Monty Hall Dilemma - Winning a GTI on a Game Show

[Timbo]

1/2 < 2/3 chances of loosing
1/2 > 1/3 chances of winning...


....I personally wouldn't change though....

That's the natural, intuitive process. But the game doesn't work to that logic. So, you'll probably lose 2 out of every 3 games.

[GTom]

Did you even play the games linked to earlier in the thread?

hahahaha.... no....... have you seen how many pages of **** there is!

[Treza360]

hahahaha.... no....... have you seen how many pages of **** there is!

Well here's the link. Play and learn: Simulator
Cheers,
Trent

[BTJ]

funny reading this thread and watching people hold on for so long to the thought that it is a 50/50 chance.

How they can come to this conclusion when they made their choice from one of three doors makes me laugh.

Top thread.

[Achtung Tiggy]

funny reading this thread and watching people hold on for so long to the thought that it is a 50/50 chance.

I can think of 2 scenarios where there is a 50/50 chance.

1. The host opens one door (goat) before the contestant makes his intial choice.
2. THe contestant picks a door. The host then opens another door without knowing where the car is, and it turns out to be a goat.

[Rocket36]

So if Australia plays Brazil in the final of the World Cup, does that mean we have a 50% chance of winning? (Since there are only 2 option:win or lose, draw is not permitted).

I hope you don't work for Tabcorp!

No - that's completely different scenario where there are quite clearly more than two possible outcomes.

Rocket, whatever the case, you've not yet agreed with those of us who have said that you should switch after the first goat is revealed because doing so doubles your chances from 1/3 to 2/3.

No - because that has nothing to do with my point, which is that where there's a choice that only has two possible outcomes, the chances are 50% or 50:50 or 1 in 2. And THAT IS A FACT THAT CAN'T BE DISPUTED. Like the toss of a coin... Pick heads or tails... The chance of being right is 50% or 50:50 or 1 in 2. Pick one of these two cards/windows/doors/etc... The chance of picking the right one is 50% or 50:50 or 1 in 2.

And as far as I can tell from your post above, you still believe you have a 50% chance of winning if you stick with your original choice and this, as has been explained in many different ways, is not correct.

Where have I said that? Your choice between two options is 50% or 50:50 or 1 in 2. That's all I'm saying.

You now appear to be suggesting that you would be willing to switch even though you believe your odds would not change.

If you believe that, why would you switch?

WTF? How do I appear to be suggesting that? If you start using the term ODDS, is in odds of winning this particular game, then that's different as ODDS take into account more than the choice of two options.

If I was playing this particular game, I would switch (obviously). That doesn't change the fact that out of two options, the chance of picking one is 50% or 50:50 or 1 in 2.

As you claim there is no benefit to switching, I cannot understand why it was unreasonable for anyone to infer that you would be disinclined to go against your initial instinct.

Not claiming that.

After all, if you believe the odds will not change whether you switch or not, most people would probably be more disappointed if they switched and lost than if they stuck with their initial choice and lost.

Whatever the case, as you seem to be firmly of the belief that the odds are 50% after the first goat has been revealed, your position on switching would seem to be more influenced by whimsy than by reason.

Like I said, ODDS of winning THIS SCENARIO is completely different to the mathematical fact that having two possible outcomes, no more, no less for anything means that the chance of picking one is 50% or 50:50 or 1 in 2.

funny reading this thread and watching people hold on for so long to the thought that it is a 50/50 chance.

How they can come to this conclusion when they made their choice from one of three doors makes me laugh.

1 of three doors would be 33.3% or 1 in 3. But the people talking about 50% or 1 in 2 aren't talking about three doors. They're talking about two.

No short of drawing diagrams for people, I can't be much clearer that all that. If you still don't understand that the second choice is a is 50% or 50:50 or 1 in 2 chance of picking one or the other then there's nothing that will explain it to you.

[schoona]

I can think of 2 scenarios where there is a 50/50 chance.

1. The host opens one door (goat) before the contestant makes his intial choice.
2. THe contestant picks a door. The host then opens another door without knowing where the car is, and it turns out to be a goat.

Thats just moving the goal posts. Pointless....

[BTJ]

Rocket,

after the door has been opened revealing a goat, leaving two doors to open your chances are not 50;50 if you stay with your original choice. Youve got that don't you?

Youre original choice was a one in three chance and if you stay with your original door, you stay with your original odds.

Put it like this.

3 doors.

You pick one.

Host opens the first door. Goat.

Host doesnt ask you to switch or stay and opens another door. Goat

You have a car behind your door! Yay!

1/3 correct?

Same again

3 doors.

You pick one.

Host opens the first door. Goat.

Host asks you switch or stay? Stay

Open the second door, goat.

You have a car! Yay!

1/3 chance

Now play that out over time and the odds are not in your favour that you are originally going to pick the car.

It is not a 50/50 or 1/2 chance if you stay with your original choice. If you only make one decision it doesnt matter if the door is opened because you arent removing it from the odds by switching, you are playing your original guess and original odds, not a 1/2 choice.

[Rocket36]

*sigh* re-read my last post BTJ because your entire post is a waste of time... You (and others) are completely missing my point regading 50% or 50:50 or 1 in 2. The original choice when there are three options has NOTHING to do with my point.

Meh. I don't care anymore.

[Flighter]

*sigh* re-read my last post BTJ because your entire post is a waste of time... You (and others) are completely missing my point regading 50% or 50:50 or 1 in 2. The original choice when there are three options has NOTHING to do with my point.

Meh. I don't care anymore.

You appear to be falsely assuming that because there are two doors left the odds somehow shift back to 1 in 2. They do not, as this is not a case of random probability (like tossing a coin), but rather one of conditional probability. This has been mentioned in the past, but you still appear to cling to the notion that somehow a choice from 1 in 3, or 1 in 1000 or whatever miraculously becomes 50:50 as soon as all but two of the doors are opened. It therefore appears that in your case a little bit of knowledge is more dangerous than none at all, for in the latter instance you would probably be more open minded about simply trying it for yourself. I'm assuming that you haven't tried it because you haven't reported that you either (a) ran the simulator and were proven right, or (b) you ran the simulator and were proven wrong. Interestingly, this presents two outcomes too, but the odds aren't 50:50 in this case either, as the odds you were proven right are zero.

Then again, maybe the Monty Hall problem is another of those global conspiracies everyone else is in on and the joke is on you.

[Rocket36]

So Flighter is another person that can't read... I'M NOT TALKING ABOUT THE ODDS. Jesus Christ!!! People need to STOP making assumptions and actually READ what I'm saying.

As for your smart arse under tones and attempts to belittle me Flighter, well I'm sure you feel way better about yourself but you won't get me to bite. Back in your box.

[Dubya]

If I was playing this particular game, I would switch (obviously). That doesn't change the fact that out of two options, the chance of picking one is 50% or 50:50 or 1 in 2.

If Rocket believes that the chance of picking the right door is 50%, can he set out his rationale for switching?

If the 50% chances he claims he has of picking a winner amongst the two remaining doors is relevant (as he seems to be claiming), why should it be obvious that he would switch from his first choice?

By his rationale, this would seem to give him the same 50/50 chance of winning.

67%ers state that one's chances of winning double (ie from 33% to 67%) if you switch.

If all Rocket is saying is that one has a 50% chance of picking right when given two options, I do not know why something so axiomatic needs to be stated in the context of the Monty Hall Dilemma.

Everyone knows about 50/50, however such a probability does not arise in the normal playing of the game, as has been explained in many different ways.

So it remains a bit of a puzzle as to why Rocket feels compelled to labour such an obvious point as 50/50 when it is irrelevant in the context of the Monty Hall Dilemma.

I can assure Rocket, the reason people are resisting his line about 50/50 is not because they do not understand it, but rather because it is wholly irrelevant to solving the Monty Hall Dilemma, as they have tried to say.

Perhaps Rocket can explain once and for all:

- Why he believes 50/50 has anything to do with the Monty Hall Game; and

- Why it should be obvious that he would switch,

as the two concepts are logically irreconcilable.

[Timbo]

All Rocket is saying is that the probability, given two doors, is 1 in 2 or 50:50, which I think we all understand, and is not in contention. Why he continues to say that, given it is totally irrelevant to the probabilities associated with the dilemma, escapes me!

[Rocket36]

The second choice in the context of the chances of picking something out of the two options is ALWAYS 50/50 regardless of the circumstances that surround that choice.

The second choice in the context of winning in this scenario is quite different, is based on odds and NOT chance, and it's NOT something I have been basing my point on AT ALL.

All Rocket is saying is that the probability, given two doors, is 1 in 2 or 50:50, which I think we all understand, and is not in contention. Why he continues to say that, given it is totally irrelevant to the probabilities associated with the dilemma, escapes me!

Since people were using the term CHANCE a LOT I have been trying to point out that chance is always 50/50 in this scenario when referring to the second choice. The odds of winning are completely different.

[Dubya]

Since people were using the term CHANCE a LOT I have been trying to point out that chance is always 50/50 in this scenario when referring to the second choice. The odds of winning are completely different.

Based on that one could infer that Rocket believes that the "odds" of winning, whether you switch or not, are 2:1.

But, no, when playing the game as described:

- The chance (or probability) that you will win if you do not switch is 33%.

- The chance that you will win if you do switch is 67%.

50% does not come into it in any way if you are playing the game as described. If it does come into it, why would you go against your gut instinct and switch?

Odds are the inducement a bookie would offer someone to encourage them to bet on the outcome of the game and are expressed as the inverse of the probability (or "chance") when no margin is added. However, the terms "probability", "chances" and "odds" are being used to mean the same thing and so are interchangeable for the purposes of this discussion. Only the inverse of the bookie's odds for each outcome would add up to more than one due to their profit margin (eg odds of $1.90 on the toss of a coin).

But the probabilities/chances for each possible outcome for this game will all add up to 1 (eg 33% + 67% for the first-chosen door and the other closed door, respectively).

The problem seems to be that, so far, Rocket has talked about a 50/50 "chance" (when it is 33/67 after the first goat is revealed) but has not said what he believes the completely different "odds" of winning are. But we can infer from this that Rocket believes the "completely different" (but related) odds are 2:1.

However, Rocket only gives half a view and unfortunately it is misleading as the chances/odds/probability are never 50% in this game.

Bottom line is, Rocket has said:

He has a 50/50 chance yet he would still switch, but without providing his rationale for doing so.

It simply does not make sense to say that you would "obviously" switch if you believe each choice provides an equal chance of success.