Monty Hall Dilemma - Winning a GTI on a Game Show

[Achtung Tiggy]

ALL I HAVE SAID is that when you have a choice between two options, with ANYTHING the result is one or the other, or 1 in 2, or 50:50.

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!

[NickZ]

I find the easiest way to think about it is to apply the situation to 1,000 doors. The chances of you picking the right door first up is 1 in 1000. If the host then, after you picking a door, eliminates 998 of the options, only leaving two doors left, and offers you to switch, then seeing as your initial chance of picking the right door initially was minute, then it is strongly in your favour to switch (i.e. 999 out of 1000 times you will win!)

Yes, we can talk about the dealers motives, that chance comes into play, and other variables which will ultimately have an impact. These things have a much more significant impact when the choice is only 3 doors as opposed to when there is more doors. But the Monty Hall is purely to explain that as a mathematical equation and excluding other variable factors, the odds are in your favour to switch.

It is a bit like poker, each situation there is a mathematical formula giving you the probability of a particular hand winning, but ultimately other variable factors such as luck, skill, bluffing, pressure etc impacts on the pure mathematics behind the game.

[Timbo]

Give someone two choices, get them to pick one only and they hav e a 50:50 choice. THAT is basic maths and if you don't understand that I'm sorry, go back to primary school.

[abreut]

Can someone please tell me where I've said...that I have a 50% chance of winning?

Isn't that what you are saying right here, 2 sentences later?

Therefore, the second choice has a 50% chance of picking the car. and a 50% chance of picking the goat.

If the second choice has a 50% chance of winning, then presumably the first door/choice has a 50% chance as well because there are only 2 doors to choose from and the car has to be behind one of them, 100% - 50% = 50%. So you ARE effectively saying that you have a 50% chance of winning.

Anyway you are still wrong, yes there are 2 doors, 2 "choices", 2 "options" but one of those is twice as likely to have a car behind it than the other. Just because there are 2 options doesn't mean that they have to have the same probability of winning, and in this case they don't.

[Rocket36]

Can someone please tell me where I've said I wouldn't switch and that I have a 50% chance of winning? ALL I HAVE SAID is that when you have a choice between two options, with ANYTHING the result is one or the other, or 1 in 2, or 50:50. Therefore, the second choice has a 50% chance of picking the car. and a 50% chance of picking the goat.

Give someone two choices, get them to pick one only and they hav e a 50:50 choice. THAT is basic maths and if you don't understand that I'm sorry, go back to primary school.

Oh, and I live in Melbourne at the moment.

[Dubya]

It's pretty easy to demonstrate. You have 3 options; A door, B door, and C door. When one losing door is eliminated, switching doors will result in winning 2 out of 3 times. When you don't switch you only win 1 out of 3 times. Try it!

But there are is a variable that must be eliminated first; you have to pick a door before the game show host picks a door.

That order is certainly a dependency for the contestant's odds to increase from 33% to 67% through switching.

A first choice made after the host has revealed a goat would provide equal odds for success and failure.

No word from Rocket, et al, of late...

[Coaster]

It's pretty easy to demonstrate. You have 3 options; A door, B door, and C door. When one losing door is eliminated, switching doors will result in winning 2 out of 3 times. When you don't switch you only win 1 out of 3 times. Try it!

But there are is a variable that must be eliminated first; you have to pick a door before the game show host picks a door.

[Dubya]

I'm just trying to get people to see the different was of problem solving & the differences between the pure maths & real situations.

Maybe so, but what is being debated here is the "pure maths".

Until you acknowledge the validity of the "67%" theory there is little point speculating on the influence that other factors might have on a contestant's behaviour in "real situations".

For instance, I might panic and forget the theory in the glare of the studio lights or just decide (irrationally) that I want to back my intuition.

Whatever the case, while I am here, I'll pose the following question (using the card version of the game):

Are there any 50/50s out there who would not take the opportunity to switch from:

- their first-selected card; to

- both of the other two cards,

if the facilitator (let's assume without any malice, mendacity or knowledge of which card was the Ace) offered it to them?

In other words, the facilitator is offering two chances out of three instead of just one chance out of three.

Now even if some would still not switch, would everyone not agree that the switch would double one's chances (all things being equal)?

If it be safe to assume we can all agree that two cards are twice as likely as one card to yield an Ace, can Rocket, Jarred or any other 50/50s explain the difference, if any, between:

- switching to both cards when they are both face down (as described above); and

- switching to the other two cards when one is face up and one is face down (which effectively is what happens when you switch to the other face down card in the version of the game as originally described)?

Of course, there is no difference:

We already knew with 100% confidence that one of the two other cards was a joker.

So switching to the other card after the Joker has been revealed provides exactly the same odds as switching to both cards before the Joker has been revealed.

The revelation of the Joker simply guides us to not choose that card and so we only need to choose the other card to have the same chance of winning as if we chose both.

To the 50/50s who would now say, "No, there are two cards and you are choosing between your card and one other and that is a 50/50 deal", you must remember:

Your first choice has a probability which is not varied by your knowledge of which of the two other cards is the Joker - you already knew one of them was.

Once you know which of the other two cards is the Joker, these two cards still hold a 67% chance of concealing the Ace. Knowledge of the Joker just tells you which card not to pick, which is the same as picking both.

However, for completeness I should add that it would be a 50/50 proposition if, after the Joker was revealed, the facilitator shuffled the two remaining cards behind his back and asked the contestant to guess again.

However this is not how the game is played.

The contestant is given the choice of sticking with his first card or switching to the only other face-down card.

And for the reasons stated above, this second option has a theoretical win rate of 67%.

Please let me know (via PM if you prefer) if any of the foregoing is not clear as I believe that I have exhausted all the different ways of explaining why the theorem is the way it is.

Well, not quite: I have thought of one other method based on selective X-ray vision, but I will reserve x-rays for any surviving 50/50s...

[Flighter]

I wonder if he has located a deck of cards or tried on of the simulators yet . . .

It appears he'd rather spend much more time arguing why he thinks he is right than take a few minutes to test his own theory empirically.

[Dubya]

Hmmm....I like Dubya's idea MUCH better than mine

My odds are a bit generous (Rocket would break-even if he scored 40%) but probably close to the margins casinos might work on for a game such as this, but I dunno.

My odds are a bit too generous to Rocket - if he still believes that after the Joker is revealed his odds somehow go up from 33% to 50% if he sticks with his first choice it should be a straight bet, as in two-up.

However it is difficult to imagine how anyone could not be spooked by the proposition of an even money bet on a single guess made from three options which can only offer a 1 in 3 chance of picking the Ace . . .

But there remains a number of people who believe that if you turn the other two cards over one after the other (rather than simultaneously), the odds of one of the cards being the Ace is 50% - ie each card has only a 25% chance of being the Ace instead of each sharing equally the 33% chance the card selected attracts.

I wonder if he has located a deck of cards or tried on of the simulators yet . . .

Mate, you come down and observe -- I'm sure Rocket will stand drinks for you, too

I'd be happy to participate via a 3-way phone hook up...

[Timbo]

Re: Monty Hall Dilemma - Winning a GTI on a Game Show

But Aaron, that's NOT the game! It's critical to the subsequent round conditional probabilities that the contestant makes a choice, and then the host always opens a door revealing a goat. Changing the game changes the probabilities.

[The_Hawk]

but who would be the one that removes ( and knows which ones are) the 'goat' cards? there's ya problem.

Not really...

One choice wont change, turn that first, if it's the GTI Card the non-switcher wins, then turn over the other persons card, if it's a Goat Card they would have switched to the GTI Card and won... in short the non switcher has a 1/3 chance of hitting the prize, the switcher has a 2/3 chance of being wrong, switching, then winning...

[Treza360]

The "house".
Timbo will be the house and give Rocket odds of, say, 2.5:1 on the basis Rocket is not allowed to switch (which he does not want to do as he thinks he has a 50/50 chance either way).
Rocket will accept these odds as he believe he has a 50% chance of winning if he does not switch and so believes at 2.5:1 he will come out in front.
Timbo will come out in front as Rocket will only win 33% of the time.
It would be amusing indeed if it were the prospect of losing real dollars that made the penny drop and Rocket realise his odds are stuck on 33% if he does not switch!
And if Rocket's chance is 33%, the house's chance (and the house sits on the card Rocket could have switched to) must be 67%.
QED.

That is if they play for long enough though. It is still possible for Rocket to come out ahead. That's the joy of statistics.

For example I played 100 rounds on that website linked to earlier and I came back with only 63%. Yes if you played forever the house would come out ahead but there is still the possibility that in only a "small" amount of games the player might come out ahead by not switching.

If they play for long enough though, the house will win.

Cheers,
Trent

[Timbo]

Hmmm....I like Dubya's idea MUCH better than mine

Mate, you come down and observe -- I'm sure Rocket will stand drinks for you, too

[Timbo]

but who would be the one that removes ( and knows which ones are) the 'goat' cards? there's ya problem.

We get a neutral dealer/croupier, hence running it at the casino. Rocket will be sure to tip her/him with his winnings