Monty Hall Dilemma - Winning a GTI on a Game Show

[The_Hawk]

Lies, damn lies and statistics, thats what's killing this, and lets not forget the original question. It was, if given the choice, should your switch your choice after they eliminate a goat?

So lets look at this backwards, in order to WIN the car you should pick a goat the first time (which is more likely at 2/3) then switch to the car after they eliminate one of the goats!

So you had a 2/3 chance of being wrong the first time which in turn means that in 2/3 cases you would be better off switching the second time around. So YES, you are statistically better off switching.

Again, you were MORE LIKELY to be wrong the first time which means your *statistically* MORE LIKELY to win if you switch the second time (even though the choice in isolation is still 50/50).

[Timbo]

Your argument for a 50/50 outcome is only valid if the host randomly selects a door to open or if there isn't a host at all.

Correct!

The host is always going to pick a door with a goat. That's 1 in 1. But that happens after your 3 in 1 decision, and if you do nothing, your odds remain 3 in 1. The odds of picking the door with the car, IF you make the switch decision, now become 2 in 3!

Rocket, you are trying to rationalise the natural intuition we have for this game, but it is wrong. Hard to accept, but wrong. As I said, get a partner to lay out sets of 3 cards in front of you, and track the outcome. You will discover that switching gives you the highest outcome of "the car"

[pixl]

You have to stop taking the host into account.

And you NEED to start taking the host into account. You can't just pick and chose what you want to include. The host makes all the difference, he's the one who stops it becoming a random choice -- This isn't a flip of the coin.

Your argument for a 50/50 outcome is only valid if the host randomly selects a door to open or if there isn't a host at all.

[Rocket36]

If you still not getting it, think of 1,000,000 doors. You pick a door and the host opens 999,998 all showing goats -- do you REALLY think your chances of winning are still 50/50?

Yes of course they are. Two doors left to open, one choice. THAT is a 1 in 2 chance (or 50/50 chance) of winning. You have to stop taking the host into account. You have to forget about the open doors. Two are closed, there is a prize behind one, you have to pick one. THAT is 1 in 2.

[pixl]

Ok, Rocket, imagine there are now 30 doors. 1 car, 29 goats.

You pick any door and your chance of winning the car is 1 in 30, right?

Now, the host opens 28 doors revealing goats, leaving only the door you picked and one other door closed.

Are you going to swap or stay with the door you chose? How confident are you that you picked the right door out of 30? Still think you've got a 50/50 chance of winning now? We're not talking about the fact that there are only two doors left, we're talking about the chances of you winning the car.

You have to remember that your first choice, you had a 1 in 30 chance of picking right. Which means, you also have a 29/30 chance of picking wrong!! The fact that there are only two doors left does NOT make it 50/50 since you chose a door with the initial odds of 1 in 30.

Same with the original problem, it's not about two doors left closed -- The fact remains that your first choice, you've got a 66% chance of being WRONG!! Not good odds if you want to win a car.

If you still not getting it, think of 1,000,000 doors. You pick a door and the host opens 999,998 all showing goats -- do you REALLY think your chances of winning are still 50/50?

[schoona]

But what you're failing to realise is once there's only two doors remaining it's ALWAYS a 1 in 2 chance. FORGET ABOUT THE DOOR THAT'S BEEN OPENED!!! It not longer has any baring on your chances. You have two to choose from and that makes it a 1 in 2 (or 50/50) chance you will choose the right one whether or not you switch.

Thinking micro yes, but from the outset thinking macro , it would be better to swap.

[schoona]

At the outset, the chance you would pick the door with the car is 1 in 3, and a door with a goat, 2 in 3. OK...lock that it.

A door is opened, revealing a goat. Now, here's the thing -- nothing has changed in terms of the odds of your original choice; it still stands at 1 in 3 for the car, and 2 in 3 for the goat, even though one door has been opened.

To reduce the odds that your door has a goat behind it, you must make the switch decision. It is a two decision process, and not making the switch does not alter the original odds, whereas making the switch does.

There is a lot of confusion about this because it is so counterintuitive, and people try to rationalise, as above. But, if you don't believe the explanation, try it with 3 cards, and see how you it plays out

x2

the counterintuitive-ness f it had me second guessing myself too, but after a while it lcicked

[Rocket36]

But what you're failing to realise is once there's only two doors remaining it's ALWAYS a 1 in 2 chance. FORGET ABOUT THE DOOR THAT'S BEEN OPENED!!! It not longer has any baring on your chances. You have two to choose from and that makes it a 1 in 2 (or 50/50) chance you will choose the right one whether or not you switch.

That youtube explanation takes into account the first decsion. Which is why so many people are fooled into thinking that there's a better than 50/50 chance the second time round. There isn't. When faced with two choices, you will only EVER have a 50/50 chance of making the winning choice. I'm not taking into account "variable change" like people want you to do so you agree with them. I'm simply trying to point out that it's 1 in 2 when faced with the second choice. 2 options, 1 choice = 1 in 2. SIMPLE!

[pixl]

Rocket, the thing you're forgetting is the that the game show host knows what door has the car, which means he ALWAYS has to reveal a goat door. Because we know he ALWAYS has to reveal a goat door, this is what tips the odds in our favour. ie. 66.6% of winning a car by switching.

You can't exclude the first door simply because there's two left. The first door is and has to be included simply because we know 100% that the door the host must reveal is a goat door.

The only way the it can work as a 50/50 choice in the end is if the host forgot which door has the car or if he decides to reveal a door randomly. The key word is random, if the host is randomly picking a door to reveal, then the last two doors are 50/50.

The problem explained in the TV show 'Numb3rs' and the movie '21' -- http://www.youtube.com/watch?v=5e_NKJD7msg

[Rocket36]

LOL Timbo... That's the funniest post in this thread so far. Do you really think that when there are only two options left and you have to make another decision to stay with your original door or switch, you still have a 1 in 3 chance? Or are you just being funny? Hahahaha

Because it really is this simple. There are two doors left, pick one. That's a 1 in 2 chance. REGARDLESS of whether or not you switch doors.

[Timbo]

At the outset, the chance you would pick the door with the car is 1 in 3, and a door with a goat, 2 in 3. OK...lock that it.

A door is opened, revealing a goat. Now, here's the thing -- nothing has changed in terms of the odds of your original choice; it still stands at 1 in 3 for the car, and 2 in 3 for the goat, even though one door has been opened.

To reduce the odds that your door has a goat behind it, you must make the switch decision. It is a two decision process, and not making the switch does not alter the original odds, whereas making the switch does.

There is a lot of confusion about this because it is so counterintuitive, and people try to rationalise, as above. But, if you don't believe the explanation, try it with 3 cards, and see how you it plays out

[Rocket36]

No BTJ, YOU are wrong. Think about it... You actually agreed with the 1 in 2 chance in your post by saying "although obviously there are only two doors left to open" which is a 50/50 chance of winning regardless of whether you stay with your initial choice or switch to the other door. You can't factor in the first choice as the second choice is all that counts, since the circumstances have changed, which is 50/50.

It really isn't something people should be arguing about as it's quite a simply philosophy. You have two doors to choose from and there is a prize behind one. That's 50/50. The fact that there were 3 choices to start with is completely irrelevant when it comes to the chances of winning as you can't win or lose with that first choice.

[Buller_Scott]

but that's the thing- my viewpoint from a practical standpoint, is that when door 1 has been opened, its no longer to be counted as it's now not an element in ''probability'' as it is now a certainty- a certainty that has been ruled out as NOT housing the GTI.

the maker of the riddle banks on people not understanding the dynamism of his riddle (not HER riddle, because if it was a female, there'd be many, many, many more variables, all constantly changing, and the ''certainties'' would continually revert back to being ''uncertainties''. not to mention the fact that working out the riddle would be much more gruelling, time consuming, exhausting, and depressing).

bottom line is, the riddle is dynamic- the numbers change (by way of eliminating uncertain elements) and thus, so does the probability.

look at it this way: if there was a dub-fest on, 20 raffle tickets sold for a GTI prize, and your ticket turns up a dud, but you get to switch, that means that you've still got your chance at the prize, but there's only 18 tickets to contend with. then, yours turns up dud again, but you get to switch, and now there're only 17 to contend with. so on and so forth until you get down to ten. if, at that point, the emcee says that the tickets have to be revealed, your shot at the car is now only one in ten. it is NOT eleven in twenty (namely, it is NOT the odds of [ten (the number of tickets you've revealed as failing tickets) + one (the ticket in your hand)] : twenty (the original number of tickets sold).

there are ten tickets left, you've got one shot. those are NOW the odds.

the riddlemaker is trying to fool people into thinking that, atleast with my dubfest example, that every time the person switches tickets (and a dud ticket is eliminated from the pool) that those dud tickets somehow are added to the number of tickets that the person is holding onto BECAUSE they have now been revealed to the ticketholder as certainties (certain failures). pretty logic, but it's not true. without the person knowing whether or not his ticket is the one AT the time that the emcee orders the remaining people to turn in their tickets, his chances, by way of possession of one ticket only, are precisely equal to those of the people who also hold one ticket only. the fact that he knows the numbers of ten tickets that have turned out to be failures, is not relevant to his one chance of winning a GTI when he's standing next to 9 other guys at the lectern/podium who're also holding their tickets out.

[BTJ]

Have to chime in here.

People on the side of 50:50 chance on the second choice are wrong.

You should switch.

Whilst it appears 50:50 and indeed would be if your only option was one of two, but you option is one of three. Just because the door is open does not exclude it from the odds.

3 doors. You make a choice. that is a 1 in 3 chance.

If you choose to stay with that choice it is still a 1 in 3 chance because you made that choice when there were 3 available. (although obviously there are only two doors left to open)

Your initial choice has a greater probability of being wrong than right.

[Buller_Scott]

yeah, stupid riddle books. they're trying to confuse people by sweeping "we're changing variables, mid-formula" under the carpet.

a-holes. someone should rat 'em out!