Hello, I already know the solution to this problem.  All I need is
verification from a math "expert".  Prefrably with a simple
explanation for my co-workers.

    The argument began while discussing emmy-award winning show Deal
or no Deal.  It is a game show hosted by bald Howie Mandell where
audienece members pick a briefcase which can contain .01 to 1,000,000.

    Anyway, the problem I proposed to my co-workers was this:

    There are 25 briefcases, one of which contains a million dollars.

    The Bank secretly KNOWS the million-dollar briefcase BEFORE the
audience member makes his choice.
  
    The audience member then chooses briefcase 16 (arbitrary number between 1 - 25)

    All the briefcases are eliminated but two.   One briefcase has
1,000,000 and the other briefcase has .01.

    The Bank offers 25,000 to settle the debt.

   The odds are no longer 50-50.  Since the bank knew the choice
before-hand, the odds of the bank having the winning briefcase are
24-25 and the audience member 1 - 25.

    And since the audience member cannot switch briefcases, the
audience member should ALWAYS take the 25,000.   Because the chances
of his briefcase containing 1 cent are 24 - 25.

   I realize this isn't how the game-show works. But in this
hypothetical game.  Aren't I correct?

   I have been arguing with members of my office, and everyone is against me. 

  They are all saying that since there are only two briefcases left
the odds are 50-50.  One of the guys is a statistic major at West
Chester University, PA.  So much for that crappy school.

    I know I am right.  Please validate my superiority.  LOL

   ANDY

---

Clarification of Question by ajlondonpa-ga on 05 Jan 2006 20:12 PST

After the audience member makes his choice, all the briefcases are
eliinated but the audience members original choice (suitcase 16) and
another briefcase.  And one of those briefcases must contain a million
dollars.

  And the BANK knows which briefcase contains 1,000,000 dollars before
the audience member picks.

Interesting question, and I'd be glad to see an actual answer.

I'm not a math expert, so I'm probably not seeing something, but in
your example (which, as you say, is not exactly how the game show DEAL
OR NO DEAL worked) I don't see how whether or not the bank knows
what's in the player's case changes the actual odds of what's in the
case.

That the bank knows what's in the player's case (when the player
doesn't) surely influences how large a buyout the bank offers the
player. And if the player knows that the bank knows more about what's
in the case than the player knows, then the player should certainly
take that information disparity into account when considering the
bank's offer. But neither of those things is quite the same thing as
affecting the odds of what's actually in the case, is it?

By the way, when you say "...emmy-award winning show Dealor no
Deal..." that's a joke, right? The game show only just aired in the
U.S. a few weeks ago; hardly enough time for it to have been nominated
and won an Emmy...

The game you describe is equivalent to this one that I am about to
describe, which I think will convince both you and your coworkers as
to the correct analysis leading to the best choice.

Suppose at the beginning you are allowed to pick 24 of the 25
briefcases!  Then the Bank, knowing exactly what is in each briefcase
beforehand, is required to open 23 of those 24 chosen briefcases that
contain .01 (1 cent).

Now we are down to two briefcases.  _If_ you were allowed to chose one
of the two briefcases and claim their contents, then of course you
would want to stick with the remaining 24th briefcase out of all you
chose, since if _any_ of those 24 had the million dollars, the
possibilities have been helpfully narrowed down by the all knowing
Bank.  So, despite the fact that there are two briefcases left, the
odds are not even as to which briefcase holds the million dollars.

However that is not the game we are playing.  To make the revised game
equivalent to the one ajlondonpa-ga has described, you the audience
member are instead given these alternatives:

(1) Take $25,000 to settle the game, OR
(2) Switch to the unchosen 25th briefcase.

Here we need to be careful.  Although the chances are 24 in 25 that
this last briefcase only contains 1 cent, there is indeed a 1 in 25
chance that it contains a million dollars!  The expected value of
choice (2) is therefore:

  (24/25)*.01 + (1/25)*1,000,000 = 40,000.0096

So your expected winnings, contrary to what you may think, are
actually maximized by taking the 25th briefcase (which of course
corresponds in the original formulation of the game with sticking with
the first chosen briefcase).

Now the strategy of maximizing expected winnings may _not_ be the
optimal strategy for _you_.  A great deal depends on the "utility" of
the various sums of money involved.  What are the _relative_ values of
1 cent, $25,000, or one million dollars, to you?

If the utilitarian values are directly proportional to the monetary
values, then this is known as being risk neutral.  In that case I'd
advise going with the 25th briefcase (either the briefcase in my
formulation, left out from the choice of 24 briefcases initially, or
the one briefcase you originally chose, in ajlondonpa-ga's version).

If the value of one million dollars is significantly less than 40
times the value of $25,000, eg. if it would only be say 20 times as
useful to you, then this is known as being risk averse.  In that case
I would advise you to take the sure $25,000 (because the "upside" of
the 1 case in 25 in which you get the million dollars doesn't quite
make up for the risk of losing $24,999.99).

In summary I think ajlondonpa-ga has a correct idea that the chances
of getting one million dollars by sticking with the original briefcase
are small.  However the extra value of one million dollars vs. $25,000
may (or may not) make up for taking that risk.

regards, mathtalk-ga

When you are sure you are right that's usually when you are wrong.

What the bank knows is irrelevant.  Their knowlege only results in
there being two briefcases remaining, one with the big money and one
with the little money. The banks alternative to picking, taking the
$25,000,has no bearing on the odds.  Its only purpose is to give the
picker an alternative.  You even state in your question that in the
end there are two briefcases and one has the million dollars and one
has the penny.  Don't overthink this.  The odds are even.

So, danial2d, are you 'sure you are right' that the odds are even? 
Say I have you pick a number between one and a million, you guess 17,
I tell you (after you guessed 17) that the correct number is either 17
or 142,596 and say that if you are right I'll give you a million
dollars, or if you want to settle now I'll give you $25,000, even
though the number is 17 or 142,596 your odds aren't even, the odds of
it being 17 is still one in a million.  If you had guessed 18 I would
have told you the answer was either 18 or 142,596 unless of course the
number was indeed 17 in which case I would have to say it is either 18
or 17 the odds of it being the number you guessed will always be
1:1,000,000 and the number given by the bank will be 999,999:1,000,000

Ah, Daniel - so sure, but so wrong.

The odds on your initial pick are 1 in 25.
The odds of it being in the other "pool" (of 24) is 24/25.
Boiling that pool down to 1 case just puts all 24/25 odds on that case.

The question then is, would you rather have $25k or a 1/25 chance of $1m.
Mathematically the 1/25 of $1m is best - and actually, it's what I'd
go with.  $25k is nice but not life changing.

The odds are *not* even.

Poet

For a discussion of this problem with some amusing letters to the
editor, lookup this book at amazon.com

The Power of Logical Thinking: Easy Lessons in the Art of
Reasoning...and Hard Facts About Its Absence in Our Lives
by Marilyn Vos Savant, Marilyn Vos Savant 

Click on see inside book.  Go to page 6 of the book.

daniel2d-ga wrote:

"What the bank knows is irrelevant."

In fact the knowledge that the Bank has is critical to proper analysis
of this game.  Let's imagine a game in which the Bank truly has no
knowledge of what is any briefcase.  Despite this, they proceed to
randomly open 23 of the unchosen briefcases.

Thus, in filming 23 out of 25 shows, they accidently open a briefcase
with the big money, and thereby waste (for entertainment purposes
anyway) the filming of those shows.

Nonetheless in the remaining 2 out of 23 shows filmed, there would
legitimately be a presumption of 50-50 odds on which of the two
remaining briefcases has the big money.

My point is that if the Bank had no knowledge and the narrowing down
to two briefcases occurs "randomly", then the naive analysis of equal
probabilities is correct.  The expected payoff for keeping the chosen
briefcase in that situation would be half a million dollars, far more
than the $25,000 settlement offered.

regards, mathtalk-ga

It's probably best to listen to the Google researchers who have "math"
as part of their screen names...

This is a little redundant but:

Imagine if you were picking 1 box out of a million and then, much to
your surprise, another contestant runs out and says "I'll take the
remaining 999,999".

You think you're going to have the right box 50% of the time? Wow! You
are certainly lucky my friend.

The confusion in this question comes in when the incorrect boxes are
removed by the bank and you ignore what happened at the outset and
incorrectly redefine the situation.

Take care kids.

Let's go back to the question:  according to Ajlondonpa rules, you
pick one of 25 briefcases, then one of the other 24 is selected and
you know that in one of them is a a million.  Now you have the choice
of opening the case you chose (you can't chose between the two at this
point according to Andy's rules) and finding one cent or the million,
or you can opt out and take 25 grand.
By Andy's rules, this second round is superfluous, but I guess it
corresponds with the TV show.

If the game values were 25 cents to opt out versus the possibility of
winning 10 dollars, we would all probably agree:  take the chance, $10
is 40 times 25 cents and I have a 1 in 25 chance of winning, and it's
no big deal if I forego the 25 cents.

Bill Gates and professional gamblers might still think this way when
playing for a million, but for a whole lot of people, the assured 25
grand is significant enough that they will be happy with that.
Looking at the TV studio situation, they have "won" already by being
selected to play.  An assured 25 grand is an amount they can
immediately see good uses for; a million is "in the stars", an amount
they don't relate to personally:  - it just doesn't happen; no one I
know has a million; it just won't happen:  "gimme the 25 grand."
(Mathtalk discusses this in more professional terms.)

On the pure odds, I think you should take the chance, but if you have
credit card debts and are paying off a car loan, you should buy a lot
of peace of mind and take the 25 grand.

These comments are fascinating, because they mirror the answers I am
getting from my co-workers.  Well, at least one guy (or girl) in this
bunch understands what I am talking about.

Not to sound like a snoot.  Anyway, the 25,000 prize complicates
matters unnecessarily.  Because statistically it IS better to take the
1 - 25 chance for the million dollars.  Something I hadn't thought
of...

But the main point I was trying to elaborate (especially to the
statistics major at WCU) was that just because their are two
briefcases left does NOT mean the odds are 50-50.

Have a nice day.  And Deal or no Deal will win an Emmy!! With Howie
mandell hosting, I can't think of any better show on TV.  LOL

ANDY

Besides being a frequent reoccuring topic in the Ask Marilyn colmun, a
solution also appears in the book "The Curious Incident of the Dog in
the Night-Time".
(which is a good book, you should read it..)

The crux: You should always switch.

Marilyn also took this to Monty himself, who agreed that with the way
the question is worded, you should switch, but Monty claimed he was
much more clever than to always eliminate just the duds.  So don't go
on TV thinking you'll beat him.

--joel

It all hinges on whether the eliminations are done to deliberately
leave the $1,000,000 in play or whether this happens fortuitously. If
the bank, knowing which briefcases contain which amounts, controls the
eliminations and never eliminates the $1,000,000 then it's definitely
better to switch - in 24 out of 25 games you won't have picked the
$1,000,000 case initially, and therefore the remaining case must have
it.

But if the eliminations are random, as they are on the TV show (since
they're controlled by the contestant, who obviously doesn't know
what's in the briefcases) then the final two briefcases do indeed have
equal chances of containing the amounts left in play. Why is this
different? After all, it's still true that in 24 out of 25 games the
initial case you picked doesn't contain the $1,000,000, so shouldn't
the same answer apply? The difference is that in most of those games
the $1,000,000 briefcase will be one of the ones eliminated. When you
restrict the games under consideration to only those in which the
$1,000,000 briefcase remains as one of the last pair, it is equally
probable to be in either briefcase.

Note that this would also be true if the Bank controlled the
eliminations but did them randomly. It is the likelihood of the
$1,000,000 briefcase being eliminated that matters; if things are
arranged to leave the $1,000,000 briefcase in play more often than
chance would have it, then your odds are improved by switching - or,
for this show, you'd probably want to take the money. If it's all done
randomly then it doesn't make any difference. (And for completeness,
if eliminations are arranged to increase the likelihood that the big
money is eliminated, to save on payouts, then you're better off
sticking with the case you have.)

"After the audience member makes his choice, all the briefcases are
eliinated but the audience members original choice (suitcase 16) and
another briefcase.  And one of THOSE briefcases must contain a million
dollars."

if "those" refers to only the one picked by the player and the one the
bank is forced to pick then according to that version of the rules one
of the 2 must contain the million and the odds are 50/50. I think that
that is not a correct wording of the rules intended.

If you reword it so the last pair may or may not contain the million
then you should always stay with your pick becuase the bank will NEVER
pick the million in a game where it's trying to keep it's money.

Sorry, egon-spangler, but that's not correct. In the situation you
describe, where one remaining briefcase *always* contains $1,000,000,
there's a 96% chance that it's in the other briefcase.

An example may help to clarify this. Instead of a single game, you are
now playing 25 simultaneous games. There is one game with the
$1,000,000 in briefcase 1, one with it in #2, and so on (but you don't
know which briefcase it's in in any particular game; this corresponds
to the $1,000,000 being equally likely to be in any briefcase). You
pick a number (e.g. 16), and you get that number briefcase in all the
games.

Then we eliminate 23 briefcases in each game, without eliminating
yours or the $1,000,000 in any of them. So there'll be one game with
briefcases 1 and 16 left, one with cases 2 and 16 left, etc. plus an
additional game (where briefcase 16 has the $1,000,000) which will be
the same combination as one of the others.

It's pretty obvious that in 24 out of these 25 games, the $1,000,000
is not in briefcase 16. In fact, if you find the two games with the
same pair of briefcases, you know that in the other 23 games briefcase
16 does not contain the $1,000,000; and of those two games, in one the
money is in #16 and in one it is not. So again, in total, 24 out of 25
times it is not in your briefcase.

I repeat again, all this is different to the situation where, in
general, the $1,000,000 briefcase can be eliminated, but in one
particular game it has not. In that case the odds are indeed 50-50
(assuming that eliminations are done uniformly randomly).

We have as restriction that we can not change the briefcase, so when
there are 2 briefcases left:

If we initially have chosen the 1.000.000$ briefcase, we will win.
That will happen 1 out of 25 times you play (in average), because
there is just one favourable case and 25 possible cases.

If we initially have chosen a 0.01$ briefcase, we will loose. That
will happen 1 out of 25 times you play (in average).


Now we have two strategies: Playing always or accepting the 25k always.

1) Let's suppose that we play lots of games and always run the risk to
go for the 1m$ case. In average, every 25 times we play it will
happen:
	*We will win 1.000.000$ once
	*We will win 0.01$ 24 times (negligible)

	So, we will win in average aproximately 1.000.000$ every 25 times we play


2) Let's suppose that we always accept the 25k$. Every 25 times we
play we will win in average 25.000*25= 625.000$

(SOLUTION HERE): So, if we could play lots of times the right strategy
would be the 1), running the risk and playing to the end.

If we just play once, matematically we should run the risk. But this
decision must depend on other non-mathematical factors (you may need
20.000$ urgently and can not run the risk... you know)



Once the original question is answered, I will explain you some other
similar situations:

a) Let's suppose that the bank eliminates the 23 briefcases before we
make our choice. They can eliminate whatever non-priced briefcases
they want. So when there are only two briefcases and we must choose
one, we just know that there is one priced and once non-priced and our
odds will be 1-1.


b) (The original Monty-Hall problem) Let's suppose that the bank
eliminates briefcases after our initial election, and they allow us to
change the briefcase when there are 2 briefcases left (no 25k option
here). What should we do: maintain our election or change it?

1 out of 25 times we play, we initially will have chosen the priced
briefcase -> So the other briefcase will be non-priced, and in this
situation if we change our election we will loose.

24 out 25 times we play, we initially will have chosen the non-priced
briefcase -> So the other last briefcase will be priced, and if we
change our election we will win.

So you see that if we always change our election when there are 2
briefcases left we will win 24 out of every 25 times we play.


Sorry about my english, hope I have helped. Regards.

The answer to your question will depend on how the unchosen cases are
eliminated. If the cases are eliminited by random chance, and opened
to show that they do not contain the million dollars, then the
probability that one of two remaining cases still holds the million
dollar prize will, in fact, be 1/2. Only 2 out of 25 contestants will
still have the million dollar case unopened when there are 2 cases
left.
You can test this with a deck of cards. Take all the cards of 2 suits
(say spades and clubs), giving you 26 cards. Remove one card to leave
you with 25 (say the ace of clubs). Now say that you want to pick the
ace of spades out of the deck. Remove the card of your choice and set
it aside face down. Start turning cards over from the remaining deck,
until you turn the ace of spades over. Chart how many times you do
this and how many cards you turn over before the ace of spades came
up. If you do this enough times you will find that there is a clear
ratio. You will make it to have; 24 cards left about 96/100 times or
24/25, 23 cards left 92/100 times or 23/25, 22 cards left 88/100 times
or 22/25, etc. Continue this and 8/100 times you will have made it to
have 2 cards left, and exactly one half of those times (4/100) the ace
of spades will be the one remaining card. Exactly the same thing would
apply to your breif case senario. 2/25 players would make it long
enough to choose between 2 cases and 1/2 of those who make it that far
would win the million dollars. So your probability of winning the
money with 2 cases left is 1/2.
Now, you didn't say in your question that the unchosen cases were
randomly eliminated. This applies to your "pick a number between one
and a million" example as well. If the cases (or numbers) are NOT
randomly eliminated everything changes. You cannot apply the same odds
to an event that is not random. Because of this, if the remaining
cases were eliminated because they are known NOT to contain the
million dollars the odds remain at 1/25.
You can demonstrate this with the cards as well. Again try and choose
the ace of spades from the deck of 25 face down. Have someone else
eliminate 23 of the 25 cards for you, and 1/25 times you will have
chosen the ace correctly.
If you need to prove that to your co-workers, tell them for 2 dollars
you'll let them try and win $25. Let them choose a card from your
deck, and then eliminate 23 of the unchosen cards. As long as they are
not allowed to change cards when only 2 are left, you will make double
what you pay out (provided you do it enough times). If you want to you
can offer to buy the card back for something less than a dollar, but
it's irrelevant and will cut into your profits.

There is a significant difference between this game and the Monte Hall
problem.  In the Monte Hall problem, the host (who knows where the big
prize is) is the one who chooses the door to open.   We know he will
always choose a door without the big prize.   Regardless of the door
you originally choose, he can always choose a door that has a small
prize to open.  Your original choice still is has only a 1 in 3 chance
of being correct, and the remaining door now has a 2 in 3 chance of
being correct.   Because of this, you should always change your
choice.

This situation is very different.   In this case, you choose the
briefcases to open, not the Bank (who knows where the money is). 
Because of this, as you choose more briefcases, the probability of the
case you possess being the big money increases with each discarded
case.   This is the difference between the two.

So, when only two cases remain, the chances of the money being in your
case are 1 in 2.   If one case contains $1,000,000 and the other
contains $0.01, the expected value of continuing is $500,000.   If
they are offering only $25,000, you should never take the offer unless
your personal feeling is that a sure $25,000 is better than a 50-50
chance at $1,000,000.   Statistically, you should keep the case.

And, given the option to switch to the other case, it doesn't matter
if you do or not.

You are all a stooges...How in your right mind can you think that "oh,
since we started with 25 (or 1,000,000 in one person's case) can you
possibly believe that it is not 50-50.

There are 2 cases left regaurdless of how the contestant got to that
point, there are but 2 cases left that's it, not 24, not 25...TWO. 
One case has 1 million, one case has .01...it's that simple...the odds
changes each and every time a case is picked...starting with 1 out of
25...then 1 out of 24...then 1 out of 23...this continues until you
have the odds of 1 out of 2.


That being said.  The question asked is "If the bank knows" and that
affected the banks offer then...yes I would definately take the
25,000.  This is because the bank should offer 500,000.  Because that
is the expected outcome of the trial...1 million/2 +.01/2 =500,000.005
or just half a million.  If the bank offered more that half a million
and knew the possible outcome, then go for the million.

Interestingly enough this is exactly oppisit of what you would want to
do if the bank did not know what was in the each case as the expected
(or average) outcome of each situation (with 2 cases left-not 25 or 24
or 23 etc.) would be 500,000.  If you are in that situation many, many
time with 2 cases left the average outcome should (as a result of
statistics) tend toward 500,000.

It's that simple...over time the average outcome would tend toward
500,000.  This is just like flipping a coin(evenly balanced) past
success and failures (heads and tails or 1 million's and 1 cent's)
have absolutely no baring on the future...it's still 50-50.  the long
term average WILL tend toward 50%.

Good gamblers know not to look at the rollette table sign that shows
past winners...when it states "red, black, red, black, red, black,
red"...what are you going to bet on black? or red...IT DOSEN'T MATTER
both are the same...

KMclean-ga,

OK.  Let's play!  You set up ten envelopes with, say, a cheque for
£1000 in one - which is marked so you know which it is but I don't.

We then play as follows:

I pick an envelope.
You then remove 8 envelopes that *don't* have your secret mark on it.
I'm left with 2 envelopes - the one I chose and the one you've left behind.
I can then pick which one I want.
If the one I actually open has the money, I get to keep it.

I'll pay you £500/turn to play this game...no, wait, I'll pay you £600/turn.

Let's play, say, 50 turns?

If you really believe your argument, you'll be up for this......and
get absolutely fleeced.

Let me know if you're up for it

John

KmcLean and Poet,
The problem as presented in the question does not allow you to pick
between the two remaining briefcases; you may only choose to open the
one you picked from the original 25 or take the $25,000.  Actually,
the "second round" is just a red herring.  All 24 other cases could
instead be removed and the contestant asked to choose between his
choice from the 25 cases and the $25,000.
There is no 50/50 choice between two cases with the alternate choice
of taking the $25,000.

As the asker points out, this is not the way the game is played, but
it is the rules for the question.

Okay, I've read many of these comments, and I need to clear a few
things up. First, the probabability you've all been discussing, the
monty hall problem, has NOTHING to do with "deal or no deal." In the
game, you pick one case out of 26 and are then forced to either take a
bank offer to buy your case or open up some the remaining cases,
slowly narrowing the possibilities of what's in your case. As far as I
can tell, the bank doesn't know what's in each case...or at least
their offers don't reflect this knowledge. The math behind the bank
offers is complicated, and if anyone knows what it is PLEASE post it.
What I do know is toward the end of the game, the math reduces down to
the mean value left on the board, rounded to the nearest grand. Given
what I know so far, bank offers are all smaller than this "expected
winnings" value in the beginning of the game (by how much, I need the
math formula!) until they equal the average as the game progresses. It
wouldn't be statistically prudent to accept an offer before this
happens. The bank will never make an offer higher than the average.

As for the problem itself, at first glance, it appeared to me to be
50/50, because I didn't sufficiently understand the problem. I think
we all agree that if someone told you to pick one of two briefcases,
one of which has a million, then the probablility of getting a million
is 50%. Upon further insightm that is not the case here. Instead of
flipping a coin, we're rolling one of those many sided dice from the
game "scattergories" the probability that the case you leave holds a
million is 1 in 25. If you did pick the case with 1,000,000, then the
bank left it for you, and that will occur the other 24 of 25 times. So
while I concur that "overthinking" a problem is definitely a bad thing
in the science and math fields, you still have to think enough to
sufficiently understand it.

Everybody is missing the obvious.

You get an insider to tell you and you give him/her half.

So, you finish up with a cool $½ million. No sweat.

No degree in math or psychology required - just an understanding of
how things work in the real world.