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quote:IF there were 25 cases of $1, right. But for the way the game is actually played (26 unique #s), it will do you no good to switch cases b/c eithe of your 2 remaining cases could be either remaining amount 50-50.

But at the beginning there was a 96% chance of 1$ right?

It is the fact that Monty is eliminating a door he knows is wrong and will never reveal the car that mathematically differentiates the two, right?

Every time I'm convinced I get it, I doubt myself.

Every time I'm convinced I get it, I doubt myself.

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but there are only two left and each has the same odds .. does no good to switch ...

Yes there are two left and they have the same odds. But a large factor of the case you picked originally still being there is because it's the one you picked originally. It is very unlikely you picked correctly originally, therefore switching is the smart decision

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by threeputt on 3/6/13 at 3:23 pm to NaturalBeam)

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ut for the way the game is actually played (26 unique #s), it will do you no good to switch cases b/c eithe of your 2 remaining cases could be either remaining amount 50-50.

it does not matter if the cases had unique numbers or not ... the last two cases are 50/50 for your desired result (if you get down to that point)

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What I'm saying, and maybe not explaining it very well, is that it is very unlikely that you picked the case you were trying to pick at the beginning.

Not as unlikely as eliminating 24 straight wrong cases.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by NaturalBeam on 3/6/13 at 3:23 pm to Tiger1242)

quote:My point is that it doesn't matter which case I subjectively want to pick. I've picked a case, and eliminated 24 others. There is my case left and 1 other. It doesn't matter which 2 values are left on the board (In this case, let's say they miraculously are $1 and $1M), there are equal odds of it being either.

I see your point and if you were trying to pick the $1 case originally than you shouldn't switch. I realize this is not like the Monty Hall situation and it isn't 96% chance your case is the wrong one. What I'm saying, and maybe not explaining it very well, is that it is very unlikely that you picked the case you were trying to pick at the beginning. So if you have a chance to switch with better odds, you should switch IMO

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Yes there are two left and they have the same odds.

He gets it!

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It is very unlikely you picked correctly originally, therefore switching is the smart decision

Oh. Never mind.

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It is the fact that Monty is eliminating a door he knows is wrong and will never reveal the car that mathematically differentiates the two, right?

For the Monty Hall situation yes, someone with knowledge eliminated an option he knew was wrong, so the odds are now on the favor of the one he did not eliminate

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. It is very unlikely you picked correctly originally, therefore switching is the smart decision

it is very unlikely that the OTHER case left has the million as well .. they both have the same odds of having the million

quote:this logic has converted me. i'm switching

Yes there are two left and they have the same odds. But a large factor of the case you picked originally still being there is because it's the one you picked originally. It is very unlikely you picked correctly originally, therefore switching is the smart decision

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by OneMoreTime on 3/6/13 at 3:27 pm to LSUBoo)

quote:Yeah, it seems like it would be more unlikely for the million dollar case to be out there and for you to eliminate 24 straight $1 cases than it is for you to have picked the million dollar case to start with. Don't feel like running the odds on that though.

Not as unlikely as eliminating 24 straight wrong cases.

This post was edited on 3/6 at 3:28 pm

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it is very unlikely that the OTHER case left has the million as well .. they both have the same odds of having the million

I know the odds shifted and it's 50/50, I'm just switching based on the fact that the only reason the case I picked is still around is because I picked it, it was likely I picked wrong.

I'm really only referring to the unlikely situation that the $1,000,000 was still around. I'm just assuming I didn't pick the $1,000,000 off the bat.

I'd probably take the deal anyway

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by threeputt on 3/6/13 at 3:29 pm to OneMoreTime)

quote:

Don't feel like running the odds on that though.

I hear you .. that is too hard ... nobody got time fo dat

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by NaturalBeam on 3/6/13 at 3:29 pm to threeputt)

quote:It does matter. If you started with a $1M case and 25 $1 cases, you have a 3.8% chance of picking the $1M case. If (big if) you then eliminated 24 other cases and were still left with the $1M, you still only have a 3.8% chance of having picked the $1M initially. You had a 96.2% chance of picking a $1 case.

it does not matter if the cases had unique numbers or not

If you have 26 unique cases, eliminated 24 of them, and are left with 2 amounts ($1 and $1M), it doesn't matter if you switch b/c you had only a 3.8% chance of picking $1M AND you only had a 3.8% chance of picking $1.

re: Computer Simulations from yesterday's Let's Make a Deal problem.(Posted by WDE24 on 3/6/13 at 3:30 pm to OneMoreTime)

Well I think that statement btu me is mathematically incorrect.

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Not switching is a choice with 50/50 odds a well. Your theory to switch seems to indicate that not switching isn't a 50/50 choice.

No I know it's 50/50, but when I first chose there was a 96% chance I chose wrong. So even if I've narrowed it down to where I have a 50% chance that I chose correctly I'm switching just based on the fact that when I originally chose there was a small chance I picked the right one. In reality it doesn't matter I guess

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