[yor_on]

There is no law deciding what door you choose. But as soon as you've done, in this example, you made a system out of it. The game leader opening one of the 'two' doors that's left, according to you, is in reality opening one door of three. The definition builds on the information you get from him opening that door, and the way you split it into two systems. If the odds get better by a split, and 'a later shift' then I want to see why. I can do the math, but I can't see how to describe it, other than this way.

[damocles]

Umm...

When I first joined this debate about the Monty Hall problem, I said that the main problem that I had with it was why so many authorities were saying that the odds increased from 1 in 3 to 1 in 2 if you swapped. In fact it increases from 1 in 3 to 2 in 3.
Wikipedia is one source that has the right answer:
http://en.wikipedia.org/wiki/Monty_Hall_problem
My reply #23 on this thread clearly points out (to my way of thinking) why this is the case. I simply cannot see why others cannot see it this way.

[yor_on]

so how would you define my thought example, it being the exact same amount of information, with one difference, the game leader instead of leaving the door (he opened), removing it all together. Would the switch then become meaningless Damocles?

[yor_on]

"The correct answer, that players who swap have a 2/3 chance of winning the car and players who stick have a 1/3 chance of winning the car, is based on the premise the host knows which door hides the car and will always reveal a goat but never the car. If the player initially selected the door that hides the car (a 1-in-3 chance only), then both remaining doors hide goats, the host may choose either door, and switching doors loses. On the other hand, if the player initially selected a door that hides a goat (a 2-in-3 chance), then the host has no choice but to show the other goat, and switching door wins for sure."

This reasoning tells me that I don't need to be there, and that my thought example is sound. Your link Damocles.

[dlorde]

Umm...

When I first joined this debate about the Monty Hall problem, I said that the main problem that I had with it was why so many authorities were saying that the odds increased from 1 in 3 to 1 in 2 if you swapped. In fact it increases from 1 in 3 to 2 in 3.
Wikipedia is one source that has the right answer:
http://en.wikipedia.org/wiki/Monty_Hall_problem
My reply #23 on this thread clearly points out (to my way of thinking) why this is the case. I simply cannot see why others cannot see it this way.

You're right of course; if your pick has one chance in three, the remaining option must have two chances in three. I must have been distracted with odd considerations of two door examples...

[yor_on]

Heh, I now know more of this weird Monty Hall problem than I ever wanted to know But it was very weird, and treated as information you might state that it had a hidden parameter, which to me then would be the game leader never opening the door with the car, knowing which door it was. And naturally we have to assume that the guy choosing a door in the beginning must inform the game master about which one, as it otherwise could be one containing a goat that the game master also might open. Which in that case should mean that the guy would stand before a 50/50 chance of getting it right. If now I got it right weird stuff.

[bizerl]

THat's all very well, but it doesn't tell me how I can win the $200 000 on "Deal or No Deal".

 []

[Pmb (OP)]

That's a weird one alright. Saw someone explain it with a hundred doors instead, suddenly making it make sense. You have a hundred doors to pick from, you pick one of them. Then the game leader opens 98 of the other doors showing you nothing in them but goats. Now the question becomes one of keeping your original one that you picked randomly out of a hundred, not opened, or use the last door out of 99 that the game leader opened? It's a question of odds, and you picked one randomly from a hundred closed doors, but the 'other side' of it is the one where 98 doors was opened to find nothing, one left. It's like two games, the one you had from the beginning being the hardest to guess, wheres the one the game leader had being the 'foolproof' one. In reality it can't be foolproof as it could be your door too, but imagining it as two separate games makes it easier to see the reasoning.

and it is weird as you could imagine yourself not choosing any of those doors, waiting until the game leader opened 98 of them, then having two doors left to choose between. In that case you would have a 50/50 % probability of getting the right one, as I see it. Statistics as magic?

It's simple, really. As time progresses you know more about the odds of winning. There is never a reason to change the doors.

[Pmb (OP)]

Thankyou Yor_on! your example just highlights my reasoning! Because you only had a 1 in 100 chance of getting the first door right, then you are 99% sure if you change doors that you will be right after the game leader has opened 98 of them

That's not how probability works. Take a guess out of the 100 doors. Your probability of guessing right is 1/100. A door is opened and its empty. Regardles of whether you keep or change doors the probabiligy will be 1/99 of choosing the right one, and so on. This is different if you were playing the lottery. When playing the lottery always play the same number since its your goal to win in your lifetime, not merely today even if the chances of the new number you pick has the same probability of winning as any other number. Each problem is specific and needs to be addressed in each case. In the Montey Hall problem the winning door is never changed whereas in the lottery problem the number is always changed.

[Pmb (OP)]

If a cat can be half dead, and half alive,

Does it still need to be fed?

If it's Schrodinger's cat, it's both dead and alive, so you only need to feed the living version. You bury the dead version.

I once learned how to mentally calculate the day of the week of any given Georgian calendar date. It was too complicated and wasn't useful enough, even as a party trick, to remember once the novelty wore off (on a Thursday).

I disagree with these interpretations of quantum mechanics. A cat is a macroscopic animal whereas an atom is not. A cat is either alive or dead and not in a superposition of both.

Einstein was pointing this out when he asked "Is the moon there when nobody is looking?"  The answer is "Yes." Just like we know that the sun was there before life was here.

[damocles]

Thankyou Yor_on! your example just highlights my reasoning! Because you only had a 1 in 100 chance of getting the first door right, then you are 99% sure if you change doors that you will be right after the game leader has opened 98 of them

That's not how probability works. Take a guess out of the 100 doors. Your probability of guessing right is 1/100. A door is opened and its empty. Regardles of whether you keep or change doors the probabiligy will be 1/99 of choosing the right one, and so on. This is different if you were playing the lottery. When playing the lottery always play the same number since its your goal to win in your lifetime, not merely today even if the chances of the new number you pick has the same probability of winning as any other number. Each problem is specific and needs to be addressed in each case. In the Montey Hall problem the winning door is never changed whereas in the lottery problem the number is always changed.

That's not how the game works! If you choose the wrong door, the host is obliged to show you where the prize is by revealing the 98 doors where he knows the prize is not, giving you a sure pointer to the prize. So your chances of winning are 1% if you stand, but 99% if you swap.

[dlorde]

When playing the lottery always play the same number since its your goal to win in your lifetime, not merely today even if the chances of the new number you pick has the same probability of winning as any other number.

Can you explain the reasoning here? you surely have the same chance whether you change your number each time or not.

As I understand it, the only criteria for selecting a lottery number is to avoid one that other people might be likely to pick too; it doesn't help your chances, but if you do win, you're less likely to be sharing the prize.

[dlorde]

I disagree with these interpretations of quantum mechanics. A cat is a macroscopic animal whereas an atom is not. A cat is either alive or dead and not in a superposition of both.

At what point from atom to cat do you draw the line? Objects visible to the eye (40 microns, ~about 10 trillion atoms) have apparently been put into quantum superposition (see quantum microphone). I don't doubt that decoherence would rule out any animate organic creature, but what about viruses or bacteria at cryogenic temps? or a cryptobiotic tardigrade?

From the POV of size alone, it would be interesting to discover the practical size limit for a measurable duration of superposition.

The comments about Schrodinger's Cat were just whimsy.

[confusious says]

Reading my electricity bill

[yor_on]

That one is weird too, there are two ways and I don't know which one, or if both, are true, but they are contradictory. If i flip a coin and get a hundred tails then that's a very unlikely result. I would call that a very low probability. But the coin is in a way reset each time you flip it, meaning that it shouldn't care for those other results, instead treating each flip as a new instance, the other results having no bearing on the outcome. So one does not lead to the other, but statistically you should have a 50/50 probability of tail and head, flipping it in a series long enough. Maybe the question should be if there is a possibility of defining when a improbable series become so overextended, only tails, that you could expect a probability of it to come up as head? It makes me think of feigenbaum's constant this one, wondering if there is some mathematical way to define a 'overextension'?

that as common sense tell me that one thousand flips, all coming out tail, should be a very seldom seen pattern. Against it you have the equally valid point that each new flip should be counted as starting anew? Looked at that way there is no reason to choose any numbers before any other, although you might want them to be those that people don't tend to pick. On the other hand, there should be some breaking point to those tails, when they become so improbable that ??
=

Keep jumping over words, and, eh, spellings
=

One can think of it this way maybe, the pattern of a thousand tails is no more uncommon that any other combinations of set patterns like tail - tail  - head, if repeated over and over, for a thousand flips. Which then just should make one thousand tails, or heads, uncommon, because it is so easy to recognize for us. That will make the one defining it as being 'reset' with each new flip the one making most sense. What i mean is that if you count the way a pattern evolve over time, flipping a coin, that all patterns possible have a equal chance of evolve, singling out no pattern as more probable. And a pattern would then be whatever way you found head and tails arrange themselves over a thousand throws.

[damocles]

If a coin were flipped and came up tails one thousand times in a row, I would be inclined to bet on it coming up tails the next time, because I would regard the previous one thousand times as statistical evidence in favour of the coin (or the toss) being biassed.

[yor_on]

Yeah, I know Damocles Can't help it going against my basic instincts though. Which is why I don't gamble, can't trust those instincts )
=

Then again, thinking of it as patterns, the probability of it coming out 'any which way' should still be 50/50, shouldn't it, as no pattern is more probable than any other? (over 1000, and 1, flips) . They should all be equally probable as it seems to me, otherwise we find a bias? And I think that one can look at the constant 'reset', as well as the 'equally probable patterns', as a logical proof for that, the coin flipped enough times, also must give us a equal amount of heads and tails.

[yor_on]

There is one more thing though, calling a outcome randomly before the flip. Would that give you a statistically better chance to be correct, more than a 50% probability, than always calling a same, set, outcome? As calling 'tail' before each flip? And if that would be true (introducing a random call before the actual throw), why would that be?

[dlorde]

Seems to me that if it's a fair (random) coin, you'll average 50% success calling the same each time or calling randomly. Each call has a 50% chance of being right, regardless of previous calls.

[yor_on]

Yes, that's what I thought too, but it seem to have been a professor in statistics, meaning that you by introducing randomness on your side too, will get a better probability? I don't think it can be right myself, but I'm not sure?