Monty Hall Dilemma - Winning a GTI on a Game Show

[Buller_Scott]

that's right. which is why i didnt really take well to this riddle book, because i actually enjoy the math involved, and the book didnt mesh.

hey, are you no longer living in melbourne? did our ridiculous self importance scare you away? ($14 a kilo for friggin cage chicken, non-clearways on hugely busy inner cbd and other arterial roads, bike pelotons forcing cars off the road and overpriced-but-inefficient public transport?)

[The_Hawk]

Second time around is all that counts.

But regarding switching the second time around, that's only half your options. The other half is to not switch. You need to take ALL options into account if you're going to talk about your chances of winning.

If you look at the odds when not switching you still at a one in three option since you picked one from three. Yes the last two doors are 50/50 regardless of if your given the choice to switch. and take it up.

my view: it's still 50/50. the moment you open a door, it's outta the picture (from a math standpoint).

Absolutely 100% agree, it's down to a 50/50 chance on that last door... but then look at the options, if you do switch, out of the three options you could have originally chosen, TWO out of three possibilities end in a correct choice when you switch...

Stupid stats.

[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!

[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]

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.

[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.

[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]

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.

[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]

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!