OT: Statistical question

This is what I just don;t see, changing yuor choice doesn;t change teh fact that you still have a 50% chance of winning. it doesn;t matter if 1000 doors were opened before yuo still have a 50/50 c hance. Changing your answer at that point doesn't change the probability.

By revealing an

Reply to
whisky-dave
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You know the door chosen first time around had a 1/3 chance of winning, nothing the host does changes the chances for that door, so that door still has 1/3 chance, why would you ever chose that door over the other one, unless you didn't know which one was which?

Reply to
Andy Burns

Roger Mills put finger to keyboard:

Because *if* you chose a Goat (which you will have done, 2 out of 3 times) then the host has eliminated the door with the other goat and by switching you win the car. So you'll win the car 2 out of 3 times.

Reply to
Scion

If it ever comes to cutting the red or blue wire, remind me to ignore you :-)

Reply to
Andy Burns

Roger Mills put finger to keyboard:

Try this:

Suppose, instead of 3 doors, there are 1000 doors. Behind one is a car, behind all the others there's a goat.

You choose one door.

The host then opens 998 of the other doors, to reveal goats.

Is it now a 50/50 chance which door the goat is behind? Would you now switch?

Reply to
Scion

There's either alien life out there or there is no alien life out they, but whether or not we look for it is irreleivant to the likelihood of there being life. It doesn't matter how many boxes or doors are opened or how many sheep we find.

Reply to
whisky-dave

Why would you, if you know all your other guesses have been correct why change your system now ?.

Reply to
whisky-dave

Forgetting goats for the moment, when you choose a door, the chance of it containing the prize is 1 in 3. Each door has the same chance. If the gameshow host picked from the remaining two doors at random, the last unchosen door would still have odds of 1 in 3 of containing the prize.

The host *doesn't* pick at random though. The 2 doors you didn't choose have a 2 in 3 chance of containing the prize and when he eliminates one that he knows doesn't contain a prize, the remaining door still has a 2 in 3 chance of containing the prize.

Imagine 100 doors instead of which you chose one. The host then eliminates

98 of the 99 remaining doors. Would you still stick with your original choice? No, because it's then rather more obvious that the winning door was more likely to be one of the 99 that you didn't choose and if the host has eliminated all but one, it's *way* more likely the the winning door is the one you didn't choose. Not certain, but a lot more likely.

Tim

Reply to
Tim+

THANK YOU!!! I've finally got it.

Reply to
F

is the winner, thus the odds are now 0.5 for each door.

Oh, indeed. So why have you got it wrong? (I think; your explanation is very poor.)

Reply to
Huge

Precisely.

Reply to
Huge

Wrong.

Reply to
Huge

But it does. The fact that only 3 doors are involved makes it hard to see the obvious. ;-)

Imaging chosing one of 100 doors. You have a 1% chance of choosing correctly.

The host then eliminates 98 of the 99 remaining doors (that collectively had a 99% chance of concealing the prize). You're now left with two doors. Your original choice or the one left by the gameshow host.

Still confident that you've got a 50/50 chance of winning? It's *way* more likely that the winning door was one of the 99 that you didn't choose and the gameshow host has helpfully weeded out all the other non-winning doors. The door he has left has a 99% chance of containing the prize, not 50%.

Tim

Tim

Reply to
Tim+

Initially you have a one in three chance of choosing the right door. [1/3]

After one door is opened you have a choice of two doors, so a one in two chance [1/2]

But. The host had no choice of which door to open. It had to be a door to nothing.

So, of the two doors to choose on the second round; although the prior probability is 1/2, in the light of new information that the host took care to not choose one of those doors, there is a greater probability that that door is worth choosing.

Doing the sums right to calculate the new probability is the bit I find difficult, but swapping to the door the host avoided is the right choice if you want to make a choice based on Bayesian statistics.

Reply to
djc

Why are you bothering with this when you are demonstrably wrong?

Reply to
Huge

is the winner, thus the odds are now 0.5 for each door. Thus it makes no difference whether one switches doors or not. Statistical probability figures are nothing more than reflections of the data they're based on, and opening one door changes the available data. Missing that point is where so many are going wrong.

What's the collective noun for idiots? A "dennis"?

Reply to
Huge

Even the frequentalists can get this one right - it is mostly members of the public with appeals to "common sense" that get it wrong.

Of the Wiki page I think this explanation sums it up most eloquently:

Leonard Mlodinow says, "The Monty Hall problem is hard to grasp, because unless you think about it carefully, the role of the host goes unappreciated." (Mlodinow 2008) The 2/3 - 1/3 answer depends on assumptions about the host's behavior. The host knows which door hides the car, so the host will always reveal a goat and never reveal the car. If the player initially selected the door that hides the car (a one-out-of-three chance), 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 two-out-of-three chance), then the host has no choice but to show the other goat, and switching door wins.

Reply to
Martin Brown

That is correct and in 2 out of the 3 possible cases that converts the remaining host's door into a certain win. Winning 2/3 > 1/3 so you swap.

Inability to think clearly.

If there were two contestants then the host would be unable to open any door 1/3 of the time. This would also give away the fact that the contestants should immediately swap with the host to get a winning door!

If the host opens a door showing a goat then one of the two contestants has the winning door with equal probabilities.

Unless they are terminally stupid they would realise that the only reason the host doesn't open a door is because it would reveal a car!

Reply to
Martin Brown

How about a different approach : There's relatively few options in this scenario, so draw them all out.

First let's make an arbitrary choice : always swap vs always not-swap. (You'll notice I'm not giving the third one yet, which is always pick one of the remaining two doors at random)

Always not-swap : what options are there?

XXO

You pick 1 (1/3 chance), host must pick 2, no swap, you lose - 1/3 You pick 2 (1/3 chance), host must pick 1, no swap, you lose - 1/3

You pick 3 (1/3 chance), host picks 1 (1/2 chance), no swap, you win -

1/3
  • 1/2 = 1/6
You pick 3 (1/3 chance), host picks 2 (1/2 chance), no swap, you win - 1/3
  • 1/2 = 1/6

(quick sanity check = 1/3 + 1/3 + 1/6 + 1/6 = 1, so I've covered all the options)

You have a 1/3 chance of winning.

Always swap : what options are there?

XXO

You pick 1 (1/3 chance), host must pick 2, swap, you win - 1/3 You pick 2 (1/3 chance), host must pick 1, swap, you win - 1/3

You pick 3 (1/3 chance), host picks 1 (1/2 chance), swap, you lose - 1/3

  • 1/2 = 1/6 You pick 3 (1/3 chance), host picks 2 (1/2 chance), swap, you lose - 1/3
  • 1/2 = 1/6

You have a 2/3 chance of winning.

Using your knowledge of which door you picked first, and the host's knowledge of where the car is (that's the "host must pick X" bit), if you're sensible you can increase your chances of winning to beyond random.

The third strategy, which I ignored originally, which is to pick one of the remaining two doors at random, obviously gives a 50% chance of winning. The contestant who didn't see which door you picked originally only has that strategy, but why wouldn't you use the one which gave you a better chance of winning?

Reply to
Clive George

No, that's where you are going wrong.

When you make your initial choice there's a 1:3 probability that your door conceals a prize.

The probability that the prize is behind one or other of the other two doors is 2:3

The host reveals one of the other two doors. The probability that the prize is behind these doors is still 2:3. He opens a door which has a goat so we know there's no price there. The odds are now 1:3 that your door has the prize and 2:3 that the other door has the prize. So swap.

The point that everyone misses is the role of the host. The probabilities are the same before and after the reveal but the reveal eliminates one of the choices. If you kept your eyes shut and didn't see the reveal the odds would be exactly as they were. You'd get no advantage by swapping randomly to one of the other two doors but swapping once you have seen the reveal doubles your chance of winning.

Reply to
Steve Firth

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