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Should I stay or should I switch doors?

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The Monty Hall Problem is a brain teaser based on the popular game show, Let's Make a Deal. The folks at Numberphile explore the famous problem which posits if a contestant should switch doors in order to find the car amongst the goats.

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11 Open Discussions

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Avatar for Oleksii Talamanov
Lesson in progress

by Oleksii Talamanov

Let's extend this experiment. Let's imagine, there are 50 doors with the prize behind one of them, and host asks you to choose one of 50. After making your choice he opens 48 'empty' doors, and asks you, would you stay or change. And, following video's logic, there will be 49/50 chances to win, if you change. I don't think that logic os correct. Actual choice is between 2 closed doors at the end ('stay' or 'change'), third door is always empty. First stage is just a part of the show.

Avatar for Tarang Vadalikar
Lesson in progress

by Tarang Vadalikar

Its always 50-50... The strategy to switch,always, as propounded by the math logic is only true if the distribution follows a natural or normal order. The probability would indeed be concentrated if the distribution was untouched . Here, each of the doors would have equal chance of having the red car. The game twists the logic and instead of producing 'choices' , it add blanks or duds to the equation. It never really has more than two states. The variables being the 'door you chose' & the 'door you can switch to'. All the other doors are simply added and subtracted like zeros to the LHS & RHS. Thus, probability would be arrived at Number of items (1)/ Number of choices(2)= 0.5 Thus it is infact the most ideal 50/50 game ever.