re: Solve this OT

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well if all the boxes were picked at random and you get to the end with two boxes IT MAKES NO DIFFERENCE IF YOU SWITCH

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Gotcha. Yeah, the randomness changes it.

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so if you randomly throw away each briefcase until there are two remaining...then there's a 50% chance that one of the remaining briefcases has the $1MM?

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So I may be jumping the gun on this analysis but actually I think Mailman is correct in this instance.

The format in which the unknowns are eliminated, assuming you have the 'option to switch' at the end, matters. Because what is essentially happening, is you are picking two briefcases in the beginning. The one you announce, and the one you decide will be the last one opened. The fact that the 24 briefcases are opened one at a time or all at once doesn't matter.

You've got a 1 in 26 chance of picking the right briefcase, that's all there is to this problem. If my lucky number is 4, and I pick 16 instead with the intention of making 4 my second to last briefcase because I think that will somehow increase my odds by being able to switch to 4 with two cases left, that's just ignint. I still determined that 4 was going to be 'my case' in the beginning.

Now, if an omnipotent person was eliminating briefcases and you KNEW only the non-winners would be eliminated first then yes, you obviously switch. But that isn't the case.

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If you pick the cases one by one and get down to the end and the million dollar case is still in play - there is a 50% chance you have it and a 50% chance you don't.

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If the 1 Mil is still in play you should switch. It's the equivalent of you having 1 gold ball. Knowing the mil is still available and that your first pick was 1/26, you should switch.

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Knowing the mil is still available and that your first pick was 1/26, you should switch.

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Only if you know the host has to pick non mil cases. That's key because then it's not random. Keeping it random means monty hall doesn't apply.

This seems too easy.
If it's box 1 then 100% box two after grabbing gold, 0%
Can't be box three.

100+0=100/2=50

50%


This is wrong.

Three variables not two

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100+0=100/2=50

50%

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This seems too easy.
If it's box 1 then 100% box two after grabbing gold, 0%
Can't be box three.

100+0=100/2=50

50%

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Nope.

Yep. I showed my work.
. Show yours.

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I agree .. but it was stated that you should always switch cases and that makes zero sense. there is a 1/26 chance at the start that you picked the million dollar case .. or a 25/26 chance you didn't pick it .. If you pick the cases one by one and get down to the end and the million dollar case is still in play - there is a 50% chance you have it and a 50% chance you don't.

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You all weren't paying attention to my original hypothetical - the host KNOWS which case has $1MM and cannot select that case to discard it.

That's the key part.

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That's the thing. It would happen so rarely (4% I think or less).

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8%, theoretically.

If you 'decide' on your 2 last cases at the beginning, you've got a 1/13 chance that one of them is the million. So you'll get down to 2 cases ~8% of the time.