Suppose you chose door one. Monty opens door three and there is a goat. So you swap to door two. YOU WIN!!!

Suppose you choose door two. Monty opens door three. You swap and YOU LOSE!!!!

Suppose you chose door three. Monty opens door one you swap and YOU WIN!

These are the only possible senarios and in 2 of them you win.

Just forget everything you know and it makes sense

]]>1. Pick the correct door(1/3 of the time). Monty opens either of the other doors. You switch, you lose.

2. Pick an incorrect door (2/3 of the time). Monty opens the only other losing door. You switch, you win.

]]>If there are three cards to choose from we know that there is a 33% chance of winning. But, when big Monty reveals a goat door, that eliminates one possibility from the scenario, because the contestant knows not to pick that door. So then there are two doors left. Forgetting all context, one knows that 50% of all those two doors wins – switching or no, thats the new probability of winning. – but this isn’t my point of view, because this alone still doesn’t seem right to me.

I view it as an equation, if you will, in which a variable, the door chosen by the contestant, is to be identified as winning or not. In the beginning, there are three possibilities: losing door A, losing door B, and winning door; therefore, we are 33% sure of the door’s identity. When Monty reveals a losing door (A or B, like we care), there are only two possibilities left – we’ll say losing door B and the winning door. Now the equation is simplified a step and we’re 50% sure of the door’s identity. Finally, Monty reveals the chosen door (simultaneously eliminating the other door, making only one possibility), and we know (100% sure) that it’s either the winner or loser B. No switching is required because that doesnt change the number of possibilities available, it just kinda symbolizes that one of the two possibilities was chosen.

]]>Big Bob had it right. Your original choice is only right 1/3 of the time, so it makes sense to switch after Monty opens one of the other doors. The odds are 2/3 that the money is behind one of the doors you didn’t pick. Monty gives you some really useful information by eliminating one of those doors. The odds are still 2/3 that the money is behind one of the other doors, but now one of those doors has been eliminated for you.

Consider this: if it were equally good to switch or stick with your original door, then you could stick with your door and be right 50% of the time. That means that, presented with 3 doors, you can pick the right one 50% of the time. Nice work, Kreskin!

Or consider this: Instead of 3 doors, there are 100. You pick one, then Monty opens 98 of the remaining 99 doors, to reveal nothing but booby prizes. Would you stick or switch to the other unopened door?

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