Quiz time

Quote:

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Originally Posted by simon c

Just to clarify, the host knows what is behind the doors and always reveals a goat leaving one goat and the car remaining.

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OK - I will stick with my answer then... You should switch doors, as there will be a greater chance of winning due to the skewed probability caused by the host not making a random choice, where your choice is random.

Say the car is behind door 1 ...
If you choose door 2 the host will open door 3... So if you switch, you win.
If you choose door 3 the host will open door 2... So if you switch, you win.
If you choose door 1 the host will open 2 or 3 - This is the only scenario where switching causes you to lose.

So you have two chances to win if you switch versus one if you stick (and originally chose door 2).

So because the host knows what he is doing, you have double the chance of winning if you switch your original choice. (I think!).

That doesn't look right to me. You really only have a choice of two doors because the host takes one out of the picture. In your scenario, doors 2 & 3 are effectively the same door, because whichever one you pick, he will eliminate the other. The probability of choosing the correct door is always 0.5 because at the end you are left with a decision between 2 doors: keep the door you have, or pick the other one. Heads or tails. That's a probability of 0.5 of picking the right one.

It actually doesn't matter which door you pick first because you are always going to be left with one door with a goat and another with a car. If the host didn't open his door and his choice was random too, then that would be a different story.

Probability therefore doesn't help and it doesn't matter whether you keep your door or switch, the chance of getting the car is the same.

sorry silent c but yours is the intuitive answer and isn't correct - hexbaz has got it spot on, the chance of winning if you switch is twice as high as if you stick. Your point about doors 2&3 effectively becoming the same door is true, but the effect of that is to combine the probability of each door into the single door, rather than eliminating that dorr's probability.

Another way to look at it is to consider if you stuck with teh same door. When you originally pick a door, the chance of you getting it right is 1 in 3. If you always stick with that door, then the chance of getting it right remains 1 in 3 irrespective of what the host does. probability must always add up to 1 so if the chance of winning if you stick is 1 in 3, then the chance of winning if you switch must be 2 in 3.

I suggest you try it as it becomes more clear when you actually go through the motions. If you decide to always stick you'll see that the chance of winning is only 1 in 3.

Simon

Quote:

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Originally Posted by simon c

sorry silent c but yours is the intuitive answer and isn't correct - hexbaz has got it spot on, the chance of winning if you switch is twice as high as if you stick. Your point about doors 2&3 effectively becoming the same door is true, but the effect of that is to combine the probability of each door into the single door, rather than eliminating that dorr's probability.

Another way to look at it is to consider if you stuck with teh same door. When you originally pick a door, the chance of you getting it right is 1 in 3. If you always stick with that door, then the chance of getting it right remains 1 in 3 irrespective of what the host does. probability must always add up to 1 so if the chance of winning if you stick is 1 in 3, then the chance of winning if you switch must be 2 in 3.

I suggest you try it as it becomes more clear when you actually go through the motions. If you decide to always stick you'll see that the chance of winning is only 1 in 3.

Simon

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Odds of 1 in 3 as I stated are correct at the start. However, once the host has eliminated one door and invited you to either stay with your current selection or switch doors, IOW a second guess, then surely we are starting over with only 2 doors to choose from which means we have odds of 1 in 2?

Sorry duckman, it's not the equivalent of starting over. As I said this is really counter-intuitive and even though mathematically I "know" the right answer - it still feels wrong. The only way I convinced myself was to sit down and try it.

By the way - this isn't just a made up puzzle, it is based on a real game show from the sixties called "Let's make a deal" (like an early day The Price is Right) and is known as the Monty Hall puzzle as he was the host.

Lies, damn lies and statistics. I think this is a subtle trick being played by mathematics.

The fact that the host is taking one of the goat doors away means there are only two doors to choose from. It doesn't matter which door you choose first because you will always have two doors, one with a goat, one with a car. Therefore the probability of choosing the correct door first is irrelevant. This is where the error lies because you can't allow the probability of choosing the correct door first to enter into the calculation. You still have to choose between two doors and this decision eliminates the probability of getting the correct door first. If you had to stick with your original choice, then I agree the probability is 1 in 3.

The question is, now that I have two doors to pick from, which one should it be (should I switch or not?). The door that is opened by the host is a furphy, it might as well not be there at all because it has no affect on the probability of finding the car.

Say the car is door 1.

If you pick door 1, the host picks 2. You now have to choose between doors 1 & 3. Probability of choosing the correct door is 0.5
If you pick door 2, the host picks 3. You choose between 1 & 2. Probability = 0.5.
You pick door 3, host picks 2, you choose between 1 & 3. Probability = 0.5.

If it is truly random, then your decision to switch or not has no bearing on the actual location of the car. In a perfect world, half the time it would be behind the door you picked first and half the time it would be behind the other door. It's never going to be behind the third door.

This is why I have a major problem with probability and statistics in general.

silent c - there is no trick, but the answer is not 50:50. The fact that the host opens a door is not a furphy but is fundamental to the process because he knows the answer and chooses after you have made your choice.

If we have some independant volunteers (maybe an administrator) I'm happy to try this online as an experiment. We would ahve to do it a few times so it averages out.

The funny thing about probability is that you can't prove it with practical examples. The probability of tossing heads vs. tails is 50/50, yet it is possible that you can toss a coin a hundred times and get 100 heads in a row. This does not prove that the probability of tossing tails is 0.

I think that it all hinges upon whether you allow the first pick to be factored in to the equation. If you do, then you are correct. If you do not, then I am. From my studies of the subject, I think this is a matter of debate. I prefer the logical approach, which says that two unconnected actions cannot have an influence on each other. Once you have chosen the first door, where you had a 1 in 3 chance of being correct, that decision is history and has no bearing on your next decision in terms of probability.

We're never going to agree on this, so we might have to agree to disagree. It was your question, so you get the call on who is right.

But you're wrong ;)

I'm tempted to keep biting back here but I might end it with an agreement to disagree.

But I REALLY suggest that you actaually spend a bit of time trying it as it was the only way I finally convinced myself. I understand you saying that getting 100 heads in a row doesn't prove anything but in actually trying it you may find that the result surprises you.

Thanks for the discussion anyway.

Simon

It would be boring if we always agreed ;)

Don't worry, I used to have similar arguments with my stats lecturer about probability. When you're studying it, you just have to accept the rules - if you want to pass that is. But in the real world :D

Silent c, this isn't an argument about maths and the real world and about different peoples opinions. I'm really, really, really suggesting that you and your partner try it as it is impossible to convince anybody on this using logic or argument or probability etc. It's just that when you try it, it starts to become clear that switching is obviously the right answer.

Pleasetry it

OK, here is a table of possible outcomes:

Let's say the car is behind door 2.

You pick door 1 and you switch - you win
You pick door 1 and you don't switch - you lose
You pick door 2 and you switch - you lose
You pick door 2 and you don't - you win
You pick door 3 and you switch - you win
You pick door 3 and you don't - you lose

So there are three 'win' outcomes and three 'lose' outcomes for all combinations of choices. That says to me a 50/50 chance of getting it right. However, if you analyse it on the basis of whether you switch or not, there are two good outcomes for switching vs. only one for not switching.

On that basis, it would appear that you are right. In order to win when switching, you must have chosen a goat door first. There is a 2 in 3 probability of that. Assuming that you have made up your mind to switch, you have a 2 in 3 chance of winning the car.

Hmmm, something funny going on here....

it's a tricky one isn't it

Yes the mistake I made was to assume that the first choice is irrelevant. :o

I always hated stats.