Yeah you can open 1000 doors if you want; AT NO POINT IN TIME do you even have a single % probability to get what he opens. So you're saying you have a 33.3% to get Monty's goat?! I see there are two doors there, but you have ZERO % probability to get what he opens. According to the 1/3-2/3 Cult, you have a 1/3 probability to get the car, and a 4/3 probability to get the goat. If you have 2/3 probability to get the goat, where are the goats Monty opens?
@ericpeterson9110
Ай бұрын
Are you ok dude?
@TristanSimondsen
Ай бұрын
@@ericpeterson9110 Ah, another delusional member of the 2/3 Cult. Dude, the Monty Hall Problem is NOT about opening one door from three. Player has TWO sequential picks, one after Monty eliminates a no-car door. Player has ZERO probability to EVER pick Monty's no-car door. The cult likes to answer the Monty Hall Problem without Monty and player only has one pick. The Monty Hall Problem asks, does the player have a better chance to stay with his original pick, or switch TO THE REMAINING door after Monty eliminates a no-car door to pick the car. And the 2/3 Cult will run simulations that basically ask the question, what is the probability of not getting the car if you open one door from three. Put in the correct parameters, NO simulation will give you a 2/3 probability to NOT get the car with two doors to choose from. Import random def simulate_game(): doors = ['car', 'goat', 'goat'] # Shuffle the doors randomly random.shuffle(doors) # Player's first pick first_pick_index = random.randint(0, 2) first_pick = doors[first_pick_index] # Check if the player picks the car from the initial pick if first_pick == 'car': return False # Player picked the car # Host eliminates one incorrect door (goat) remaining_doors = [i for i in range(3) if i != first_pick_index and doors[i] == 'goat'] eliminate_index = random.choice(remaining_doors) # Player's second pick (switch to the remaining door) second_pick_index = next(i for i in range(3) if i != first_pick_index and i != eliminate_index) second_pick = doors[second_pick_index] # Determine if the player picked the car after switching return second_pick == 'car' # Simulate the game 100 times not_picked_car_count = 0 for _ in range(100): if simulate_game(): not_picked_car_count += 1 # Calculate probability of not picking the car probability_not_picked_car = not_picked_car_count / 100 print(f"Probability of not picking the car: {probability_not_picked_car:.2f}") Calculation: Probability of not picking the car in Pick A: P(not car in Pick A)=⅔ Conditional Probability of not picking the car in Pick B, given Pick A was not the car: After the host eliminates a door that does not have the car, the scenario reduces to two doors: one that the player initially picked (not the car), and one that wasn't picked (which could be the car). If the player initially picked a door without the car (probability ⅔), there is a ½ chance that the car is behind the remaining door after the host's reveal. Therefore, the probability of not picking the car in Pick B, given that Pick A was not the car, is: ½ Overall Probability (Law of Total Probability): Combining these probabilities using the law of total probability: P(not car in Pick B) = P(not car in Pick B∣not car in Pick A) ⋅ P(not car in Pick A) Substitute the values we have: P(not car in Pick B)=½ ⋅ ⅔ = ⅓ Conclusion: Therefore, the probability that the player does not pick the car in Pick B is ⅓, or approximately 33.33%. This aligns with the conditional probability calculated based on the initial pick and the subsequent revelation by the host. BTW, the “switch” option in the ⅔ hoax actually refers to switching BACK to the initial door you picked, not switching to the remaining door that you DIDN’T initially pick. I’ve seen simulations where the score is completely skewed due to the confusion of which door the code is scoring for “switch”. This is a clear case of reductio ad absurdum. Let me put this to bed. If you're picking one door from three and there is a car randomly placed behind one of the doors, you have a ⅓ probability to get the car and a ⅔ probability to not get the car. The Monty Hall Problem, on the other hand, is a 50/50 guess with the remaining two doors because after Monty eliminates a no-car door, you have a ⅓ probability to get the car, and a ⅓ probability to not get the car. 🤙
@ericpeterson9110
Ай бұрын
@@TristanSimondsen So you ran a simulation and found that in 50% of games the "player" picked the car initially? Maybe your PC needs checking in addition to your head?
@TristanSimondsen
Ай бұрын
@@ericpeterson9110 And you believe you have a 1/3 probability to pick Monty's door? 👌
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