and win or 50% Gm and lose)
1/2 *1 +1/2*0 = 1/2 - Monty picks Gm, (in which case there's a 50% chance you pick C and win or 50% Gf and lose)
1/2 *1 +1/2*0 = 1/2
p_win = 1/2 * 1/2 + 1/2 * 1/2 = 1/2
Still different from the 3-door case where you get to choose the first door.
Show your work for how:
Interesting discussion
I don’t want to argue, but the fact that SLM increases the probability to 2/3 seems obvious to me
The question is not the probability of a door being the winning one (it’s 50/50 when you have 2) the question is whether you have better chances if you switch.
The maths and simulation shows you have 66% chances of winning if you switch rather than stay.
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@Delta_G
Do you agree that the probability of finding a car by opening two out of three doors is 2/3?
When choosing an SLM strategy, you are actually checking two doors
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https://www.rossmanchance.com/applets/2021/montyhall/Monty.html
Try with 500 or 1000 stay (or switch)
(Disable animation)
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Might also indicate if 'the fix is in'. Statistically, across multiple years I'd be looking under the roulette wheel if the numbers didn't converge.
(back to lurk mode).
It is 2/3. If you haven't, keep looking for an an explanation that works for you.
I wa snoozing for a bit and see some more discussion. I can't see if it is yet that everyone agrees with switching, TBH if I read an explanation that says everyone else has been wrong for decades and that online simulation are invalid, I'd prolly be talked into it.
I did mention the shovel, I hope no one is digging. I mention now the tent, which if not yet should soon be folded…
Here's what worked recently for a few of my mates:
There are three boxes, car/goat/goat. Anyone can see the 1/3 probability of correct box.
Move one box to the left of the bar, and two to the right. You pick is the one box, now everyone can still, I hope, agree that there is 2/3 probability the cat is in a box on the right, along with a goat box, and still 1/3 the box you picked on the left.
When Monty opens a box on the left, he cannot possibly have changed the probability that the cat is on the right hand side. All he did was make it so we can say that if the car is on the right hand side, it must be in the unopened box.
So we switch.
Key: Monty knows which door he can open...
It is here where the 100 door open 98 goat doors version makes it clear - you'd have to have been damn lucky to have picked the right box first time.
Here's a thousand more words on the matter:
I know that won't help. The unconvinced shoukd just play an online version, or make it interesting and I'll come over and take you money.
a7