My favorite has to be the guy who dies and has a host, a sort of butler, who is able to grant any wish to the man. Whatever strikes his fancy, he can have, just for the asking. At first he is elated but as time goes on he realizes how boring such an existence is, and finally he protests that Heaven is not so great after all. His host then smiles wryly and says "HEAVEN? What makes you think this is Heaven?"

Consider this hypothetical. Suppose we take two islands in the Pacific on populate them with chess players. One island is filled with FIDE grandmasters, so Svidler is there, Carlsen, Anand, Kramnik, maybe Kasparov decides he needs a break from politics--they all are living on this island now and they have no contact with the outside world. But they sure play a lot of chess, and we track their ratings. Call this "GM Island".

The other island we fill up with mentally deficient people from an asylum, many of them have barely enough wits to even learn the rules of chess. Not to be cruel, but these people are morons by definition, so let's call it "Moron Island".

Now we reset everybody's rating so they all start at 1500 and they have to re-earn their rating. After many years, many chess tournaments, may matches, we continue to track the ratings of everybody on both island.

Do you think that "GM Island" will have higher ratings than "Idiot Island"? Of course not!! They will have the same distribution. Both island will have people rated 2700-2800, both islands will have people rated 1000-1200, and there would be the very same bell-shaped distribution that we always see. That's right--we could see a Peter Leko rated 2000, we might see a lessor GM rated 1500, and at the very bottom of the the scale we would see a FIDE Grandmaster with a rating of 1200.

Meanwhile on Moron Island, there will be the "king of the idiots", with a 2800 rating. He might employ some trick like e4/Qh5/Bc4 every single game, and set up trivial mating traps that his idiot opponents walk right into. Meanwhile people who need to be constantly reminded how the "horsie moves" will have 2200 ratings.

The point here is, that ratings are not a measure of objective chess strength. In any closed system, the players who win the most games will have enormous ratings, perhaps in the 2800 range. The worst players will have low ratings, perhaps 1000 or even lower.

<I adore that puzzle. It's as close to a genuine paradox as you will ever find. I nickname it the "Double or Half Paradox." If you are holding an envelope contain X dollars, and the other envelope contains either double or half of X, then the expectation value of switching envelopes is

(0.5)*X*2 + (0.5)*X/2 = X*1.25

Ergo, it's 25% more profitable to switch. This makes sense to me, since every gambler knows that "double or nothing" is a fair bet--"double or half" is unheard of and clearly favors the bettor. So this begs the question, if it is always profitable to switch, why didn't you just pick the other envelope in the first place?

To make matters worse, once you've switched envelopes, you can use the exact same logic to conclude that it is profitable to switch back! And switch again, and again, and again...

...
So our intuition tells us, you neither gain nor lose by switching. My computer program confirmed that our intuition is correct. The math, on the other hand, dictates otherwise! Could math be flawed?>

It is a bit thought-provoking. :-)
Here are some of the thoughts it provoked from me:

The situation, as stated, suggest that you might have a few as two envelopes, and all you know is that one has double the money as the other (but you don't know which is which). One is in your hand, and the other in on the table, and you have the choice of swapping the one in your hand with the one on the table.

I don't think that the 'expectation value' of 1.25 really compels you to swap envelopes. The reason is that your choice is between (1) keeping the envelope that you have or (2) swapping it for the one on the table, and the expectation value for the envelop on the table is the same as the expectation value for the one you have in your hand.

Now consider the 3-envelope case:

Again, you have one envelope in your hand that contains X dollars. But now there are 2 evelopes on the table, and you know one of them contains 2X and the other contains 0.5X. Your choice is to either (1) keep the one you have, or (2) swap it for one of the ones on the table.

Now, the expectation for keeping your envelope is X and for swapping it is 1.25X. This does compel you to swap envelopes! After you swap, the table will now have (1) your original envelope, and (2) a *new* envelope (exchanged by some "bookie") that contains either half of your new envelope (if you picked the 1/2X envelope) or double your new envelope (if you picked the 2X envelope).

In this case, expectation compels you to swap again, even though you might just get your original envelope back (note: you might as well flip a coin to pick which envelope to take from the table). If you keep swapping envelopes, probability says that on average, you have equal chances of gaining money or losing money. *However*, the upside potential for winning is much bigger than the downside.

Example:
Suppose you swap envelopes in this manner 6 times. In the worst case, you took the 0.5X envelope each time and ended up with (1/64)*X. In the best case, you took the 2X envelope each time and ended up with 64X. The expected winning would be calculated using the binomial distribution as follows:

#times picked 2X envelope, # of ways, probability, net result

0, 1, 1/64, (1/64)*X
1, 6, 6/64, (1/16)*X
2, 15, 15/64, (1/4)*X
3, 20, 20/64, X
4, 15, 15/64, (4)*X
5, 6, 6/64, (16)*X
6, 1, 1/64, (64)*X

So the expectation is multiplying the probabilities times the results, and adding them up:

(1/64)*(1/64) + (6/64)*(1/16) + (15/64)*(1/4) + (20/64)*1 + (15/64)*4 + (6/64)*16 + (1/64)*64

Which comes out to about 3.185.

I think that that the expectation of 1.25 actually does compel you to swap.

But after the first swap, you now have the option to swap again. Does the expectation of 1.25 still apply? Does it still compel you to swap again?

No. The problem with the 'paradox' is that it presumes that you've forgotten that you swapped. Consequently, it assumes that you might actually double your money twice (or cut it in half twice).

But that is incorrect application of probability. On the 2nd time you swap, you have a "joint probability" problem. That is, the probabilty of an event *assuming a given condition*. The condition is that the first swap occurred, and it changes the probabilities for the 2nd swap.

The first time you swapped, you don't know whether you've doubled your money, or cut it in half, but the SECOND time you swap, you know 100% that you'll merely do the opposite of what you did the first time. That is, if you got 0.5X the first time, you'll get 2X the 2nd time. If you got 2X the first time, you'll get (0.5X) the 2nd time. In either case, you end up with (0.5)(2X) = X.

You might suggest that in the 2-envelope case, that after each time you swap envelopes, that our bookie friend comes along "re-randomizes" the envelope on the table. This means that each time, you really can double or half your money, e.g. in 2 swaps you might double your money twice, or half it twice.

<But after the first swap, you now have the option to swap again. Does the expectation of 1.25 still apply? Does it still compel you to swap again?

No. The problem with the 'paradox' is that it presumes that you've forgotten that you swapped. Consequently, it assumes that you might actually double your money twice (or cut it in half twice).

But that is incorrect application of probability. On the 2nd time you swap, you have a "joint probability" problem. That is, the probabilty of an event *assuming a given condition*. The condition is that the first swap occurred, and it changes the probabilities for the 2nd swap.>

I think you're onto something here. Similar to the famous "Monty Hall Problem" you cannot ignore information that has been dropped in your lap. The fact that you know for certain that switching merely undoes your original switch changes everything.

However, I think you're still embroiled in the paradox when you write this:

<I think that that the expectation of 1.25 actually does compel you to swap [the first time].>

Part of the baffling nature of this paradox is that even the ORIGINAL SWITCH does not truly confer an advantage to the person receiving the envelope.

I proved this point to myself when I wrote a computer program to receive a million envelopes and ran it twice, first picking an envelope at random for all million trials, and then by switching one time for all million trials. As intuition tells us, you gain nothing by switching.

<Suppose it's your birthday, as a special present for you I wanted to give you want I know you want the most.... an envelope stuffed with cash! However I know that you love to gamble so I also wanted to turn this into a gambling game. Since it's your birthday I rigged the game so it's impossible for you to walk away a loser.

I labeled one envelope "A" and the other envelope "B". I grabbed one envelope at random and inserted a sum a money. In the other envelope, I put exactly twice that sum of money. Now, friend, go ahead, pick an envelope.>

She says "I will take envelope A."

<I said, great! Now you can open this envelope and keep what's in it. But if you want, I'll give you a chance to change your mind. Would you like to switch it for what's in envelope B?>

She said "No thanks, I'll stick to my original guess."

I tried to explain to her the mathematical reasons for why switching is profitable, but she'd have none of it. She said "I have a 50% chance to pick the bigger envelope. If I switch I still have a 50% chance. So I'll stick to my intuition about envelope A and skip all the head-games."

Suppose instead of dealing with "some unknown amount of money" in the envelopes we stipulated that one envelope contained $10 and the other contained $20, each time, every time.

Then it's obvious that the action of switching confers no advantage. 50% of the time you will increase your $10 to $20, the other half you will reduce your $20 to $10. So in effect, the action of switching is just a straight coin-flip wager for $10. You either win 10 bucks or lose 10 bucks with an equal chance of each. As any gambler should know, wagering even money on coin-flips is neither a winning nor losing proposition in the long haul.

click for larger view

Analysis by Rybka 2.3 mp 32-bit (18-ply):

1. = (0.03): 1.e4 b4 2.Ne2 e5 3.exd5 e4 4.Bxe4 Nxe4 5.Qxe4 Nf6 6.Qd3 Bf5 7.Qb3 Nxd5

2. = (0.00): 1.Ne2 Qe7 2.h5 Nxh5 3.Rxh5 gxh5 4.Bxh7+ Kh8 5.0-0-0 e5 6.dxe5 Nxe5 7.Rh1 Nxf3

3. = (-0.02): 1.h5 Nxh5 2.Rxh5 gxh5 3.Bxh7+ Kh8 4.0-0-0 Qe8 5.Bd3 e5 6.dxe5 Nxc5 7.Be2 f6

4. = (-0.09): 1.a4 b4 2.Ne2 a5 3.Ng3 e5 4.Be2 Ng4 5.0-0 Qe7 6.b3 Ndf6 7.Ra2 e4

5. = (-0.23): 1.Be2 Re8 2.0-0 e5 3.b4 a5 4.a3 Bb7 5.Rfd1 Qc7 6.Re1 Ng4 7.g3 e4

6. (-0.26): 1.0-0 e5 2.Be2 Qc7 3.Nh2 Re8 4.Rfe1 b4 5.Na4 Ne4 6.Nf3 a5 7.h5 Bb7

7. (-0.26): 1.Nh2 e5 2.0-0 Qc7 3.Be2 Re8 4.Rfe1 b4 5.Na4 Ne4 6.Nf3 a5 7.h5 Bb7

8. (-0.30): 1.0-0-0 Ng4 2.Be1 e5 3.Be2 e4 4.Nh2 Ndf6 5.Nxg4 Nxg4 6.h5 Bd7 7.Kb1 Qf6

9. (-0.33): 1.Rc1 Re8 2.0-0 e5 3.dxe5 Nxe5 4.Nxe5 Rxe5 5.Ne2 Ne4 6.Be1 Bd7 7.b4 a5

Suppose I put two envelops on the table. One has X dollars in it and one has (X + 5 dollars) in it.

If you open yours and find $33, then the EV of switching is zero, but your guys argument about not-switching because there is a 50% chance you already have the best envelop is absurd. Its just the examples you gave dont exemplify the absurdity!

According to the same logic, you wouldn't switch an envelop that has x with one that has 50% probability of 5000x and (1/50)*x

If you open envelop and it has $30,000 and you are 100% certain that the other envelop has with equal probability either $6000 or 1.5 million dollars, you'd walk away and say "well there's a 50% chance I already have the richest envelop"

of course not!

<I proved this point to myself when I wrote a computer program to receive a million envelopes and ran it twice, first picking an envelope at random for all million trials, and then by switching one time for all million trials. As intuition tells us, you gain nothing by switching.>

Your computer program is FLAWED!!!!! You misstate the "1 switch version" of the paradox (the 1 switch version is the only "valid" paradox)

My computer program was written out in hand.

I receive an envelop with $5 and switch it one time. I ran this experiment 1,000,000 times using "keep the envelop" and 1,000,000 for switch.

These are the outcomes

$5 and keep

=outcome=
1,000,000 outcomes of $5
EV = +0

Next experiment

= $5 and switch =

=outcome=

600,000 outcome of $2.50
400,000 outcome of $10.0

EV = +.50

Even when I sabotage my luck, I still have +EV

The Real EV is actually +$1.25

The one-switch variation doesn't let anyone make a statement that goes "well if you switch twice..." which is the easy way out.

You wrote your computer program in such a way that the 2nd envelop was DEPENDENT on the first envelop, when in fact the 2nd envelop per the stipulations have 50% equal probability of being greater than or less than the first one.

<As intuition tells us, you gain nothing by switching.> This is because your encoded your intuition into the program. You must completely say that the envelop you are switching for has 50% probability of 2x and 50% probability of half-x not probably as you did "50% probability of envelop 1 and 50% probability of envelop 2"

By this method if you are first dealt envelop 2, (which happens 50% of the time in your program) you ALWAYS lose EV by switching, and is the same amount of EV that is GAINED by switching from 1 to 2.

Contestant A and B are racist people of opposite race and purely want to maximize their payoff.

Person A opens envelop and smiles, knowing that he can either lose 98% of it, or multiply it by 50.

Person B opens envelop and smiles, knowing that he can either lose 98% of it, or multiply it by 50.

The game is financed by a saudi prince trillionaire who has $80,000,000,000,000 dollars. Additionally, the rules of the game is that the envelops are decided by this method. The "richest envelop" has equal probability of every amount between 0 and $80 trillion (every envelop will have a check with at least 10 decimal places so that no hints can be inferred by an amount such as $97.17) when the game is begun.

Player A opens his envelop and sees $5 million. He has a 50/50 chance of swapping to gain $250 million or could lose 4.9 million.

Now answer BOTH these two hypotheticals.

Player B holds $100,000. From B's perspective, what is his EV if he doesn't know what player A has.

Player B holds $250 Million. From B's perspective, what is his EV if he doesn't know what player A has.

Already your "intuition" will start telling you that player B should intuitively know that to swap a low case and keep a high case. This is simply incorrect because of a 3rd type scenario.

Each player is allowed to make a deal that "100% indemnity gets automatically paid to the loser" thus if Player B profits from the transaction, he gives back 75% of the difference, which will reduce risk on both sides and make both more eager.

Now consider this 3rd type. They agree to switch one time and split all the money 50/50.

That way a person who has $5 million before the flop will either walk out with $2.5 million or $125 million which is obviously better than taking $5million without having a payoff ratio of 75-to-1 on the part that he actually risks which only "loses" 50% of the time.

1. There are two envelopes, one with X dollars and one with 2X dollars.

2. First case--you have the one with X. If you flip a coin either you will hold (50% probability) or switch (50%). Your expectation is .5*X+.5*2X = 1.5X

3. Second case--you have the one with 2X. Your expectation (hold or switch) is .5*2X+.5*X = 1.5X

4. It is exactly the same in each case no matter what X happens to be.

The math proves that you always switch (given the one-switch variation). But honestly though, I do know the "answer" which is why I'm able to play the wordings so good.

You guys keep assuming it is a matter of "you start out with either envelop #1 or envelop #2"

That is like saying you can open your envelop and you have 100% certain outcome, so long as you know which envelop you have, or what the 2 envelops contain.

It is "take double or half" or "I'll keep whats in my envelop". False dichotomy is believing that if you have one envelop out of only 2 possible envelops, then by this stipulation you don't really have a 50/50 chance of doubling or halving, but instead it is "do you switch envelop #1 for envelop #2 or viceversa"

If you hold envelop #2 and there is $6, then by definition envelop #1 is $3, but there must also be an envelop with $12 if you are holding $6 whose existence is equally guaranteed as the $3 envelop (too many pessimists here).

The real answer involves setting it up as a probability function of infinitely many binomial coeffecients who's EV does in fact equal the EV of the results.

But clearly, you guys are engaging in false dichotomy aka false dilemna.

When you hold X dollars there are 3 outcomes... half-x, double-x, or keep your envelop.

To say that you have the larger envelop, you would actually have to say

there is a 50% probability that you hold 2x, and if you switch to x, your EV is -x, but there is also 50% probability that you hold 4x, and if you switch to 4x your EV is +2x.

Note that it is a "double or halve" or keep. There is serious error that when you say "if you have the larger envelop and switch to the smaller envelop" it already has false dichotomy in the first 6 words!

If you double or halve the amount in your envelop, what is the payoff to "switch". What is the payoff to "keep". The answer is not "it depends on if you are holding the larger or smaller envelop. It depends on "will I double or halve the money in my envelop".

This post was not word tricks, its 100% truthful. Forget the envelops.

Imagine a saudi royal prince who will double your bet if you roll a 4,5, or 6 on a fair dice roll. Or he'll keep half your bet (and return half of it) if you roll 1, 2, or 3

I think you're confusing the 2-envelope case (which <hms123> answered) with the 3-envelope case (which I discussed in my first post above).

In the 3-envelope case, you should swap your envelope for one of the other two, for the reasons you say (being that the EV is 1.25).

But the 2-envelope case is equivalent to simply picking one of two envelopes, where you know that one has twice as much money as the other. If we let the lower one contain X and the higher 2X, then the EV is 1.5X.

In the 2-envelope case, you are making the same mistake I was, by thinking that there were equal probabilities for doubling (2X) or halving (0.5X) your orginal money (X).