Brain teaser for you

[chumdawg]

If you are tired of expending energy figuring out how to get Shaq to Dallas, then wrap your brain around this one:

You are a contestant on a game show. There are three doors you can choose from. Behind two of the doors is...well, let's make this somewhat Mavs related...behind two of the doors is Antoine Walker. That's right, he's in two places at once. Behind one of the doors is Shaquille O'Neal. You get the idea. Two of the doors have losing options, and one has the grand prize.

You get to choose a door. Now see, the game show host knows which door contains the grand prize. After you make your selection, the host will open one of the doors you did not choose--revealing, of course, a losing option. He will then give you a choice: you can stick with your original choice, or you can switch to the other door.

The question is: are you better off staying put, or are you better off switching? Or does it make a difference?

[u2sarajevo]

Originally posted by: chumdawg
If you are tired of expending energy figuring out how to get Shaq to Dallas, then wrap your brain around this one:

You are a contestant on a game show. There are three doors you can choose from. Behind two of the doors is...well, let's make this somewhat Mavs related...behind two of the doors is Antoine Walker. That's right, he's in two places at once. Behind one of the doors is Shaquille O'Neal. You get the idea. Two of the doors have losing options, and one has the grand prize.

You get to choose a door. Now see, the game show host knows which door contains the grand prize. After you make your selection, the host will open one of the doors you did not choose--revealing, of course, a losing option. He will then give you a choice: you can stick with your original choice, or you can switch to the other door.

The question is: are you better off staying put, or are you better off switching? Or does it make a difference?

It only makes a difference if when he opens the other door and Shaq is sitting there... Therefore I know I am stuck with Walker and I blow my head off. [img]i/expressions/moon.gif[/img]

[Drbio]

Stay stay stay. You originally had a 33% chance of Shaq. Now that one choice is revealed you are at 66%. If you switch you fall into the 50% chance.


Then you blow your head off because we all know that Mavs luck has traditionally given us Walker quality results.

[chumdawg]

Right on, Doc, that you had a 33% chance of choosing the Big Fella in the first place. But that means that you had a 66% chance of NOT choosing him, which is why you should switch. Because if in fact you didn't choose him in the first place, then the host is going to leave him there for you when he opens one of the other two doors to reveal Antoine. In other words, if you didn't choose him in the first place, you are guaranteed to get him if you switch.

If you switch, the only way you lose him is if you landed him in the first place (33% chance). But yeah, with the Mavs luck that's about what would happen.

[sike]

i'm not sure this is logical...the variables change as the problem changes….in the problem regarding three doors, of course you had a 33% chance of guessing rightly…but now that there are only two doors it is not logical to assume that you can have any higher than a 50% shot at getting it correct. If at any point the problem comes to a two door proposition where either door could be the answer…you are stuck with a 50% percent situation. No matter how many doors or what percentage you began with, the situation you find yourself in is a one out of two…this giving you a 50% chance….when the variables change, the math must change. When you eliminate a door you have changed the math..thus changing the problem. I am no mathematician, so if I am proved incorrect I can live with it. Then again I may not understand the question….is the host a shady character??? Can he move Shaq once the selection has begun??? My head hurts… [img]i/expressions/face-icon-small-wink.gif[/img]

[madape]

It's 50/50 no matter which door you choose. You have the same odds if you switch as you do if you don't.

[chumdawg]

What makes this such a teaser is that it's so counter-intuitive. Normally you would look at two doors and think the odds are 50/50. But they would only be 50/50 if BOTH doors were arrived at randomly. That's not the case here. Only the door you chose initially got there randomly. The host "selected" the other door based on his knowledge of where Shaq was.

Basically, the host cleared off all the other wrong choices and left you one door only. Either that door contains Shaq (since all the other Antoines have been cleared away), or your door contains Shaq. Shaq has to be in one place or the other.

So the chances are the same as they were in the beginning. Either you picked Shaq (1-in-3 chance) or you didn't pick Shaq (2-in-3 chance). If you switch doors, then the 2-in-3 odds are now on your side.

It's easier to grasp if you make it 1,000 doors instead of 3. You pick one door from 1,000, and then the host "clears off" 998 doors by opening them to show you Antoines. Now there are two doors left. If you stick with your first choice, then you are banking on your random 1-in-1,000 odds that you started with. But if you switch, then it's 999-in-1,000 in your favor.

[twelli]

the chances are 50/50, unless of course you have bribed the host beforehand promising him Dirk, than you will know which door to choose

[sike]

Originally posted by: chumdawg
What makes this such a teaser is that it's so counter-intuitive. Normally you would look at two doors and think the odds are 50/50. But they would only be 50/50 if BOTH doors were arrived at randomly. That's not the case here. Only the door you chose initially got there randomly. The host "selected" the other door based on his knowledge of where Shaq was.

Basically, the host cleared off all the other wrong choices and left you one door only. Either that door contains Shaq (since all the other Antoines have been cleared away), or your door contains Shaq. Shaq has to be in one place or the other.

So the chances are the same as they were in the beginning. Either you picked Shaq (1-in-3 chance) or you didn't pick Shaq (2-in-3 chance). If you switch doors, then the 2-in-3 odds are now on your side.

It's easier to grasp if you make it 1,000 doors instead of 3. You pick one door from 1,000, and then the host "clears off" 998 doors by opening them to show you Antoines. Now there are two doors left. If you stick with your first choice, then you are banking on your random 1-in-1,000 odds that you started with. But if you switch, then it's 999-in-1,000 in your favor.

ok..the game host is dirty and the doors are not random.....but if your option in making a choice is one of two doors...then despite previous options your current option is a ONE OF TWO...this current option, the immiedate pick before you is a fifty-fifty option....all other previous percentages in this door game change....NOW YOU HAVE A 50% CHANCE...

[madape]

I get it now. I agree with chumdawg. We know nothing more about the door we originally picked. It was a 1/3 chance when we picked it. The fact that one door is revealed does not change the fact that our original pick is still 1/3.

We know much more about the OTHER closed door. The current open door was chosen based on our original choice. If we choose the other closed door, we have a 50% of getting Shaq.

[Usually Lurkin]

It's true. It doesn't make sense to me, but I think it's true.

I ran a simulation in R. Here's the percentage correct for staying and switching:
[1] "stays:: " "0.321138211382114"
[1] "switch: " "0.665354330708661"

still doesn't make sense to me. Either I'm a better programmer than statistician, or Chumdawg is wrong.


stays <- 0
switches <- 0
staywin <- 0
switchwin <- 0

for (x in 1:1000)
{
stay <- sample(c(0,1),1) ## choose 1 or 0 to make a stay or switch trial

cp <- c(1,2,3) ## possible choices of doors
cp2 <- c(0,0) ## will hold the possible choices of the 2 doors after elimination

shaqat <- sample(cp,1) ## this is where shaq is (either 1,2,3)
choice1<- sample(cp,1) ## this is your first choice (either 1,2,3)


## now, host eliminates a non-shaq door
if (choice1 == 1) {
if (shaqat == 2) cp2 <- c(1,2) ## choice is now between what you have and where shaq is
if (shaqat == 3) cp2 <- c(1,3) ## choice is now between what you have and where shaq is
if (shaqat == 1) cp2 <- c(1, sample(c(2,3),1)) ## what you have now is where shaq is, host eliminates 1 of the other 2 doors
}

if (choice1 == 2) {
if (shaqat == 1) cp2 <- c(2,1)
if (shaqat == 3) cp2 <- c(2,3)
if (shaqat == 2) cp2 <- c(2, sample(c(1,3),1))
}

if (choice1 == 3) {
if (shaqat == 1) cp2 <- c(3,1)
if (shaqat == 2) cp2 <- c(3,2)
if (shaqat == 3) cp2 <- c(3, sample(c(1,2),1))
}

## choose to stay
if (stay == 1)
{
choice2 <- cp2[1] ## the first cell of cp2 is always choice1
stays <- stays + 1 ## total up the number of times you stay
if (shaqat == choice2) staywin <- staywin+1 ## total up the number of times staying wins
}

## choose to switch
if (stay == 0)
{
choice2 <- cp2[2]
switches <- switches +1
if (shaqat == choice2) switchwin <- switchwin+1
}

## calculate and display win percentages for stay and switch
print(c("stays:: ", staywin/stays))
print(c("switch: ", switchwin/switches))

## stay will win 33% of the time
## switch will win 66% of the time
}

[sike]

if the choice is between two doors, dispite previous choices....there are still two doors and your one choice. One out of two....you know that you have a one out of two chance now....if i am wrong I dont know why.

of course they revealed an empty door....when the information changes so does the percentage....i understand the explination, but i still see a choice between two doors...and one choice....

[Usually Lurkin]

sike,

when you make your first choice, you will be wrong 2 out of 3 times.
Each of those times you are wrong on your first choice, switching on your second choice will make you right. That is, 2 out of 3 times you play the game.

when you make your first choice, you will be right 1 out of 3 times.
Each of those times you are right on your first choice, switching will make you wrong. But that's only 1 out of 3 times.

[sike]

Originally posted by: Usually Lurkin
sike,

when you make your first choice, you will be wrong 2 out of 3 times.
Each of those times you are wrong on your first choice, switching on your second choice will make you right. That is, 2 out of 3 times you play the game.

when you make your first choice, you will be right 1 out of 3 times.
Each of those times you are right on your first choice, switching will make you wrong. But that's only 1 out of 3 times.

but you can never know this....they certainly are not going to tell you when you are right and wrong..so all you see is the door you chose, the open door which of course is empty, and the other yet to be chosen door....I understand UL, but the person making the choice cannot know if his original choice was right or wrong....because of course they will only reveal a door that is wrong. I still dont see how the uninformed player of this game could possibly know anthing other than the door they opened was the wrong door...is of course the door they will open. Unless all the while, and this may be what i was missing, you guys are saying that if the original pick was wrong they would have simply revealed the right pick and our game player would lose....

[chumdawg]

Wow, UL, that's great work! And yes, Madape, you said it much better than I did. That's the key: you don't know anything more about your own door, but now you know a lot more about the other doors. It's not that the host was "shady." Rather, it's just that he is going to reveal to you a door that doesn't have the grand prize. And his revealing that to you doesn't change anything about your own choice.

You might hear this puzzle phrased in a different way. There are three prisoners: Tom, Dick, and Harry. They are on death row, and the prison has decided that one of them will be executed in the morning. The prison warden tells the guard which one will be executed, but the guard is instructed not to tell the prisoners who it will be. Well, Tom is freaking out. He knows his chances of dying are one-in-three, but he wants to know something more. He begs the guard to tell him something, anything. So the guard says "Dick will not die in the morning."

Did Tom's chances of dying just go up to 50%, or are they the same as they were in the beginning?

[sike]

Originally posted by: chumdawg
Wow, UL, that's great work! And yes, Madape, you said it much better than I did. That's the key: you don't know anything more about your own door, but now you know a lot more about the other doors. It's not that the host was "shady." Rather, it's just that he is going to reveal to you a door that doesn't have the grand prize. And his revealing that to you doesn't change anything about your own choice.

You might hear this puzzle phrased in a different way. There are three prisoners: Tom, Dick, and Harry. They are on death row, and the prison has decided that one of them will be executed in the morning. The prison warden tells the guard which one will be executed, but the guard is instructed not to tell the prisoners who it will be. Well, Tom is freaking out. He knows his chances of dying are one-in-three, but he wants to know something more. He begs the guard to tell him something, anything. So the guard says "Dick will not die in the morning."

Did Tom's chances of dying just go up to 50%, or are they the same as they were in the beginning?

this must be different....because in this case, there is no choice and the outcome is set....

[chumdawg]

Exactly, Sike. The outcome is already set. And for that reason, it doesn't change things at all when the guard "removes" one of the initial options.

The way the problems are similar is to imagine the "outcome" of the prisoner problem as being the same 1-in-3 random chance that the game show contestant had when he first chose a door. Tom might be the guy who is going to die, and he might not be. The contestant may have chosen Shaq with his first choice, and he may not have. The 1-in-3 odds are the same.

[sike]

i give..... [img]i/expressions/face-icon-small-wink.gif[/img]

[AnMan21]

This is the Monty Hall - Let's Make a Deal dilema. What chum is saying is correct. he hit it even closer on the head when he pushed the amt of doors up to 1000. the chances of you actually selecting the prize are 1/1000 on the first go round. if they reveal 998 losers, you know one of the last two doors has the prize - either the one you picked, or the other closed door. every person would switch knowing the odds of picking the 1/1000 winners first is unlikely. i was a probability/stats major in college and we studied this to no end...the thing about only having 3 doors, is the doubt that creeps into the mind thinking "I may have picked correctly."

[chumdawg]

Right on, Anman. Our brains can play tricks on us sometimes.

Since you were a stat major, you may be able to help me with this. I've never understood that one problem that goes something like: how many people do you have to gather in a room to guarantee a 50% chance that someone shares your birthday. I forget exactly how it goes. Do you know the one I am talking about?

[AnMan21]

That one goes something like this:

The probability of an event happening equals one minus the probability of it not happening:
P(S) = 1 - P(F) where S: success and F: failure

Let: P(N) = the probability that no one has the same birthday
and let P(S) = the probability that AT LEAST two people share the same birthday

So it stands to reason that P(S) = 1 - P(N)

Therefore, you need to find P(N) - the probability that no one has the same birthday.
Start like so: Amy enters a room. The probability that she has a unique bday is 365/365. Now enters Joe. In order for their bdays to be different his bday must fall on one of the 364 other days of the year. the probability these unrelated events happen is the product of these fractions 365/365*364/365 = 365*364/365^2 = P(N) If another person enters they must have a bday coming from one of the 363 remaining days. So if you continue this process you end up with a P(N) = 365*364*363*...(365-k)/365^k. where k = # of people in the room. Subtracting your answer from one leaves the probability of success. If I remember correctly I think 23 people in a room results in slightly over .5 or 50% probability. I believe as you approach 50 people in a room, the probability approaches like 96% or so...

Long winded answer, but I hope that helps.

[chumdawg]

Thanks for the explanation. That sounds like what I remember. You wouldn't figure that only 23 people would give you a 50% chance!

[AnMan21]

I surely wouldn't go betting on it. But I'm an actuary and I try to avoid risk whenever possible - hahaha

[Drbio]

Let us not forget who nailed the solution huh?

[img]i/expressions/face-icon-small-wink.gif[/img]

[mavsman55]

I have never been more confused in my life than I am right now... [img]i/expressions/face-icon-small-confused.gif[/img]

[sike]

Originally posted by: mavsman55
I have never been more confused in my life than I am right now... [img]i/expressions/face-icon-small-confused.gif[/img]

Try studying infralapsarianism/sublapsarianism/supralapsarianism..if you want to keep your head spinning [img]i/expressions/face-icon-small-wink.gif[/img]