View Full Version : game with a not too obvious solution
When someone asked me to solve this game, I first thought there was no solution. Later on I realized there was, and I was surprised...
The ante for this game is $1 for both you (I) and your opponent (II). II takes two pieces of paper and writes secretly on both of them a whole number between minus and plus infinity. II puts the papers face down on the table. I take one of them (my dicision), and have a look at it. I may now either keep it and give the other one to II, or give it back to II and take the other one.
The one with the highest number wins the pot.
Is there a strategy that makes the EV for I strictly positive? If so, give the strategy or an example of such a strategy.
As you suspect, the opposite of what you expected is true: there is...
Next Time.
pudley4
10-09-2003, 03:40 PM
Always switch if you pick a negative number. Never switch if you pick a positive number.
That's not a right observation!
Since your opponent may choose the numbers himself, he only has to pick two numbers with the same sign and there it is: EV=0, wich means no strictly positive EV.
Next Time.
1800GAMBLER
10-09-2003, 09:59 PM
If he filled the size of the page with numbers and it's a minus choose the other one. If he did it with a positive keep it.
If he filled 3/4 of the page and then looked tired or bored of this game (unsurprising) follow the above pattern.
1800GAMBLER
10-09-2003, 10:05 PM
Seriously though. Since you have no idea if is going to be tricky with the signs you know he isn't going to be tricky with the length of the numbers.
Hence if you see it and think he writen something which is longer than his average and the number and it is positive then go less. If it's above average length and minus go higher.
Meanwhile you have no EV if you think the number is smaller than his average.
Copernicus
10-10-2003, 07:28 AM
If you use words like "If you think", then the answer cant result in a strictly positive EV, which I cant find, and somehow doubt exists!
1800GAMBLER
10-10-2003, 07:48 AM
After playing this game once you can get an average of the length of his numbers. First game go by half the length of the card.
Othernote:
If you think it's 70% a player has a worse hand and,
If you think it's 30% a player has a better hand.
Then it's EV when you bet the river. That can be 'strictly' EV by grouping together all the possible hands.
Hadn't time to overthink all of your suggestions, but most of them depend on analysis of the opponent. I am looking for a solution that on EVERY imaginable opponent results in a strictly positive EV.
Hint: you may have to use a calculator and/or a random number generator before you make your action.
Maybe this will help you all.
Next Time.
irchans
10-10-2003, 10:12 AM
switch with probability f(x) where f(x) is strictly decreasing, 1>f(x)>0, and x is the amount written. For example f(x) = 1/(1+exp(x)).
Copernicus
10-10-2003, 10:38 AM
If II generates his numbers randomly, and given that the tails are infinite, I cant think of any probability function that will have an EV other than 0. No matter how large or small the first number he picks is, there is still a "50/50" chance that the next number will be greater or smaller. If II can have a strategy where even he cant predict the relationship between the two numbers, how can I do better than 0 EV?
If the tails were finite but very large then a f(x) like you are proposing might work.
[ QUOTE ]
switch with probability f(x) where f(x) is strictly decreasing, 1>f(x)>0, and x is the amount written. For example f(x) = 1/(1+exp(x)).
[/ QUOTE ]
That's a correct solution!
Let me prove it for the ones who don't believe it.
How would you describe a strategy for I?
The only dicision to be made is wether to switch or not, and the only information you have is the number z, the number on your note.
Since this problem is not solvable for pure strategies, I must have a probability for switching based on the number z.
Hence my strategie is a function f(x) from (-Inf,+Inf) to [0,1].
Denote the two numbers by x and y, assuming x<y.
My first pick will be either one of them, both with probability .5.
I win in these two situations:
1. I pick x and I choose not to switch, probability: .5*(1-f(x).
2. I pick y and I choose to switch, probability: .5*f(y).
Addition of these probabilities results in
probability I win equals (1+f(x)-f(y))/2.
For a strictly positive EV I need my winning probability to be over .5.
(1+f(x)-f(y))/2 > 1/2 <=>
f(x)-f(y)>0 <=>
f(x)>f(y)
So any function with f(x)>f(y) for all x>y and 0<=f(x)<=1 for all x in (-Inf,Inf).
f(x) = 1/(1+exp(x)) indeed is such a function.
The function I came up with was f(x)=.5-arctan(x)/pi.
Note that on most opponents you could inprove the effect of your function if you scaled it a little.
For instance, in these functions, replace x by x/m for say m=10, 20... Have a look at it.
Next Time.
Copernicus
10-10-2003, 01:02 PM
I believe this is a dead even strategy, not +EV (using the exp version). Ive tested it for 50 number combinations, and it is always .5. Give me a scenario where it is +EV.
PokerNoob
10-10-2003, 01:15 PM
How is this different than two signed random numbers of a given precision being generated? Always switch if your number is negative. Reason: there exist more numbers on the the positive side of the number line plus those between the chosen negative number and zero than negative number to negative bound, even if the bound is infinte.
Or did you forget to mention that we must play this game an infinite number of times?
Copernicus
10-10-2003, 01:21 PM
There are not more numbers on the positive side...infinite is infinite, there is no "more infinite"
PokerNoob
10-10-2003, 02:14 PM
Okay, how about this. You can write down any real number between 0 and +infinity and any rational number between 0 and -infinity.
Seriously, how does "switch with probability" have any meaning? Its a binary choice. You can't .7th switch.
[ QUOTE ]
Okay, how about this. You can write down any real number between 0 and +infinity and any rational number between 0 and -infinity.
Seriously, how does "switch with probability" have any meaning? Its a binary choice. You can't .7th switch.
[/ QUOTE ]
Switch with probablitiy p means that there is a probability p for the event that you switch, and a probablity 1-p for the event that you don't.
A realization of this could be like: take p=f(x).
Then use a random number generator to generate a random number between 0 and 1 (use Unif(0,1)), denoted by w.
If w<=p then switch, otherwise don't!
I hope I have been clear.
Next Time.
squiffy
10-10-2003, 03:45 PM
Well, can you explain your solution in layman's terms. I don't understand the functions or numbers. But Copernicus's observations seem intuitively obvious to me. You have no information about how your Opponent is selecting the two numbers. So they may as well be two random numbers. You are only seeing one of them. Which one you choose seems completely random. And you have no basis for determining what relation the number you select bears to the one you have not selected. If the numbers were finite perhaps it would be a different story. But didn't you specify positive and negative infinity.
Copernicus
10-10-2003, 04:00 PM
much clearer, and it works. I think you can use the same technique to contact the dead.
PokerNoob
10-10-2003, 04:05 PM
[ QUOTE ]
much clearer, and it works. I think you can use the same technique to contact the dead.
[/ QUOTE ]
Or at least Schrödinger's Cat.
Copernicus
10-10-2003, 04:11 PM
The reason it works is that the technique for choosing is biased...it produces a decision to muck with a higher frequency when the card is low, and a decision to keep with a higher frequency when the card is high. (though there is a technical problem in evaluating numbers in all but a fairly narrow range).
Heres an example:
<font class="small">Code:</font><hr /><pre> x f(x)
8 0.00033535
1 0.268941421
pick
8 1
1-f(x) keep 0.99966465 0.731058579
f(x) discard 0.00033535 0.268941421
winning prob (top left bot right): 0.634303036
losing prob (bot left top right): 0.365696964
</pre><hr />
(The sums of the diagonals have to be divided by 2 to keep the total probability at 1).
doormat
10-10-2003, 04:15 PM
Ok, I write down a number on a piece of paper. Then I flip a coin - heads I write a larger number on the other piece of paper, tails I write a smaller one. To my way of thinking, no matter which paper you choose and no matter what number you see on it, it is a random equal chance that it will be high or low because of the coin flip. So exactly what is your function that will predict how my coin flips?
doormat
Bozeman
10-10-2003, 05:13 PM
Yes, this is true, but he will switch more often with the lower # than with the higher number!
squiffy
10-10-2003, 07:48 PM
Told my college roommate about the problem. He was a Phi Beta Kappa math major and studied Real Analysis in college, which dealt a lot with one to one mapping of infinite number lines to say points on the circumference of a circle, etc. His approach was to explain that you can simplify the problem by showing that the fact that the number choices stretch from negative infinity to positive infinity does not really affect the basic problem. So you can convert the problem into one where the numberline is finite, say from 0 to 100. And that problem is easier to solve intuitively.
But that really is the hardest step, realizing that the fact that the number choices are infinite does not change the problem.
You can still take a number and assign a probability to it so that you hold a higher number a larger percentage of the time than any lower number.
This makes sense to me. Though I certainly cannot follow the math.
[ QUOTE ]
Well, can you explain your solution in layman's terms. I don't understand the functions or numbers.
[/ QUOTE ]
I am not familiar with the term layman's terms. But I will try to give it a less mathematical try...
The two numbers are written, and you hold one of them.
Since you want the biggest one, you would like to switch with the smaller and keep the bigger one.
But you don't know wich one you got.
Agree that it would be good if you switched the small one strictly more of the time than you would switch the bigger one.
So all you have to do is to add a probability to switch to each number, such that for every two distinct numbers
the biggest one is least likely to be switched.
Now mathematics is unevitable for adding such probabilities to these numbers.
If you look at the above, you could easiliy see that you need a strictly decreasing function that maps {..,-2,-1,0,1,2,..} onto (0,1).
Maybe now you can look at the posts and plot one of the appropriate functions. See if you understand.
Hope I have been clear.
Next Time.
Bozeman
10-10-2003, 11:11 PM
"that maps {..,-2,-1,0,1,2,..} onto (0,1)."
Actually, there is no such function, since (0,1) is a different order of infinite than {...,-1,0,1,...}. You map onto a subset of (0,1)
Craig
aces961
10-11-2003, 05:15 AM
This may be being quite picky regarding the problem, but the case where player II puts the same number on both pieces of paper is not ruled out. In this case the players would split the pot every single time and the ev for player 1 would be zero which is not strictly positive.
[ QUOTE ]
"that maps {..,-2,-1,0,1,2,..} onto (0,1)."
Actually, there is no such function, since (0,1) is a different order of infinite than {...,-1,0,1,...}. You map onto a subset of (0,1)
Craig
[/ QUOTE ]
You might be right there. I don't know exactly every English term for the math,
I'm from the Netherlands, but it is a function from {..,-2,-1,0,1,2,...} to (0,1).
And indeed no surjection of such exists. Maybe the term 'onto' I used implies surjection.
Next Time.
[ QUOTE ]
This may be being quite picky regarding the problem, but the case where player II puts the same number on both pieces of paper is not ruled out. In this case the players would split the pot every single time and the ev for player 1 would be zero which is not strictly positive.
[/ QUOTE ]
You're right, I thought I already stated that, but I did not.
[ QUOTE ]
The ante for this game is $1 for both you (I) and your opponent (II). II takes two pieces of paper and writes secretly on both of them a whole number between minus and plus infinity.
[/ QUOTE ]
These numbers indeed must be distinct, otherwise it evidently results in a EV=0 strategy for II.
Next Time.
irchans
10-11-2003, 08:49 AM
This "game with a not too obvious solution" is similar to the envelope problem (posted 9/29/02 by me).
[ QUOTE ]
Two envelope questions to think about:
1) A kind man tells you that he has two envelopes each with money in them. He says that you can choose either envelope, open it, and keep the money in that envelope, or trade the contents of the opened envelope for the unopened one. Is there any strategy that has a higher expectation than randomly choosing, opening, and not trading?
2) A millionaire is funding a game between you and another person. The other person is given 1000 one dollar bills and two identical envelopes. He is instructed to put any integral dollar amount up to $333 in the first envelope and put exactly twice that amount into the second envelope. He shuffles the envelopes and presents them to you. As in 1) you get to choose one envelope, open it, and then decide whether to trade for the unopened envelope. If you end up with the envelope containing less money, he gets $100. If you end up with the envelope containing more $, he gets nothing. What is a good strategy for you? What is a good strategy for him? What is his worst strategy?
[/ QUOTE ]
How do you create a hyperlink to an old post?
The similarity of the first problem is obvious, but there are some things I miss about the second.
[ QUOTE ]
A millionaire is funding a game between you and another person. The other person is given 1000 one dollar bills and two identical envelopes. He is instructed to put any integral dollar amount up to $333 in the first envelope and put exactly twice that amount into the second envelope. He shuffles the envelopes and presents them to you. As in 1) you get to choose one envelope, open it, and then decide whether to trade for the unopened envelope. If you end up with the envelope containing less money, he gets $100. If you end up with the envelope containing more $, he gets nothing. What is a good strategy for you? What is a good strategy for him? What is his worst strategy?
[/ QUOTE ]
Do I get to keep the money in the envelope I picked?
Does II return the remaining of the 1000 dollar bills to the millionaire?
I suppose II cannot choose $0 for otherwise his payoff is not completely defined?
I am solving the game under these assumptions right now.
[ QUOTE ]
How do you create a hyperlink to an old post?
[/ QUOTE ]
I do not know.
Next Time.
irchans
10-11-2003, 10:24 AM
>Do I get to keep the money in the envelope I picked?
Yes.
>Does II return the remaining of the 1000 dollar bills to the millionaire?
Yes.
>I suppose II cannot choose $0 for otherwise his payoff is not completely defined?
Yes.
Does anyone know how to link to an old 2+2 post?
[ QUOTE ]
Does anyone know how to link to an old 2+2 post?
[/ QUOTE ]
I'll give it a try, this link should bring you back to your envelopes question...
your question (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=368794&page=0&view=ex panded&sb=5&o=1&vc=1).
And it works!
Right click on the post you want to be linking to and copy the shortcut.
While composing your post, click URL and paste the link.
Next Time.
BruceZ
10-11-2003, 07:05 PM
Does anyone know how to link to an old 2+2 post?
First find the old post using search (link at top). There seems to be a problem using your display name to find all your posts, so you may have to get creative here by searching for key words within your post. Then once you find a link to your post, do whatever you have to do in your browser to bring up the URL of that link (long string starting with http). In IE, I right click on the link, and bring up Properties. Then highlight the url (long multi-line thing starting with http ) and copy it by holding the control key and the "C" key at the same time (in IE). Then to insert the link in your post, click on "URL" next to the Graemlins. It will prompt you for the address. At this point, paste the link by holding the control key and "V". It will try to provide the initial http://, so be sure to delete that if you are already copying that from your link. Alternatively, you can just copy and paste everything to the right of the http://, and this is faster. Then it will prompt you for a title that you want the link to say. Type in whatever you want, and it will then insert the link in your post where your cursor was. There will be a lot of text inserted, and you can just type your post around that. For all this to work, you have to have "using UBBCode" selected from the drop down menu (above the Greamlins) if it is not already selected. That's it. Sounds complicated, but once you get used to it, the whole thing takes a few seconds. Someone else suggested using "copy shortcut" instead of properties, so you can try that too if that is an option for you in your browser and operating system.
Bozeman
10-11-2003, 08:09 PM
there is a link on your "My Home" page to all your posts, if that is what you're looking for, Bruce
BruceZ
10-11-2003, 08:25 PM
That works much better than the search feature, thanks.
Bozeman
10-12-2003, 06:58 PM
Oops, I misremembered how to get to the right page, it is not "My Home", it is the profile page that you get to by clicking on a user name.
Sorry,
Craig
BruceZ
10-12-2003, 08:40 PM
I knew what you meant, and now I know what's wrong with the search script:
http://forumserver.twoplustwo.com/dosearch.php?Cat=&Forum=All_Forums&Words=BruceZ&Ma tch=Username&Searchpage=1&Limit=25&Old=allposts
Once I hack in the correct user number as Words=197 instead of Words=BruceZ it works fine. Apparently dosearch.php can't handle usernames correctly. It only returns 1 page unless it gets a user number.
Legato
10-13-2003, 08:13 AM
[ QUOTE ]
2) A millionaire is funding a game between you and another person. The other person is given 1000 one dollar bills and two identical envelopes. He is instructed to put any integral dollar amount up to $333 in the first envelope and put exactly twice that amount into the second envelope. He shuffles the envelopes and presents them to you. As in 1) you get to choose one envelope, open it, and then decide whether to trade for the unopened envelope. If you end up with the envelope containing less money, he gets $100. If you end up with the envelope containing more $, he gets nothing. What is a good strategy for you? What is a good strategy for him? What is his worst strategy?
[/ QUOTE ]
Hmm this is my try:
His worst choice must be putting any odd number between 167 and 333 in the small envelope since that will make me do the correct decision every time.
Further any even number larger than 168 in the smaller envelope must also be bad since I will always correctly keep the larger envelope when I get that. And the nearer 168 the worse it is since that will make me more inclined to correctly switch when I get the "small" envelope.
Assuming he follows this I know that the maximum I can get in the big envelope is 332. Thus the big envelope will contain $4-$332 for an average of $168. The small envelope will contain $2-$166 for an average of $84. The average of both envelopes will be $126. Thus I would switch at anytime I get less than $126.
But assuming he know I do this, his strategy changes and he can put 126-166 in the smaller envelope since that will make me never change and him get his $100 50% of the times which must be the best he can hope for? Hmmmm /images/graemlins/smile.gif
But then assuming I know that he knows I will change strategy once again....
Is this really solvable, doesn't it all depend on what you can assume that the counterparty knows and understands?
thylacine
10-15-2003, 06:32 PM
I did not read thread so I may be duplicating.
Suppose your strategy is, for each integer n, you keep with probability p(n), and swap with probability 1-p(n). Your opponent knows this.
If your opponent happens to pick number i and j, and i<j, then you see one of them with probability 1/2, then keep or swap as above. Your EV is
E(i,j)=(1/2)[p(i)+1-p(j)]-(1/2)[1-p(i)+p(j)]=p(i)-p(j)
So you can always have a positive EV (assuming i,j are required to be different) provided you choose
p(i)>p(j) whenever i<j.
This is certainly possible, e.g. choose p(n) that is strictly decreasing and bounded between 0 and 1.
On the other hand your opponent can choose your EV to be as small as possible, by choosing appropriate i,j. That is, for any e>0 no matter how small, your opponent can choose two numbers i<j such that E(i,j)=p(i)-p(j)<e.
thylacine
10-15-2003, 06:42 PM
[ QUOTE ]
I did not read thread so I may be duplicating.
[/ QUOTE ]
Now I checked. It's already below.
[ QUOTE ]
But Copernicus's observations seem intuitively obvious to me.
[/ QUOTE ]
They weren't so obvious in 1500.
~D
mostsmooth
11-07-2003, 09:42 PM
forgive me if this has been stated already, but write down a number, choose the paper he wrote on an dlook at it, if the number you wrote is higher, switch
What kind of number will you write down?
Clearly, if this is always same the number N, this will not work.
So - if there is a solution to the game of this nature - your number must be a sample of some
random distribution.
Please, do reply if you think you can think of one - and we will see if it works.
Next time.
mosta
11-09-2003, 09:18 PM
well I told MY roommate about this problem and he
is a Berkeley math PhD who worked under a
Fields Medalist (so there!) and his response
is that this is really not a "winning" strategy
because asymptotically your EV is within epsilon
of zero. The intuition of the strategy is good--
that you want to switch more for smaller numbers
and less for bigger numbers, but these functions
you use to map +/- infinity to (0,1) very quickly
converge to zero or one. They only do anything
over a relatively small range--relative to an
infinite set. And that range is negligibly small:
Note, you suggest rescaling to "improve" the strategy--
in fact rescaling shows that the strategy doesn't
work because you can rescale away your winnings. Ie
the "solution" is not scale invariant, and therefore
is not a solution. So for your strategy:
It's not that easy--and it's because your range
of numbers is unbounded.
[ QUOTE ]
[...] this is really not a "winning" strategy
because asymptotically your EV is within epsilon
of zero. [...] They only do anything
over a relatively small range--relative to an
infinite set. And that range is negligibly small [...]
[/ QUOTE ]
I believe I already stated that - when your opponent knows what you are doing - he can bring
the EV as close to zero as he likes, but never exactly zero.
Hence, no matter what the EV is strictly positive.
Sure, you can say that your edge is negligibly small, but the question was:
Is there a strategy assuring you of a strictly positive EV?
And still, the posed solution does just that.
If you define a "winnning strategy" as a strategy with EV >= w for some w greater than zero
and all the possible strategies of your opponent, then - of course - you are right.
Next Time.
mosta
11-10-2003, 06:50 PM
[ QUOTE ]
I believe I already stated that - when your opponent knows what you are doing - he can bring
the EV as close to zero as he likes, but never exactly zero.
Hence, no matter what the EV is strictly positive.
Sure, you can say that your edge is negligibly small, but the question was:
Is there a strategy assuring you of a strictly positive EV?
And still, the posed solution does just that.
If you define a "winnning strategy" as a strategy with EV >= w for some w greater than zero
and all the possible strategies of your opponent, then - of course - you are right.
Next Time.
[/ QUOTE ]
maybe so. I don't know. I'm not sure that "indistinguishable from zero" counts as "winning". And I don't think your opponent would need to know your strategy. We have to drop the background assumptions that we're talking about numbers that actual people would probably come up with (like between +/- 1 million)--if the numbers are truly picked from all the whole numbers, without bias, then the strategy reduces to always switch below and arbitray point and never switch above it (because the p of landing in the range where your functions does anything is indistinguishable from zero)--and don't fall into the fallacy of thinking: well if we always switch for values below some arbitrary negative number then we get all the postitives and a finite amount of negatives--infinities don't work that way.
thylacine
11-11-2003, 11:57 AM
Mosta, get your Berkeley friend to explain it to you.
thylacine
11-11-2003, 12:04 PM
Here's a way of addressing a point about this. (Actually the answer is already in this thread a few times in various forms.)
Question: is there a(n appropriately defined) Nash Equilibrium to this game?
irchans
11-11-2003, 08:49 PM
As far as I can tell Mosta and his friend are correct to say that for every envelope picking strategy, there exists an envelope stuffing strategy that can bring the picker's expectation arbitrarily close to the expectation for not switching. (Some strategies are guaranteed to always do better than not switching, but not "measurably" better as pointed out by Mosta.)
So how about if we modify the question and
assume that we know the envelopes contain between $1 and $1 trillion dollars.
Then I think there is a picking strategy that guarantees the picker has an expectation gain of at least $0.25 above the expectation for not switching.
bigpooch
11-12-2003, 06:44 AM
Of course if you choose the switching function f(x) = 1/x
it would meet the requirement. Forgive me if I have
already sent a similar post (didn't see it!). Then the
expected gain would be precisely 1/4.
irchans
11-12-2003, 07:59 AM
bigpooch,
Switching with probability 1/x is kind of neat even if there is no restriction on the size of the envelopes. If the smaller envelope contains $1 or more, it gains $0.25 in expectation over the strategies of "always switch" or "never switch".
thylacine
11-12-2003, 12:24 PM
Mosta and friend are just realising what was already said earlier in this post.
But what is this `envelope stuffing'? Could you define this problem so we can have a go at it? Thanks
irchans
11-19-2003, 01:17 PM
Sorry for the delayed response, I was out of town. Let me try to define the game more precisely.
Player 1 (the stuffer) chooses an amount between $1 and $500 million dollars (Let's assume a whole dollar amount.) and puts it in an envelope. Then he puts twice that amount in another envelope. So the stuffer's strategy is purely a probability distribution over the numbers between 1 and 500,000,000. (A list of 500 million non-negative numbers that sum to 1.) The envelopes are randomly mixed.
Player 2 (the picker) opens one envelope and looks at the amount inside. Then he chooses whether to keep the envelope or switch. The pickers strategy can be described by a list of 1 million switching probabilities. (i.e. one probability for every possible amount.)
The questions is "What strategy guarantees the most expectation for the switcher?" and "What strategy guarantees the most expecation gain over not switching?". I don't know for sure the answers to these questions.
thylacine
11-19-2003, 03:02 PM
If both play optimally, then I believe the expected gain over not-switching is
$0.250000009313225781
ropey
11-20-2003, 05:46 PM
I am thinking about this problem using no math, rather, common sense...and I can't believe that there is an answer to this.
There will always be an infinite amount of numbers below a chosen number, and an infinite amount of numbers above a chosen number. Thus, there cannot be positive EV.
Bozeman
11-20-2003, 05:55 PM
Just more proof that common sense is not relevant to infinity
ropey
11-20-2003, 05:57 PM
what do you mean?
Bozeman
11-20-2003, 07:17 PM
There are lots of non-intuitive things about infinity, like the fact that there are infinite integers, and just as many rational numbers, but more real numbers. Have you read the thread on the 3 envelope paradox in the General Theory Forum (warning, most of what is said there is flat wrong).
Craig
bigpooch
11-21-2003, 12:42 AM
the term "onto" is just as you expect: it means a
surjection; the previous poster did not understand
the proper use of "onto" in mathematical lingo
thylacine
11-21-2003, 09:29 AM
Irchans' problem is FINITE. He said amounts $1 to $5000000000.
$5000000000 is FINITE!
Tailgunner
11-21-2003, 02:31 PM
[ QUOTE ]
1) A kind man tells you that he has two envelopes each with money in them. He says that you can choose either envelope, open it, and keep the money in that envelope, or trade the contents of the opened envelope for the unopened one. Is there any strategy that has a higher expectation than randomly choosing, opening, and not trading?
[/ QUOTE ]
I'll pick what's behind pipe number one.
*thud*
::takes both envelopes::
[ QUOTE ]
2) A millionaire is funding a game between you and another person. The other person is given 1000 one dollar bills and two identical envelopes. He is instructed to put any integral dollar amount up to $333 in the first envelope and put exactly twice that amount into the second envelope. He shuffles the envelopes and presents them to you. As in 1) you get to choose one envelope, open it, and then decide whether to trade for the unopened envelope. If you end up with the envelope containing less money, he gets $100. If you end up with the envelope containing more $, he gets nothing. What is a good strategy for you? What is a good strategy for him? What is his worst strategy?
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I'd pick the fatter envelope every time.
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