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nomadking · 06-12-2022, 21:28

With the 3 doors scenario, if you pick correctly first time, then the host has TWO doors to pick from. If you pick one of the 2 the wrong ones, then only ONE to pick from. Overall there are 4 possible end results.
There is a prize door and your initial choice of door. There is a 50/50 chance they are one in the same. That is the question you are being asked, no matter how many doors you start with.

The mistake has been made in treating all non-prize doors as a single outcome.
Prize is behind A.
Combination 1 : Pick A, Host opens B, stick wins
Combination 2 : Pick A, Host opens C, stick wins
Combination 3 : Pick B, Host opens C, swap wins
Combination 4 : Pick C, Host opens B, swap wins
If you're going to treat 1&2 as a single combination, then you have to treat 3&4 as a single combination. That still brings you around to a 50/50 chance.
Now 4 starting doors, prize is behind A
Comb 1 : Pick A, opens B&C, stick wins
Comb 2 : Pick A, opens B&D, stick wins
Comb 3 : Pick A, opens C&D, stick wins
Comb 4 : Pick B, opens C&D. swap wins
Comb 5 : Pick C, opens B&D. swap wins
Comb 6 : Pick D, opens B&C. swap wins
50/50 again.
Would you when throwing a pair of dice, treat all non-double six combinations as single combination?

06-12-2022, 21:28	#31
nomadking cf.mega poster Join Date: Apr 2004 Location: Northampton Services: Virgin Media TV&BB 350Mb, V6 STB Posts: 8,152	Re: What are the odds of winning? With the 3 doors scenario, if you pick correctly first time, then the host has TWO doors to pick from. If you pick one of the 2 the wrong ones, then only ONE to pick from. Overall there are 4 possible end results. There is a prize door and your initial choice of door. There is a 50/50 chance they are one in the same. That is the question you are being asked, no matter how many doors you start with. The mistake has been made in treating all non-prize doors as a single outcome. Prize is behind A. Combination 1 : Pick A, Host opens B, stick wins Combination 2 : Pick A, Host opens C, stick wins Combination 3 : Pick B, Host opens C, swap wins Combination 4 : Pick C, Host opens B, swap wins If you're going to treat 1&2 as a single combination, then you have to treat 3&4 as a single combination. That still brings you around to a 50/50 chance. Now 4 starting doors, prize is behind A Comb 1 : Pick A, opens B&C, stick wins Comb 2 : Pick A, opens B&D, stick wins Comb 3 : Pick A, opens C&D, stick wins Comb 4 : Pick B, opens C&D. swap wins Comb 5 : Pick C, opens B&D. swap wins Comb 6 : Pick D, opens B&C. swap wins 50/50 again. Would you when throwing a pair of dice, treat all non-double six combinations as single combination?