re: Solve this OT

This is incredible.

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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.

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If you end up with a $1 case and a $1MM case, what difference does it make if the host threw away non-winners or you just lucked the frick out and happened to throw away non-winners? Nothing happens during those rounds but throwing cases away. If the host throws away trash cases, that just rigidly sets the odds in the final showdown. If you randomly toss the cases one by one, the chances are constantly shifting, but each time you DON'T throw out the $1MM case it comes closer and closer to the game where the host throws out non-winners. By the time you get to the end, it's exactly the same as if the host had thrown out cases he knew were not winners, the only difference is that you just got there by luck. At any point, you could have thrown away the million and that is a different game. But you didn't throw out the million, so it's the same game.

Regardless of who selected the cases to be thrown out, if you choose one case at the outset, there's a 1/26 chance your case had the million and a 25/26 chance the million was in the other 25 cases. If the million is still in play at the end, how you got there is inconsequential. The other cases still had a 25/26 chance of having the million.

EDIT: Yeah, I fricked this one up. Point and laugh.

ok

eta: answer me this big timer .. you have to cases left $1 and $1,000,000 .. you are saying that you have 1/26 chance of having the $1,000,000



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what are your odds that you have the $1 case

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If the million is still in play at the end, how you got there is inconsequential.

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Although correct, the "how you got there" is important to the statement of the problem.

If one of the last two cases is guaranteed to be the $1MM, then the odds are as you describe.

If you choose a case, then 24 truly random cases are eliminated, the odds of either case being the $1MM are 1/13.

You are welcome to google why you are wrong on this one.

This link has the math

In a true monty Hall, the host ALWAYS opens the door with a goat. That isn’t the case here.

if you have a 1/26 chance of having the million - what are your odds of having the $1 case


ITS THE SAME ODDS BTW

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Regardless of who selected the cases to be thrown out, if you choose one case at the outset, there's a 1/26 chance your case had the million and a 25/26 chance the million was in the other 25 cases. If the million is still in play at the end, how you got there is inconsequential. The other cases still had a 25/26 chance of having the million.

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Because the host could have been lucky.

I once thought like you with regards to Monty Hall - I didn't understand why it mattered if he knew or got lucky - but I do now.

I know it seems counterintuitive, but how you arrived at the $1 case and $1MM case matters. If he was lucky, switching doesn't matter. If you were ALWAYS going to end up with $1MM in one of the two unopened cases, you should always switch.

I believe you are the idiot. If the contestant is picking two cases to be there at the end, and the contestant picks which cases are eliminated up until the end, and the contentant can pick whichever case they want to keep they have a 50/50 shot of picking the right one.

The key here is that you don’t know what you picked first. And you aren’t guaranteed that every opened subsequent case doesn’t contain 1 mil until 2 are left. That’s the key. With no guarantee, the 1 mill case could be anywhere and as you open them, the percentages distribute evenly among the remaining cases. If there are 2 left, given the randomness, it’s 50/50 for the two remaining values.