[Q] Bayesian explanation of the Monty Hall paradox?
Bob is on a game show and he can choose between three doors. One door has a car behind it, the others have goats. Say Bob picks door 1. The show host then opens door 3 to reveal a goat. (The show host knows what’s behind all the doors.) He then asks Bob if he wants to switch to door 2 or stick with door 1. Should Bob switch?
I know that switching the door can be shown to be the better choice 2/3 of the time, but how do we explain this in a Bayesian way? This is my attempt but it's an incomplete explanation: The fact that the host picked door 3 gives you nontrivial information about door 2, because we know that the host does not want to open the door with the car behind it (because that would make for a boring show). You should use that information to update your belief/probability that the car is behind door 2. But why do we specifically double that probability, going from 1/3 to 2/3? How does the extra information we received exactly double the probability? Edit: I'm looking for an intuitive explanation rather than a blind application of Bayes Rule. Though as pointed out by a comment this approach does not necessarily need to be called Bayesian
ELI5: Does the Monty Hall Problem depend on the host knowing which door has the car? If the host has no idea, how does that affect the probabilities and should you still switch doors?
Does the Monty Hall Problem work with two contestants?
Ok. So Monty Hall problem from somewhere online: “Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? Believe it or not, it’s actually to your benefit to switch: If you switch, you have roughly a 2/3 chance of winning the car. If you stick to your original choice you have roughly a 1/3 chance of winning the car.” But if we had two players, both playing against eaxhother; do they both have a 2/3 chance of winning the car? So in the above example, Person A picks door 1, Person B picks door 2, the host tells them Door 3 does not contain the car but they can switch, do they both have a 2/3 chance of winning if they switch? Does playing against each other vs they can both win make a difference? Does playing independantly not knowing the other persons choices make a difference? If they both do have a 2/3 chance of winning if they switch, how can this be possible does probability not need to equal 100%?
End Of Show Deals On The Old Monty Hall "Let's Make A Deal"
On the old version of Let's Make a Deal, with Monty Hall, after the Big Deal of the Day, with picking 1 of 3 doors was finished, if there was time, Monty used to go into the audience and offer to trade common items for cash. Maybe I'm remembering this wrong, but I swear, once he told a guy he would trade him, I believe it was $50, for every penny he had on him. $50 was not an insignificant sum of money in the late '60's. This guy started pulling out rolls of pennies, at least 5 or 6, with a big smile on his face. Monty looked like he wanted to pass out. The guy ended up getting more than what the top prize on the Big Deal of the Day was worth. I was only 10 or so at the time, but it was so vivid. Any other oldtimers remember this, or perhaps a similar Hall-audience deal?
Another Monty Hall problem: There are four doors and one prize, what is the probability the prize is behind door B?
These details matter: There are four doors, A, B, C and D. One conceals a prize. Monty Hall knows which door conceals the prize. You choose door A. But it has not yet been opened. Monty announces "I will choose one of the remaining doors (i.e., not A) and open it. The door I choose will not conceal a prize." He opens door D. Now the only doors left unopened are A, B and C. What is the probability the prize is behind door B? I will describe a solution when the poll closes. View Poll
In the Monty Hall problem, I understand why the probabilities on the revealed doors collapse to zero. However, why do those probabilities only add onto the unchosen door? Why do they not equally distribute to the chosen door? Is it something to do with the difference between being chosen and not chosen? Thanks in advance!
Hello, I've tried googling and searching for an answer to my question but I can't find an answer and my maths is too rusty to trust my intuition. In the Monty hall problem, I understand that switching is always the better option as it increases your odds. I also understand that Monty can't pick the winning door and must reveal a goat. My question isn't about refuting that or asking for help to understand/prove it. I have two scenarios I'm curious about and I understand both change the problem and therefore the conclusions that can be made, but I just want to know for my own peace of mind. A. In the original Monty hall problem, if an outside actor with no information replaces the original contestant what are their odds of winning after monte opens a door? I.e is it a 50/50 chance for said actor given they did not pick from 3 possible doors? Are they making an independent random choice, or does the original contestant's restrictions upon monte alter the odds? Similarly, if the monte hall problem is performed with 100 doors, what are the independent actors' odds of victory when presented with the final choice? B. In the monte hall problem, are the odds of winning different from the odds of what's behind each door in a random selection? I know this isn't the original problem's remit, but I have a feeling a lot of confusion people have pertaining to the problem comes from some of these assumptions and how people abstract the choices in their heads. Sorry if this is an obvious question.
Was recently thinking about the Monty Hall problem again
I recently found myself having to explain the Monty hall problem to someone who knew nothing about it and I came to an intuitive reasoning about it, however I wanted to verify that reasoning is even correct: Initially, the player has 1/3 probability of getting the car on whatever door they pick. Assuming that’s door 1, the remaining probability amongst doors 2 and 3 is 2/3. Assuming the host opens door 2 and shows it as empty, the probability of that door having the car is immediately known to be 0. That means door 3 has 2/3 - 0 = 2/3 probability of having the car. So that’s why it’s better to switch. I’m aware there’s a conditional probability formula to get to the correct answer, but I find the reasoning above to be more satisfying lol. Is it valid though?
Hi, I'm looking for puzzles that are similar to the monty hall problem. Anything where the solver is asked to make a decision that then bears on the chance of success. The Monty Hall problem is fantastic because it's so counter-intuitive but I would be happy with any puzzles that involve the solver having to make a decision based on the state of play (nothing too complex--but roughly the difficulty of a "mate in 2" chess problem). Any help?
The Monty Hall problem was discussed at length over on /confidentlyincorrect with many many Redditors taking a crack at explaining why switching doors when offered the chance improves your odds. Ran the simulation in the code below a million times, then ran that a million times and it consistently produces that switching your choice when Monty shows a loosing door among 3 doors improves your chances of winning from 1/3 to 2/3. This makes more sense if you consider 100 doors and let the host open 98 of them. Staying with your original choice locks you into 1/100 odds, switching improves those to 99/100
def montyhall(doors=3, monty_doors=1, iterations=1000000): result={'stay':0, 'switch': 0} for x in range(iterations): # setup doors (default 3), put a car behind only 1 doorlist=list(range(1,doors+1)) choice_correct = random.choice(doorlist) # contestant chooses any of the doors, with no knowledge of whats behind contestant_choice_1 = random.choice(doorlist) # monty starts with the the same set of doors, with knowledge of what's behind available_to_monty=doorlist.copy() # ... and with some rules he must follow available_to_monty.remove(contestant_choice_1) # can't choose the same door as the contestant if choice_correct in available_to_monty: # can't reveal the winner right away available_to_monty.remove(choice_correct) for y in range(1,monty_doors+1): # Monty opens as many doors as directed (default 1) monty_choice = random.choice(available_to_monty) doorlist.remove(monty_choice) # remove that door from what the contestant can choose available_to_monty.remove(monty_choice) # remove that door from what Monty can choose # now the contestant has a decision to make decision = random.choice(['stay', 'switch']) if decision == 'stay': # final choice is the first choice contestant_choice_final = contestant_choice_1 else: # pick another door doorlist.remove(contestant_choice_1) contestant_choice_final = random.choice(doorlist) if contestant_choice_final == choice_correct: result[decision]+=1 # add 1 to the number of games won for that choice print(f"doors: {doors}") print(f"stay: {(result['stay']/(result['stay']+result['switch']))*100:.5f}%") print(f"switch: {(result['switch']/(result['stay']+result['switch']))*100:.5f}%")
Why I don’t agree with the Monty Hall problem. [D]
Edit: I understand why I am wrong now. The game is as follows: - There are 3 doors with prizes, 2 with goats and 1 with a car. - players picks 1 of the doors. - Regardless of the door picked the host will reveal a goat leaving two doors. - The player may change their door if they wish. Many people believe that since pick 1 has a 2/3 chance of being a goat then 2 out of every 3 games changing your 1st pick is favorable in order to get the car... resulting in wins 66.6% of the time. Inversely if you don’t change your mind there is only a 33.3% chance you will win. If you tested this out a 10 times it is true that you will be extremely likely to win more than 33.3% of the time by changing your mind, confirming the calculation. However this is all a mistake caused by being mislead, confusion, confirmation bias, and typical sample sizes being too small... At least that is my argument. I will list every possible scenario for the game:
pick goat A, goat B removed, don’t change mind, lose.
pick goat A, goat B removed, change mind, win.
pick goat B, goat A removed, don’t change mind, lose.
Can the logic for the Monty hall problem be used to take advantage of the “50/50” lifeline in who wants to be a millionaire.
Say you had zero clue about an answer and but note down “A” in your head. Answers B and C are taken away. Are your chances higher if you switch to answer D.
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