Finally someone picked the $1 million case yet she made the deal for $210,000. Still not bad financially. When she sold the case there was $400, $1000, and $1 million still on the board.

Do you think anyone will will the million?

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The only way it could possibly happen is if the remaining cases are the million, and something like $400,000 and $500,000. That way the person might figure he's getting a big six-figure payday anyway, so why not go for it. I cannot believe that a reasonable person would go for it if the options were 400, 1000 and a million like you outlined.

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I'm sorry I missed it. If she had gone one more round and kept the mil in play, probability says you always switch the last two cases (the "Monty Hall problem")...that would have been a *heartbreaker* here.

Originally posted by Mr ShhI'm sorry I missed it. If she had gone one more round and kept the mil in play, probability says you always switch the last two cases (the "Monty Hall problem")...that would have been a *heartbreaker* here.

I believe the Monty Hall statistical problem only applies when there's a Monty Hall (who knows what values are in what cases and will always chose a known "losing" case).

I have to admit I was geeky enough to think about this during early episodes when I started arguing with the on-screen probability graphics...then I thought about it and realised they were probably right after all.

After she made the deal, Howie asked her what her next case was. She chose the $1000 case. Howie then said that the next offer would have been $500,000.

But yeah, the kid was cute.

This was the first time in the history of DOND that someone picked the million dollar case.

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Originally posted by CRZI believe the Monty Hall statistical problem only applies when there's a Monty Hall (who knows what values are in what cases and will always chose a known "losing" case).

Yup...can't believe I neglected the underlying assumption of the problem. I never bothered to double-check though since a lot of the times it was shown that switching would have been better.

A few weeks ago, a woman from my town was on there who happened to work for the Special Olympics. She did a really cool thing and donated 1/2 of her winnings to the charity.

Originally posted by CRZI have to admit I was geeky enough to think about this during early episodes when I started arguing with the on-screen probability graphics...then I thought about it and realised they were probably right after all.

After the first two times we watched this, my wife told me to stop talking to the TV. I am operating under the assumption that each contestant has a 1 in 26 (3.85%) chance in selecting the million dollars. The odds don't go up after they start opening the cases. If they are down to 10 cases and the million has not been opened, the odds are 1 in 10 that the million is in that case, but the contestant's odds of selecting the right case were and are still 3.85%.

If you do the odds, don't you have to consider the following?

1/26 times they pick the correct case initially and would be incorrect to change, with a 0/10 probability of choosing the million.

25/26 times they pick the wrong case and at the time they select from the 10 cases remaining, they have a 10% chance of being correct.

So 1/26 of the time they have a 100% chance of being right by not changing and a 0% chance if they do change, and 25/26 of the time they have a 0% chance of winning if they don't change and a 10% chance of winning if they do. (1 * 100) / 26 or 3.84% chance VS (25 * 10) / 26 or 9.61% chance

I think that averages out better if you change, but maybe my logic is wrong.

This is assuming that the 1 million case has not been eliminated prior to the choice.

Do they even have this choice or are we just guessing here?

(edited by Guru Zim on 28.4.06 1546) Ignorance is bliss for you, hell for me.

I believe you can only switch when it gets down to just two cases left.

And CRZ is right, you have to have the "Monty Hall" that knows where the winning and losing choices (and he must choose based on that knowledge) are for the statistical results of that problem to apply. Since the player's choices are random (or at least without the knowledge of where the $1M is located), the probability of a bad result is equal to the number of bad choices over the total number of choices. Previous choices has no bearing on the new choice (unlike the Monty Hall problem, where he is making his choice based on the choice the player made).

Another way to look at this, say Howie was going to be "Monty Hall" and he has decided that he is going to reveal 25 losing cases and then let you choose keeping the one who have already picked or the one he hasn't open. 25 of every 26 times this is done, Howie would have to pick around the winning briefcase because it would still be up there. Only 1 out of every 26 times could he just picked any 25 he wanted because you have the right briefcase.

On the other hand, with the current rules, the likelihood that you picked the right briefcase from the start is exactly the same as the likelihood that you picked the wrong briefcase but were lucky enough to avoid the winning briefcase when narrowing those down to one (1 in 26). So it becomes a 50/50 proposition if you are lucky enough to get that far.

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Only 1/26 of the time will you choose the correct briefcase in the beginning. 25/26 you will choose the wrong one.

At the end, if you have two cases, it breaks down like this:

1/26 time (let's call it 3%) you have the correct case and switch to the incorrect case.

25/26 times you have the incorrect case and switch to the correct case.

If you get down to two cases, you have much better odds of having picked the wrong one to start with than you do of having picked the right one. You should always switch.

Where the odds come into play really matter only at the point that you are positing getting down to the final case without picking the million $$$ one assuming that it isn't the one that you have. But - if you break it down to only the beginning and end points, I think I'm right.

Can you give me math to back up the 50/50?

When there are only two cases left, you don't need the Monty Hall knowledge because there are only two options - good case or bad case.

Let me clarify the Monty Hall part... haven't you recreated this with your intermediary choices? I mean the situation only occurs at the last case, so isn't it the same result no matter if a human monty hall eliminates them all or if you do it yourself via an exremely improbable series of picks?

(edited by Guru Zim on 28.4.06 1848) Ignorance is bliss for you, hell for me.

Originally posted by Guru ZimCan you give me math to back up the 50/50?

Okay, math.

There are two cases.

Odds you have the case with what you want: 1 in 2 (50%) Odds you have the other case: 1 in 2 (50%)

When there are only two cases left, you don't need the Monty Hall knowledge because there are only two options - good case or bad case.

You are arguing FOR 50/50, you realise that?

Let me clarify the Monty Hall part... haven't you recreated this with your intermediary choices?

No, because Monty knows what's in every case and you don't. Monty knows if you have the million or not, you don't. Monty will always pick a "losing" case but you are picking "randomly."

I mean the situation only occurs at the last case, so isn't it the same result no matter if a human monty hall eliminates them all or if you do it yourself via an exremely improbable series of picks?

No, because you don't know what's in the cases. You are not picking with any knowledge of what's in any of the cases. Monty knows where the million is and picks accordingly, but you don't know and are picking purely by chance.

I hope I've been clear. If not, you don't have to believe me but you BETTER believe Wikipedia!

Originally posted by ges7184On the other hand, with the current rules, the likelihood that you picked the right briefcase from the start is exactly the same as the likelihood that you picked the wrong briefcase but were lucky enough to avoid the winning briefcase when narrowing those down to one (1 in 26). So it becomes a 50/50 proposition if you are lucky enough to get that far.

(edited by ges7184 on 28.4.06 1849)

This is where I disagree. I think you are looking at previous choice, not me.

No matter what the probability was of reaching the final two cases, once you are there you have to consider the initial choice and the final choice.

Even though it is extremely unlikely that you picked around the winning case if there are only two left, it is more likely that you didn't pick the right case to start with than than that you did. 97% chance wrong, 3% chance right.

I think those odds are still in play at the end. You have a 3% chance that you chose correctly, and a 97% chance you didn't. Intermediate choice doesn't matter -- there are only two cases. So - you can choose the only option left from the 97% pool or you can keep your 3% chance.

See, this is why my wife tells me to stop talking to the TV.

You had a 3% chance of picking the right case at the beginning. It's still 3% when you are down to two cases. Guru is right.

But, since you randomly eliminated the other 24 cases to get to the final two, wouldn't the odds of the other case having the million be 3% at the beginning, too? Thus, the 50/50 proposition at the end?

Now my head hurts. I am still going with Guru on this one.

Originally posted by piemanSee, this is why my wife tells me to stop talking to the TV.

You had a 3% chance of picking the right case at the beginning. It's still 3% when you are down to two cases. Guru is right.

But, since you randomly eliminated the other 24 cases to get to the final two, wouldn't the odds of the other case having the million be 3% at the beginning, too? Thus, the 50/50 proposition at the end?

Now my head hurts. I am still going with Guru on this one.

If you think it's 50/50, you're right. Unfortunately, Aaron doesn't think it's 50/50. He also has one less math degree than I do.

The probability that the contestant picks the 1M case with his first choice (and thus in possession of) is 1 in 26. The probability that the contestant picks the 1M case with his second choice is 1 in 26 (25/26 x 1/25). The probability that the contestant picks the 1M case with his third choice is 1 in 26 (25/26 x 24/25 x 1/24). And on and on until you get the 26th and final briefcase, with the probability of picking the 1M is 1 in 26 (25/26 x 24/25 x 23/24 x .... x 2/3 x 1/2 x 1). Therefore he is no more likely to find the 1M with his 26th choice as he was with his previous first choice.

So now he's left with two choices. Given that 1M was not in the 24 cases selected after the first selection and before the final selection, what is the probability that the 1M is in each case. In both cases, it is 1 in 2... the probability of the 1M case being in any given case/Probability of the 1M case being in any two given cases or (1/26)/(2/26).

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