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#### RE: Would you switch?

in LeoFinancelast year

I would say it would always be a 50/50 chance of picking the right door. In first try it would be 50/50 between winner pick or non-winner, even if one options has two doors left.

Once you pick the wrong door, it is still 50/50 chance to pick the winner between remaining doors. Opened door becomes irrelevant at that point.

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last year

How could it be a 50/50 chance when you are picking one out of three on the first try?
You are a Python dev, program it out, you will see it isn't 50/50 in either situation.

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last year (edited)

Because ultimately everything comes down to 0 vs 1, True or False.

Especially in this case, since I can open only one door at a time, my options are:

a. Door 1 vs (Door 2 + Door 3)
b. Door 2 vs (Door 1 + Door 3)
c. Door 3 vs (Door 2 + Door 3)

Whichever I choose, I face with 50/50 chance.

The only difference is, with the first try I have 50% chance to get the correct one, and 50% to get another opportunity to try. With the second try, I have 50% chance to get the correct one, and 50% to lose.

But also, coins only have two sides. I would toss a coin to decide for me. And for that reason I need to make it 50/50.

P.S. I misunderstood the puzzle at first. But I would still use the coin toss, and treat each step as 50/50 chance. So can't tell if I would switch or stay until I flip the coin.