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a "Price is Right" game
Matthew Leverton
Supreme Loser
January 1999
avatar

I like to throw the same teaser puzzle at people I know:

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You are on a game show, playing a game that consists of three doors and one prize. You have to guess which door contains the prize. After you pick a door, the game show host opens up a door (of the two that you did not choose) that he knows does not contain the prize. He then asks you, "Do you want to keep your original choice or switch to the other closed door?"

Most people (math "nerds" excluded) stubbornly believe that the odds are 50-50 for each of the two doors that are closed. But really, you have 2/3 chance if you switch and only 1/3 if you keep the door. The game show host is essentially saying, "Would you rather pick one door or two doors?" But most people stick with their first choice, instead of picking the other two - because they only see the puzzle as 50-50.

--

Yes, it's old, and has even been mentioned here before. I bring it up because a friend of mine (whom I stumped with the puzzle) had the "Price is Right" on today, when a certain game came on...

There were four "doors," and three of them had prizes. The contestant would get all three prizes if she correctly picked all three spots at random. If she missed any, she got none of them. So after picking three doors at random, Bob Barker opened up two correct doors that the contest had picked, leaving two more closed.

Now, as I'm sure you've guessed, the contestant had the choice: would you like to switch your initial pick to the other closed door or keep it the same? Bob even offered the contestant $500 to stay with her original choice. Thinking the odds were 50-50 while keeping $500 no matter what, she stayed.

And she lost. Now I wonder if Bob had asked her: "Do you want $500 plus a 25% chance of $6000 or $0 with a 75% chance of $6000?", what she would have done?

--

Someone here, I think perhaps Evert, has said several times that people are bad at understanding or relating with natural probabilities. This is just a classic example...

If any of you have some puzzles that stump people (and cause them to stubbornly argue their case), I'd like to hear them. 8-)

HoHo
Member #4,534
April 2004
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We had a discussion about that same quiz on some other forum. It took areound 250 posts and several example programs with millions of iterations to make people believe it is not 50:50 chance.

We also had a very long discussion about if 0.9(9)=1 or not, that took about 200 posts to make things clear.

These threads are here:
http://foorum.hinnavaatlus.ee/viewtopic.php?t=175463&highlight=999
http://foorum.hinnavaatlus.ee/viewtopic.php?t=175550&highlight=kolm+ust
I doubt more than 2-3 people in a.cc could read these threads though :P

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Hard Rock
Member #1,547
September 2001
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We had that brought up in class a while ago, and one of our TA's explained it.

I've completely forgotten the logic we were told then. :D

[edit]
Okay I remember now....
I'll be honest though, it's something you need to think about :D

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ReyBrujo
Moderator
January 2001
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I remember at school, a certain point nobody passed in a test. There are three guys, and three doors. Only one has a prize. The first guy comes and chooses a door. He has 1/3 of probabilities of choosing the winner. What is the probability for the second guy to choose the correct door?

Most (if not all) replied 1/2, but that would be true if there were only 2 doors at the beginning. Since there are three doors, you need to expect the previous guy to fail, which makes things even.

Quote:

We also had a very long discussion about if 0.9(9)=1 or not, that took about 200 posts to make things clear.

Oh, interesting, the 0.999... article recently reached featured status at Wikipedia (the scale would be start, stub, article, good article and featured article).

--
RB
光子「あたしただ…奪う側に回ろうと思っただけよ」
Mitsuko's last words, Battle Royale

FMC
Member #4,431
March 2004
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Quote:

Most people (math "nerds" excluded) stubbornly believe that the odds are 50-50 for each of the two doors that are closed. But really, you have 2/3 chance if you switch and only 1/3 if you keep the door. The game show host is essentially saying, "Would you rather pick one door or two doors?" But most people stick with their first choice, instead of picking the other two - because they only see the puzzle as 50-50.

err... i don't understand why it's not 50/50 :P

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Inphernic
Member #1,111
March 2001

How about:

http://www.duckiehorde.net/whee.jpg

Onewing
Member #6,152
August 2005
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Quote:

Do you want $500 plus a 25% chance of $6000 or $0 with a 75% chance of $6000?

Which is also like betting $500 dollars that you are going to get $6000 with a 75% chance. What I mean, is if you were playing cards and you knew you had a 75% chance of winning a pot of $6000, would you bet $500? Believe it or not, that's still kind of iffy to me. If you win, you increase your profit by 600%, but that 25% that you might lose is still a big gap in my opinion. Now, if the pot was like $10,000 or higher, I'd probably go for it. And if I were rich, I'd go for the $6000 or even $501. :P

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Hard Rock
Member #1,547
September 2001
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Quote:

err... i don't understand why it's not 50/50

It's becuase you have a greater chance of picking an empty door then one with the prize.

[edit]
You can also look at Inphernic's diagram.

The green->red is when you land on the right one but move away. the red->green is when you pick the wrong door and move to the right one.

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ReyBrujo
Moderator
January 2001
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I believe most people don't switch because, if you stay and the prize is in the other door, you have bad luck, but if you switch and the prize was in the door you originally chose, you are an idiot. Psychology, I would call it.

--
RB
光子「あたしただ…奪う側に回ろうと思っただけよ」
Mitsuko's last words, Battle Royale

Simon Parzer
Member #3,330
March 2003
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Essentially you have two scenarios here:

1) The player has chosen the right door (chance = 1/3), that way by switching he is going to loose
2) The player has chosen the wrong door (chance = 2/3).
The showmaster has to choose one door to remove. He can't choose the door already chosen by the player and he can't choose the one with the prize. So he MUST choose the second wrong door. In this case you win by switching no matter what.

I believe this explanation is the most understandable.

Matthew Leverton
Supreme Loser
January 1999
avatar

Quote:

Which is also like betting $500 dollars that you are going to get $6000 with a 75% chance.

I'd take those odds all day long. I play twenty times. I lose $500 five times. I gain $6000 fifteen times. $87,500 in my pocket. :P

Of course you only play once, not getting the benefit of winning over time. But also, you aren't losing $500 because you never had it to begin with. By that logic, if you keep the $500 (and miss) and don't get the $6000, you went home a $5500 loser.

Quote:

err... i don't understand why it's not 50/50 :P

Heh, that's why I love this puzzle. Put it this way:

Scenario 1
You pick door 1.
Host says, "Would you rather open doors 2 AND 3 instead?"

What would you do? You'd pick doors 2 and 3 if you weren't stupid. Obvious, eh? Change it slightly:

Scenario 2
You pick door 1.
Host says, would you rather open doors 2 AND 3 instead? For suspense, I will open an incorrect door first.

What would you do? You'd still pick doors 2 and 3 if you weren't stupid. Change it slightly:

Scenario 3
You pick door 1.
Host opens door 2, revealing that it is empty. He asks, would you like to open door 3 instead?

Now people scratch their heads and think it's 50-50, but it's the same thing as the first two options.

If that doesn't help, think of it this way: you have 2/3 chance of getting wrong initially. If you swap, you'll get it right. Thus 2/3 chance of winning if you swap. On the other hand, you have a 1/3 chance of getting it right initially. If you don't swap, you'll be right 1/3 of the time.

Quote:

I believe most people don't switch because, if you stay and the prize is in the other door, you have bad luck, but if you switch and the prize was in the door you originally chose, you are an idiot. Psychology, I would call it.

Yes, but most people think it's 50-50. If they knew they had better odds, they would switch. However, in that context I agree that they psyche themselves out by having this "gut feeling" that they were correct the first time, because lo and behold, there are two doors left standing - including the one they initially picked!

Anyway, submit your puzzle ideas! .999...=1 is good, but hard to get the common person to have interest. I like to confuse people.

Onewing
Member #6,152
August 2005
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Quote:

By that logic, if you keep the $500 (and miss) and don't get the $6000, you went home a $5500 loser.

True, I guess I'd have to go and buy the lower-model PS3 and wallow in shame.

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Billybob
Member #3,136
January 2003

I think most people don't know the host has set parameters for what door he opens, they think it's all random.

Matthew Leverton
Supreme Loser
January 1999
avatar

I don't think so. The problem specifically states, he opens a door "that he knows does not contain the prize". People are told it's not random. It wouldn't make any sense if the host didn't know. He could open up the right door. :P They still think that it's 50-50.

And anyway, whether the host knows or not is not relevant to the fact that opening up two doors is better than one. And everyone would open up two doors if asked in a straightforward manner.

Billybob
Member #3,136
January 2003

In the puzzle or on The Price is Right?

Matthew Leverton
Supreme Loser
January 1999
avatar

In both cases it's made clear that the host knows. And in both cases, it would be dumb if the host didn't. The host could turn up the right door (in the puzzle) or the wrong door (in Price is Right), thus ruining the game.

Inphernic
Member #1,111
March 2001

Quote:

If that doesn't help, think of it this way: you have 2/3 chance of getting wrong initially. If you swap, you'll get it right. Thus 2/3 chance of winning if you swap. On the other hand, you have a 1/3 chance of getting it right initially. If you don't swap, you'll be right 1/3 of the time.

The way I see it:

Pr(A) = Pr(winning by picking correct and sticking to it) = 1/3
Pr(B) = Pr(winning by picking wrong and swapping it to correct) = (1/2*1/3)+(1/2*1/3) = 1/3
Pr(C) = Pr(winning at all) = Pr(A)+Pr(B) = 2/3

Quote:

you have 2/3 chance of getting wrong initially. If you swap, you'll get it right.

Pr(pick wrong #1) = 1/3
Pr(pick wrong #2) = 1/3
Pr(swap to correct after picking wrong #n) = 1/2

Pr(win by swapping from wrong #1) = (1/3)*(1/2) = 1/6
Pr(win by swapping from wrong #2) = (1/3)*(1/2) = 1/6

Pr(win by swapping from wrong #n) = (1/6)+(1/6) = 2/6 = 1/3

In short, swapping all the time would not lead to a 2/3 probability of winning, but the probability of winning is 2/3 if all contingencies are taken into account.

Or did I misunderstand the initial problem/what you were trying to say? ..wording is so much fun. ;)

Matthew Leverton
Supreme Loser
January 1999
avatar

If you pick wrong and swap, you always get it right by virtue of the game host opening up the incorrect door prior to swapping. Thus, in context of the game, you always have 2/3 chance of getting it right after a swap.

See the table on <http://www.matthewleverton.com/doorprize/> for clarification.

ReyBrujo
Moderator
January 2001
avatar

First game, stayed with my original option, won. I retire. :P

--
RB
光子「あたしただ…奪う側に回ろうと思っただけよ」
Mitsuko's last words, Battle Royale

Fladimir da Gorf
Member #1,565
October 2001
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You're all wrong, the show host just slips the prize quickly in his pocket when he opens the "wrong" box. You loose!

EDIT: Unless you've chosen the right box to begin with, of course...

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Peter Hull
Member #1,136
March 2001

My wife and I went to a party last weekend. There were eight other couples there. At the beginning everyone circulated, shaking hands with each other. No-one shook hands with their partner, of course.

Later on, I asked everyone how many people they had shaken hands with, and, amazingly, everyone had shaken hands with a different number of people.

How many people did my wife shake hands with?

(This was in the Guardian this weekend)

Pete

FMC
Member #4,431
March 2004
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At first glance it seems easy:
in case there are 8 couples, there are 16 people, you shake hands with 14 (-1 you, -1 your partner)

Clearly my answer is wrong :P

[FMC Studios] - [Caries Field] - [Ctris] - [Pman] - [Chess for allegroites]
Written laws are like spiders' webs, and will, like them, only entangle and hold the poor and weak, while the rich and powerful will easily break through them. -Anacharsis
Twenty years from now you will be more disappointed by the things that you didn't do than by the ones you did do. So throw off the bowlines. Sail away from the safe harbor. Catch the trade winds in your sails. Explore. Dream. Discover. -Mark Twain

Hard Rock
Member #1,547
September 2001
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Actually there would be 18 people

(8x2(other people) + 2(you +wife))

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Rick
Member #3,572
June 2003
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Quote:

Now, as I'm sure you've guessed, the contestant had the choice: would you like to switch your initial pick to the other closed door or keep it the same?

At this point, with 2 doors left, and 1 is good and 1 is bad, doesn't it become 50/50?

Obviously it wasn't 50/50 when she started, but it became 50/50 right? I mean he gave her a chance to switch, which really makes it a new game. Now the game became, she has 2 doors. 1 has a prize, the other doesn't.

Maybe my thinking is wrong, but it seems logical to me.

========================================================
Actually I think I'm a tad ugly, but some women disagree, mostly Asians for some reason.

ReyBrujo
Moderator
January 2001
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Friends don't shake hands between each other, pat him in the back, kiss her in the cheek, etc. Since your wife is pretty friendly, I guess she shook nobody's hands :P

--
RB
光子「あたしただ…奪う側に回ろうと思っただけよ」
Mitsuko's last words, Battle Royale

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