Monty Hall Dilemma - Winning a GTI on a Game Show

[Mr.B]

66.6%

http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

As a motivating example behind the discussion of probability, an applet has been developed which allows students to investigate the Let's Make a Deal Paradox. This paradox is related to a popular television show in the 1970's. In the show, a contestant was given a choice of three doors of which one contained a prize. The other two doors contained gag gifts like a chicken or a donkey. After the contestant chose an initial door, the host of the show then revealed an empty door among the two unchosen doors, and asks the contestant if he or she would like to switch to the other unchosen door. The question is should the contestant switch. Do the odds of winning increase by switching to the remaining door?

The intuition of most students tells them that each of the doors, the chosen door and the unchosen door, are equally likely to contain the prize so that there is a 50-50 chance of winning with either selection. This, however, is not the case. The probability of winning by using the switching technique is 2/3 while the odds of winning by not switching is 1/3. The easiest way to explain this to students is as follows. The probability of picking the wrong door in the initial stage of the game is 2/3. If the contestant picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game. Thus, if the contestant switches after picking the wrong door initially, the contestant will win the prize. The probability of winning by switching then reduces to the probability of picking the wrong door in the initial stage which is clearly 2/3.

Despite a very clear explanation of this paradox, most students have a difficulty understanding the problem. It is very difficult to conquer the strong intuition which most students have in this case. As a challenge to students who don't believe the explanation, an instructor may ask the students to actually play the game a number of times by switching and by not switching and to keep track of the relative frequency of wins with each strategy. An applet has developed which allows students to repeatedly play the game and keep track of the results.

[cme2c]

WOW 7 days and 10 pages

I'd just like to add that 4 out of every 3 people don't understand statistics.

And most people have more than the average number of feet.

[team_v]

As for Schrödinger's cat -- let's face it -- it's as dead as the parrot in Monty Python

Maybe in your universe buddy!



50:50 is the right answer

[Timbo]

My last word

Actually, mathematicians tend initially to have the greatest problem with this..er..problem, because they throw themselves into standard probability thinking. But it's a game, with rules, so conditional logic comes into play. It is critical to understand the rules as these affect the choices and the probabilities. If you rush in, you'll be fooled by applying your own probability rules (Scott, did you ever have an issue at school reading the question incorrectly? I did )

SpilledPrawn -- no-one in their right mind, least of all an engineer -- is going to obey your rules

As for Schrödinger's cat -- let's face it -- it's as dead as the parrot in Monty Python

[Dubya]

That's just because there's only 10 kinds of people in the world. Those that understand binary, and those that don't.

.... and those who understand the Monty Hall dilemma (and back the "house") and those who don't (and back their luck).

In a simple card version of the game, when there is no option to switch, the "dealer" has a 67% chance of winning. By switching you effectively take their position and assume their odds.

Is that the sound of pennies dropping?

You see Monty Hall presents two challenges in one:

- understanding it; and (the greater challenge),

- explaining it to those who don't,

which is worth it for the pleasure of seeing realisation dawn.

[Rocket36]

I'd just like to add that 4 out of every 3 people don't understand statistics.

That's just because there's only 10 kinds of people in the world. Those that understand binary, and those that don't.

[JustCruisn]

WOW 7 days and 10 pages

I'd just like to add that 4 out of every 3 people don't understand statistics.

[Dubya]

Regardless of what ANYONE says, the second choice is ALWAYS going to be 50/50. That is mathematical FACT and anyone who disagrees with that is simply wrong. When choosing between two things, whatever they are, it's ALWAYS 50/50, or 1 in 2.

Well, it is 1 in 2, or 50/50 for anyone who enters the room or tunes in after the dud door has been opened and when only two closed doors remain, presuming the new entrant doesn't know which door the contestant chose initially.

Or, as someone suggested, if there were 30 doors to start with and someone entered the show or tuned in after 28 dud doors had been opened (and did not know the contestant's initial choice) they, but only they (or someone with the same information), would have a 50/50 chance of choosing the GTI, as they have only two doors to choose from. But when the contestant chose, their chances of success were 1/n, where n is the number of doors (presuming only one GTI on offer).

But a contestant's chances of winning do not change as each dud door is opened: in the above examples they remain either 1/3 or 1/30.

But they are 50/50 for a new entrant who does not know what the contestant initially chose. For he has only two options, the contestant had 3 (or 30).

So expressed another way:

Should the new entrant back the contestant's first choice or choose the only other option?

Well, the contestant's choice has odds of 1/n whereas the odds that the other remaining closed door has the GTI behind it are n-1/n.

But as others have said, if this does not gel, apply the theory by having a friend lay out an Ace and two Jokers and try picking the Ace in the same fashion.

If you do not change your initial choice after one of the jokers is revealed you will "win" 33% of the time.

But if you switch your odds are increased by a factor of n-1.

[Buller_Scott]

No, it was a mistake on my part. The quote is actually attributable to Schoona, so I apologise for the misquote (see, I'm human, and am not afraid to admit it).

lol it's all good, no need to apologise- it's a dub forum! no bad blood.....

[Flighter]

lol flighter, good one! (i really, really hope you were joking, right? that you thought i posted that comment about the finite 100m point?)

No, it was a mistake on my part. The quote is actually attributable to Schoona, so I apologise for the misquote (see, I'm human, and am not afraid to admit it).

[Buller_Scott]

lol flighter, good one! (i really, really hope you were joking, right? that you thought i posted that comment about the finite 100m point?)

anywhoo.

damn straight it's obfuscation. as for the card thing- i am actually considering trying this at home, like, a THOUSAND times.

the only problem is, folks, im more than happy to try it, but i dont understand how it will prove the 2/3 theory correct.

here's why: my sister will have three cards. she will show me one card that she KNOWS for a fact is NOT my card. now, why in the eff does the fact that she's simply showing me a card that she knows is a dud, lend itself to MY chances, especially when i know for a FACT that she must, as according to the riddle, show me a dud card anyway?

if im going to be doing this a thousand times, knowing that every time, she will be showing me a dud card first, because it's her function in the riddle, then, quite simply, knowing the outcome of the first card to be revealed as a consistently certain failure, im simply going to tell her not to bother displaying the first card to me-she HAS to show me a dud first card, we both know this, so why bother?

so, if we stop bothering with the first card, and she simply asks me if i want to switch the card i've pegged as mine from the remaining two, why the hell are we now bringing a third card into the game in THEORY as adding to the elements in probability, AND YET we're not even bothering to pick a third (first shown) card out of the deck because we both know that the rules say it has to be a failure?

so how are my chances 2/3 again? in real life? with a deck of cards? actually doing this for real? <<--- that was directed more in the general direction of everywhere/one.


hey schoona, the horny dude will never ever reach the try line. google asymptote hyperbola and f(x)=1/x.

you can get closer and closer and closer, but technically, you will never actually touch the goal line.

[Flighter]

i KNOW probability, and any comprehensive mathematical explanations in this thread are REAL, and CANNOT be disproved by some playing cards from the local $2 shop. my explanations have a basis in mathematics, in mathematical probability.

You can prove it by trying it out, but it sounds like you are disinclined to. If you are so certain that you are right, why not do it and post your results to prove your case? Is it a case of "I could do that but I don't want to?"

Also, look up the term "empirical evidence" sometime (wikipedia will suffice), and consider it in relation to your statement about how useless those $2 cards are in this case.

[Flighter]

[QUOTE=schoona;478697][QUOTE=cme2c;478679]

You only move half the distance yes, but the 100m is a finite point, therefore you will reach it.

I think that statement says a lot about your mathematical ability.

[schoona]

[QUOTE=cme2c;478679][QUOTE=Buller_Scott;478661]

IF the host revealed that OUR DOOR had a goat and invited us to change, then our chances are resolved to 50/50 (only this removes the dependancy of the original choice)



I, too did 4 Unit maths. My wife has done statistics to 3rd year university. Yet she still plays poker machines. I have given up pointing this out, as I dislike pain.

Anyhow, the probabilities are fixed with the contestants choice. Whatever the choice, the host can open a door which hides a goat. It is obfuscation. Smoke and mirrors. 3 card monte.

On the subject of theory and practice.

An engineer and a physicist, both male and heterosexual, are placed at one goal line of a football field. At the other goal line opposite each is a naked, desireable woman.

They are each told that a whistle will sound. At each whistle, thay can move half the distance to the other end of the field. Each time they can halve the distance again, and so on.

So assuming a 100m field, you would be 50, then 75, then 87.5 m and so on.

The whistle blows. The physicist doesn't move. The engineer runs to the half way line. The physicist explains that as he can only halve the distance each time, he can never reach the other goal line. The engineer is given this information, agrees, but says "I can get close enough".

Try it with $2 shop cards. The Monty Hall problem, that is.

You only move half the distance yes, but the 100m is a finite point, therefore you will reach it.

[cme2c]

[QUOTE=Buller_Scott;478661][QUOTE=Spilledprawn;478589]

IF the host revealed that OUR DOOR had a goat and invited us to change, then our chances are resolved to 50/50 (only this removes the dependancy of the original choice)

isnt that precisely what happens in this particular scenario, that the host knowingly selects the first door to be opened as one which houses a goat, THEN he asks if you want to switch?

a couple of thread pages ago, there was some discussion in which the function of the host was established as intentionally opening a door which he knew housed a goat, before asking the contestant if he wants to switch.

I, too did 4 Unit maths. My wife has done statistics to 3rd year university. Yet she still plays poker machines. I have given up pointing this out, as I dislike pain.

Anyhow, the probabilities are fixed with the contestants choice. Whatever the choice, the host can open a door which hides a goat. It is obfuscation. Smoke and mirrors. 3 card monte.

On the subject of theory and practice.

An engineer and a physicist, both male and heterosexual, are placed at one goal line of a football field. At the other goal line opposite each is a naked, desireable woman.

They are each told that a whistle will sound. At each whistle, thay can move half the distance to the other end of the field. Each time they can halve the distance again, and so on.

So assuming a 100m field, you would be 50, then 75, then 87.5 m and so on.

The whistle blows. The physicist doesn't move. The engineer runs to the half way line. The physicist explains that as he can only halve the distance each time, he can never reach the other goal line. The engineer is given this information, agrees, but says "I can get close enough".

Try it with $2 shop cards. The Monty Hall problem, that is.