# Conversation and Fun > Just Conversation >  Friday morning brain teaser

## chm2023

You are on a game show.  Very simple premise:  you are shown 3 envelopes.  One has $1 million check, the other two have $1 checks.  You are asked to pick one.  You do so, say "A".  The MC, who is entirely disinterested in the outcome, then opens envelope "B" and shows you it contains a check of $1.  He then asks you if you want to keep your original pick, or change to "C".

What should you do?

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## walt

> You are on a game show. Very simple premise: you are shown 3 envelopes. One has $1 million check, the other two have $1 checks. You are asked to pick one. You do so, say "A". The MC, who is entirely disinterested in the outcome, then opens envelope "B" and shows you it contains a check of $1. He then asks you if you want to keep your original pick, or change to "C".
> 
> What should you do?


Punch the MC. :angry:

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## hcjilson

Open your envelope! 
hj

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## chm2023

Everybody's a comedian....:p 

Let me re-state:  you can stay with A or change to C, only one of those choices.  What do you do?

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## walt

> Everybody's a comedian....:p 
> 
> Let me re-state: you can stay with A or change to C, only one of those choices. What do you do?


Fake a heart attack, then hire Edward's law firm and sue for $10 million for toying with your emotions

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## ikon44

seems to me that the first time you chose you would have a probability of

67% of making the incorrect choice, you now have a 50/50 chance and as the
original odds were in favour of you being wrong, you should change your choice
of envelope.

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## hcjilson

Therin lies the rub......the MC has no ax to grind-why would he want you to make another choice. A unknown is whether or not the MC knows where the million is. Open the envelope!

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## Lee Prewitt

9 times out of 10, your gut reaction is correct.  Keep your envelope and open it.

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## Judy Canty

I agree. Open the envelope...unless it's one of those multiple choice things where "c" is the best guess.  Oh no, I hope there's a right answer or I'm going to spend way too much time worrying about that other envelope..:hammer:

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## Pete Hanlin

I side with ikon44, take the other envelope.  You were willing to make a choice when you had only a 33% chance- now you've gone up to 50%.  That makes no sense to me, but it sounds mathematical and all so...

Take what's in envelope C!

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## Blake

Now that you're left to choose between just two envelopes, the previous round is meaningless.  You have a 50% chance of being right whether you stay with A or change to C.  Switching envelopes will neither improve nor hurt your chance of winning the big money.  There is no mathematical/statistical way to ensure a win - it's left up to chance (unless your brother-in-law stuffed the envelopes :bbg:  )

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## chm2023

You should change to "C". When you choose "A", you have a 33 1/3% chance. Now what if you were given the chance from the get-go to choose two envelopes instead of one? 

This is what is happening, all the other stuff is smoke. Effectively, the MC is now giving you a chance for "B" AND "C"--him showing you the contents of B is completely irrelevant, you knew at least one of the other two you didn't pick had to be a loser. So by changing, you will win 2 out of 3 times or increase yours odds to 66 2/3%.

It seems counter-intuitive, as is a lot of probability! Try it and it will become a lot more apparent.

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## hcjilson

You are only giving us probability......I opened my envelope and inside was a check for a cool million........I want to take this opportunity to say thanks for allowing me to play the game, and retire early!

hj

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## chm2023

If you're happy, I'm happy!!!!!

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## Blake

Okay, it's been a while since I took statistics, so I'm still trying to wrap my mind around this one.  Seems to me that if I have a choice of three items, I have a 33.3% probability of selecting the right one, but if I only have to choose between two items, the probability is 50%.  I guess what I have a problem with is that it happens in two stages.  In the second stage envelope B is irrelevant, bundling it with the chosen envelope neither helps nor hurts your chances since it is a known loser. 

Maybe I'm overanalyzing it... definitely spending too much time thinking about it!

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## chm2023

But that's the common mistake: all the envelopes are relevant. Because B is the wrong choice does not mean it doesn't "count"--it still represents one of the three chances.  You are evaluating your chances based on the entire sequence.

Think of it this way: Say you pick A and stay with it; you have a one in three chance of winning. If you pick A and change; you lose if the prize is in A, but you win if it is in B or C----MC shows you losing C, you have the winning B; MC shows you the losing B, you have the winning C. Ergo, your chances of winning if you switch are 2 out of 3.

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## walt

Do you take the bus to work, or your lunch.

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## Blake

"Think of it this way: Say you pick A and stay with it; you have a one in three chance of winning. If you pick A and change; you lose if the prize is in A, but you win if it is in B or C----*MC shows you losing C, you have the winning B*; MC shows you the losing B, you have the winning C. Ergo, your chances of winning if you switch are 2 out of 3."

_How can I have a winning B if B is known to be a loser?_  Once B is known, it doesn't matter if you pair it with A or C (the two still unknowns).  In effect, I could just as easily say I have chosen A and B.  Therefore, using your logic I have a 2/3 chance of winning, which would go to 1/3 if I switch... but I still say it's 50/50 given the scenario you suggested.

Now, if one was given the opportunity to choose 2 of 3 UNKNOWN envelopes, your logic would clearly hold.

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## walt

> Now, if one was given the opportunity to choose 2 of 3 UNKNOWN envelopes, your logic would clearly hold.


Or if you're into the old man's whiskey again.

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## chm2023

> "Think of it this way: Say you pick A and stay with it; you have a one in three chance of winning. If you pick A and change; you lose if the prize is in A, but you win if it is in B or C----*MC shows you losing C, you have the winning B*; MC shows you the losing B, you have the winning C. Ergo, your chances of winning if you switch are 2 out of 3."
> 
> _How can I have a winning B if B is known to be a loser?_ Once B is known, it doesn't matter if you pair it with A or C (the two still unknowns). In effect, I could just as easily say I have chosen A and B. Therefore, using your logic I have a 2/3 chance of winning, which would go to 1/3 if I switch... but I still say it's 50/50 given the scenario you suggested.
> 
> Now, if one was given the opportunity to choose 2 of 3 UNKNOWN envelopes, your logic would clearly hold.


There are 3 possibilities: the winner is A, B or C. Pick one. Say A.

When the winner is A, you keep your pick, you win.
When the winner is B, you keep your pick, you lose.
When the winner is C, you keep your pick, you lose.

Are you with me so far? Therefore, keeping your original pick, you win 1 in 3 times.

Ok, same set-up, only this time, given the opportunity, you change. Ready?

You pick A

When the winner is A, I show you losing B or C, you change your pick, you lose.
When the winner is B, I show you losing C, you change your pick, you win.
When the winner is C, I show you losing B, you change your pick, you win.

Therefore, changing your original pick, you win 2 in 3 times.

Sit down and play this a couple of times, it will become obvious pretty quickly.

This is one of those things that is hard to grasp; I remember taking probability in school and the professor blowing our minds with the following: you are one of one hundred people who are each given an envelope. 99 say "LOSE", 1 says "WIN". 50 people open theirs, and each one says "LOSE". What are the odds you have the "WIN"? The answer of course is 1 in a hundred, but everyone tends to think it's 1 in 50.

For further mind-numbing see http://www.brainyencyclopedia.com/en...l_problem.html

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## Blake

Initially, the probability of A, B or C being the winner are equal.  Prob(A)=Prob(B)=Prob(C)=1/3.  Then, the contents of B are revealed.  If B is a winner, Prob(B)=1, and Prob(A)=Prob(C)=0.  If, as in this case, B is a loser, Prob(B)=0.  Since Prob(A)+Prob(B)+Prob(C)=1, Prob(A)=Prob(C)=1/2.  You keep using examples where B wins, when it obviously doesn't.  At this point, who in their right mind would pick B?

Let's say you're on Who Wants to be a Millionaire.  The million dollar question comes up, and you haven't got a clue as to the answer.  You select "C", figuring you've got a 1 in 4 chance (assume you don't lost the previous amount if you're wrong).  Regis then reminds you that the 50/50 lifeline is still available, allowing you to throw out two wrong answers.  If "C" is one of the ones removed, your probability of success is zero.  But what if, say, C and D remain.  By your logic, you should switch to "D", as you claim the probability of winning will increase to 75%.  But there are now only two viable options, meaning your probability of success with either choice is 50%.

It's not that I don't understand your solution - I simply disagree with it.

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## chm2023

First, the Millionaire game is not played the way you describe, but if it were, the same would apply. (The issue here of course is that presumably you may know the answer which eliminates the element of chance). 

I don't know what you mean that I keep saying "B" wins. No, that is not at all what I said--I clearly, and in slow motion, point out that A, B and C each are the "winner" one in three times. As you can see below from my previous post.

"There are 3 possibilities: the winner is A, B or C. Pick one. Say A.

When the winner is A, you keep your pick, you win.
When the winner is B, you keep your pick, you lose.
When the winner is C, you keep your pick, you lose."

Probabilities do not change!!!!!!! If I pick A and stay with it, I will win one in three times. You can show me B as wrong or C as wrong, and it makes NO difference. Probability is set from the time the game involves 3 envelopes. I can't make it plainer. 



"It's not that I don't understand your solution - I simply disagree with it." Mathematics are not a matter of opinion. 2 plus 2 are four. 

Again, and for the last time, go run through the process and tally your results--I don't understand why you won't do this, it's very simple and will give a definitive answer you can understand. Or look at the attached proof. 

Or if we can figure out how to do it on the Board, would you be willing to put your money where your mouth is?;) ;)

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## Blake

Okay, I finally see your point.  It's very subtle.  I didn't consider that if the host opens B and it wins, game over.  

Oh well, even I get it wrong sometimes...  :Nerd:

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## Amit

cant be asked to go into depth. change to c...or kill the mc, take ALL envelopes and run away to the bahamas, with ur million. use the $2 from the other 2 envelopes to buy a coke, have a smile and shut the hell up. :)

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## RT

CHM2023:

Seems like you're assuming some information that you didn't acutally post.  Your last post assumes that the MC knew whether or not B was a winner.  Go back and read the original question.  The distinterested MC opened B.  It didn't say that he opened B because he knew it was a loser.  It just says he opened B.  You say "You can show me B as wrong or C as wrong...".  But that's not what happened.  You showed B.  There is no evidence in the original phrasing that indicates that you would have showed C under any circumstances.  For all we know, if B was the winner, you would have showed it, which would have made for a boring brain teaser!

The nature of this problem is really semantic. The question comes down to whether or not any new information was truly given when B was revealed.  Many people interpret the purposely ambiguous statements to mean that new info was revealed (envelope B, selected at random is not the winner), and that the probability is now 50%.  Others interpret it to mean that no new info was revealed (i.e. the MC was showing a known loser and was not showing another envelope at random).  Neither case is supported by the original phrasing, which is intentionally vague about how the MC came to reveal B.  It DOES matter whether B was revealed because it was known to be a loser, or if B was selected randomly out of the set of envelopes not chosen (B and C).

If the MC knows the contents of B and C, and then chooses to reveal B because he knows it is not the winner, the odds are 1/3:2/3, and you should switch.  If the MC always reveals B and never C, without knowledge of the contents of the envelopes, then new information has been added (B is not a winner), and the odds are 1/2:1/2.  Switch or stay, it doesn't matter.  However, the phrasing of the original question doesn't indicate the level of the MC's knowledge, only that he is "disinterested".

And yes, I WAS a math major.  Math is math, and numbers are numbers, ONLY when you understand ALL the assumptions behind them.

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## chm2023

The MC always shows you a losing envelope, which is what I think you are pointing out.  (The "problem" can't play out if he shows you a winning envelope.)   However, given the initial post and the way the problem is stated, you have enough info to know you should switch.  Read the proof, it's tedious but pretty simple.

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