Had quite a discussion over this at my Monday night poker game.

Here's the deal: In the movie "21", the professor picks a student and says, "Pretend you're on a game show. There's three doors and you can pick one. One door has a new car behind it and the other two have a donkey."

Lets say you pick door #1. You have a 1 in 3 chance of being right.

Door #2 is revealed and it contains a donkey.

You're given a chance to keep door #1 or change to door #3. Is there a statistically correct answer to this?

According to the movie, you change doors. They say that eliminating one of the doors changed the odds in favor of the 'originally' unsellected door. A couple guys at poker argued that it's now a simple 50/50 shot.

If there are still 3 doors, the odds are the same.

Please don't make me give examples where people know what is right, they know what is not right, and still make the wtong choice.

The odds only change if the person doing the selecting is smart enough to eliminate door 2 altogether, leaving it a choice between 1 and 3.

In the game show example it is an easy choice. People often make the wrong choice when the roads to the benefits are not as easy or fun.

Jeff - I agree with the guys at poker saying that with one door eliminated, it's now a 50/50 shot. I understood why they said to change doors in the movie, but it's still comes down to a 50% win/50% lose situation. I'd stick with my original door!

I'm still strugling with the math on this.

It seems like there could be something to the fact that eliminating one door does in fact swing the odds. If they eliminated one door before you chose your; yes, it would be a 50/50 shot.

I can't help that feel that having one door held out of the choices to eliminate, swings the odds in favor of the other door.

Sorry, I didn't see the movie.

Are they saying that door #1 was a 1 in 3 chance, and since door #2 was assumed to be eliminated that door #3 is now a 50-50. And now they are concluding that a 50-50 is better than the 1 in 3 odds? If so, that's just stupid.

Regardless, if you are to assume that #2 is eliminated, the two remaining are 50-50. (remember that a previous spin of the wheel, previous roll of the dice, or previous flip of the coin does not change the odds of what is in front of you.

On the first post, (since I didn't see the movie) I thought the point was supposed to stress how you have to keep an accurate count all the time no matter how meaningless it might seem for each and every card.

That's not stupid because a 50% shot at winning IS better than a 33.3% chance, but that's not what they're saying.

The student chose door #1. Door #2 was then revealed as a loser. The movie claims that door #3 is now a better than 50/50 favorite. I'm not sure if they're saying rather than going to 50/50 it went to 33.3/66.6 but they're definately saying that the odds are now in favor of door #3.

I understand that this is a movie and many times movies depict things that aren't true but this makes some sense to me. I'd just like to know what is actually statistically correct. DID having door #1 'set aside' when one loser was revealed actually sway the odds in favor of door #3?

If that donkey only has two places left to hide, that sounds like a 50-50 chance to me. I guess I'll have to see the movie to see how they spun it, but I might forget this discussion before it's available at BlockBuster<g>.

Yes, you are correct that 50% is better than 33%. I took about 8 years worth of college math is 3 years. I haven't retained very much of it in those 30 something years since. But I am familiar with various levels.

I resereve the term stupid, but for the sake of this discussion, let's just use "theatrical". What is theatrical about this situation is how they came up with their conclusions. Are there other cicumstances in the movie that haven't been revealed here? It's 50-50 until they can show one has an edge. I'm guessing it is a theatrical edge. Perhaps they went to our local schools to learn math? I understand math is not their forte.

You say it makes sense to you. How? I'm willing to learn. Would it help to use another another analogy that brings your point to light, other than a game show?

Hi! My name is Tim and I'm Eric&Ellen's son. I'm a math major at Purdue, and this is a very common problem in probability textbooks.

Referring to "Introduction to Probability" by Douglas G. Kelly, copyright 1994, pages 110-112. This is, in fact, the Monty Hall Paradox. If you calculate the conditional probability, you discover that the probability of door 1 winning continues to be 1/3rd while the door 3 winning probability is 2/3rd.

The key is assuming the host knows which door is the winner, and the door that he opens will be neither the door you chose, nor the one with the good prize. If you pick door 1, and the prize is actually behind door 1, then the host can open either door 2 or door 3. If you pick door 1, and the prize is actually behind door 2, then the only door the host can open is door 3. Similarly, if you pick door 1, and the prize is actually behind door 3, then the only door that the host can open is door 2.

Really, then, there are only four ways this can go:

Prize behind 1, host opens 2. (call it P1H2)
Prize behind 1, host opens 3. (call it P1H3)
Prize behind 2, host opens 3. (call it P2H3)
Prize behind 3, host opens 2. (call it P3H2)

We all agree that the chance of the good prize being behind any particular door is 1/3, so the probabilities for P2H3 and P3H2 are each 1/3. For the case of P1, assuming the host opens one of the other two doors at random, then P1H2 and P1H3 each equal 1/3 * 1/2 = 1/6.

Here are our probabilities now:
P1H2 = 1/6
P1H3 = 1/6
P2H3 = 1/3
P3H2 = 1/3

Now, suppose that the host opens door 2. So we now know that the prize is behind either door 1 or door 3, and we can rule out the P2H3 option with probability 1/3. Since the host opened door 2, we can also throw out the P2H3 option with probability 1/3. This leaves us with only two possible outcomes: P1H2 and P3H2, each with probabilities 1/6 and 1/3, respectively.

Using the rules of conditional probability, we get the following: 1/6 + 1/3 = 1/2.
Probability of P1H2 = (1/6)/(1/2) = 1/3.
Probability of P3H2 = (1/3)/(1/2) = 2/3.

So, the movie is correct. The chance that the prize is behind the 3rd door is actually higher.

Tim

yes,that was just the way i was going to explain it<G>

Thanks Tim. Eric and Ellen didn't raise a dummy. This is new math for me since I went to college in the mid 70s.

I can buy the fact that if the host knows the answer and reveals a known booby prize as a part of the equation, we have a controlled situation that can alter the playing field.

There still seems to be holes. Since I didn't see the movie, I don't know how it was presented or what the rules were.

Hole 1. If door 2 was picked randomly and revealed a donkey, wouldn't we back down to a 50-50 shot?

Hole 2. What if there were 3 contestants. Contestant A picked door 1, contestant B picked door 2, and contestant C picked door 3. Door 2 revealed a donkey and contestant B was eliminated.
Does contestant C have better odds than A?
What if A and C both wanted to change with each other? Would both of their chances increase?
Do the odds change if the host eliminated B/2 selectively or if contestant B just happened to open his door first?
I'm thinking 50-50.

Hole 3. What if the host doesn't remember the correct door? If he makes a random choice, he is no longer a controlling factor so we are back at an even playing field.

It "seems" that from a contestants perspective, looking forward, the odds don't change. From a known end, working backwards with manipulation, the odds can be controlled, yet I'm short of being convinced a player should change. I haven't quite absorbed the present tense of making a decision versus the future tense of controlled odds.

I'm gonna have to sleep before I decide how I feel about Hole 2 and it's variations.


Thanks again Tim. I appreciate your explaination and lesson in new math.

Note to Eric and Ellen: I deleted your other post in the What If topic that "appeared" to be a reply to this topic and contained some of the same info provided by Tim. It appeared to be in the wrong topic and the deletion of the note was just general housekeeping. This was not censorship of any kind. If there were any points in that note that you want to repost, feel free.

Tim! Thanks for this explanation! I graduated from college in 1978 and although I wasn't a math major, I was very good at it and took some math courses as electives to bring my average up (after a bad sophomore year).

Your lesson in probabilities was really interesting and brought back a lot of memories. The "Monty Hall Paradox" - fascinating! I totally understand this now. On the surface, it's not apparent but mathematically, it makes sense.

You're obviously a bright young man. Good luck with your studies at Purdue and in your future endeavors!!

So Tim, have you thought of playing Blackjack?

Thanks for jumping in here and explaining something I obviously couldn't.

Tim, I only had one year of math in college back in the 60s (and hated it<g>), so I need one more bit of explanation please ....

When you say there are only four ways this can go, those four scenarios are based on the fact that we already know door 2 is out. I'm fine with that, BUT when you discuss the case of P1, you still list P1H2 and P1H3 with equal weight, even though we know the host didn't open door 3 - that part has me confused. How can those two be equally weighted when we already know P1H3 didn't happen? I have the same basic question about P2H3 and P3H2 being equal in the new probabilites since we know the prize isn't behind door 2. The only possibilities this "old math" mind can come up with are P1H3, P1H1, P3H1 and P3H3.

Thank you for a fascinating post - and thanks to Eric & Ellen for drafting you into this discussion.

Hi, guys. Tim here again. To make life easier for all of us I just registered with TalkVegas for myself.

JMT said:
[quote]

Welcome to TalkVegas, Tim. We appreciate your input and look forward to more discussions on this and other subjects.

JMT said:
That

I will take a vowel, Vanna .....

And I am a retired Architect...... :rolleyes:

I'll take Potent Potables for a thousand, Alex.
Ellen

I'm sorry I posted something that actually made us think. I promise I won't do it again.

But on second thought, it got us a new member. And one who's smart at that!