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magicjohn2278 Special user Isle of Man UK 536 Posts 
It is probably more correct for me to ask, "If the value of X changes according to a function and you switch what are the chances of applying the function to X compared to the chances of applying the inverse function to X."


Steve Martin Inner circle 1119 Posts 
John, I understand what you are saying, but the bottom line in all this is how much of an INCREASE (plus) or DECREASE (minus) you make with your money.
When you double your money, you are simply adding its value to what you had. When you halve your money, you are subtracting half its value from what you had. So it is ultimately a matter of making (adding) and losing (subtracting) money. I don't think this lies at the heart of the problem. If Tomas can answer my question about the 2/3 of Z, he will satisfy me that his solution is correct!
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Steve, almost right, but assuming that you do switch the final outcome was predetermined when you selected the first envelope (X). If you selected the larger amount you lose X/2, if you selected the smaller you gain X  but unfortunately (mathematically speaking) X/2 is equal to X  there is no point switching.


Steve Martin Inner circle 1119 Posts 
John  X/2 does not equal X
Tomas  if Z is the maximum total of both envelopes, surely it is impossible for X to ever be *greater* than 2/3 of Z. Equal or less than, yes, but not greater. What ya say to that, ya little rascal?!
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

TomasB Inner circle Sweden 1143 Posts 
Hehehe, I already told you that was an error. This is what it should sound like.
I just need a good way to make the random variables Y and Y/2 so their sum always is less than Z. That means that Y+Y/2<=Z which means that 3/2*Y<=Z which means that Y<=2/3*Z. Let's just form W=2/3*Z and follow the reasoning in my post: Choose a Y randomly that is less than W and put in one envelope. Put Y/2 in the other. The sum is now automatically random but less than Z. We are back in the same case I described where there are double as many cases where we find X <= W/2 than we find above W/2. Ok? /Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060329 10:37, Steve Martin wrote: Steve in this case if you have X and switch for X/2 your loss is equal to your gain had you had X and switched for 2X. 1/2=1 

TomasB Inner circle Sweden 1143 Posts 
John, please don't say that, because I'd gladly pay a dollar to flip a coin and you pay me 2 dollars if it shows heads and you pay me 50 cents if it shows tails.
/Tomas 

Steve Martin Inner circle 1119 Posts 
Ah, yep  sorry, Tomas, I hadn't appreciated from your earlier reply that it was an error...
I see it now. We set a combined maximum total of Z. For every value we choose (for the first envelope) that is between 1/3 of Z and 2/3 of Z, there is a second value (for the other envelope) that is between 0 and 1/3 of Z. AND  for every value we choose (for the first envelope) between 0 and 1/3 of Z, there is a second value (for the other envelope) that is between 0 and 1/3 of Z. THEREFORE, we generate twice as many possible values between 0 and 1/3 of Z, as we do between 1/3 of Z and 2/3 of Z. And that is the key which gives rise to the revised expectation equation: 2/3 * (0.5 * X/2 + 0.5 * 2X) + 1/3 * (1 * X/2 + 0 * 2X) = X You little beauty! Full credits to Tomas and Andrei for being utterly brilliant  I now understand this!
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060329 10:55, TomasB wrote: Not the same situation.. I'll play if you give me half what you have when I win and I double what you have if I lose.  My expectation of losing is zero! .. as losing double is the inverse of winning half! They are the same! 

TomasB Inner circle Sweden 1143 Posts 
Quote:
On 20060329 11:20, magicjohn2278 wrote: Ok, I hold one of my own dollars. When the coin shows tails I pay you 50 cents and when the coin shows heads you pay me what? Please explain the rules. /Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Right, you start off with say $1000, If I win you give me half what you have, If I lose, I will double whatever you have.
If on the first go, I lose, I give you $1000, you now have $2000, on the next turn, if I lose, I give you $2000.... But if on the first go, I win, you give me just $500, and If I win again, you give me just $250 I can't offer better than that! You are a sure winnner! 

TomasB Inner circle Sweden 1143 Posts 
I see. Funny game, just a random walk where I can never loose more than 1000 dollars but can gain an infinite amount. I can in other words decide that I want to stop playing when I'm above 1000 000 dollars. Since it's a random walk that will eventually happen and the game does not have a stop because my money can only be halved but never gone completely.
Not sure why you posted it here though. In your game you can never find 687 dollars at any stage, for example, since you just move in discrete steps that are very uneven in size...which is very different from the envelope game. /Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
No it isn't it is the same. You can start with any amount you wish 687 dollars or whatever... the rules won't change.
So what is your expected gain over say 100 games? (or to make it easier, 10.) 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060329 12:31, TomasB wrote: The envelopes vary in identical steps, <...1/2...1...2...> 

Andrei Veteran user Romania 353 Posts 
Expected gain or no expected gain, he'll just play and play and play until his earnings exceed an amount favourable to him, assuming, of course, that he can do this within his life span. It takes time to flip coins, after all.
Andrei 

magicjohn2278 Special user Isle of Man UK 536 Posts 
The point I am trying to make is if you have envelope X, and switch, you will either double or half what you have. With equal probability.
If X = 10 it is worth the risk of losing 5 against the possibility of gaining 10. But it only looks that way because we imagine that the other envelope might contain either 5 or 20. It is the wrong way of looking at it. It contains either half or double what we hold. We are imagining there is a potential range of envelopes X/2..X...2X.(which there isn't.) We say that doubling X is better than losing half X because we convert X/2 and 2X to numbers that we are used to dealing with. We think we know that losing half what you have has less negative effect on our expected gains than doubling what you have. It doesn't  the effects are the equal and opposite. If I can show this, then our original thinking is proved wrong. It;s too simple to ues our two envelope model for this.. you would just say, "well what did you expect!" when I show you the answer, so I have to make the game a little more complicated!  But it IS the same game. I give you an envelope containing X, if you wish, I will switch it for another which contains either half or double what you hold... then I offer you the same deal, over and over again. "Winning" is switching for the envelope containing double the value of your current envelope. "Losing" is switching for the envelope containing half the value of your current envelope. We are going to win two games then lose two games, obviously, as we know that when we lose we only lose 1/2 of what we have, and when we win we double what we have we will show a profit at the end. Your starting envelope contains say $40, you switch and win envelope with $80 in. You switch and win an envelope containing $160. Two wins. From this position, you are going to lose 2 games.. Switch and lose you now have $80. Switch again, lose and your envelope contains $40. We won two games, we lost two games, we have exactly the same as we started with. What went wrong? 

Steve Martin Inner circle 1119 Posts 
Here is a summary of the solution in words (credit to Tomas and Andrei for getting us to this point!) 
If you take every possible pair of envelopes, every value that is equal to or less than onethird of the maximum combined total of both envelopes, will appear twice as often as every value that is greater than onethird of the maximum combined total of both envelopes. Therefore, when you choose one of the envelopes, it is twice as likely that when you swap, you will go either up or down, as it is that when you swap you will go down. Hence, when you swap your envelope containing X, you expect to get: 2/3 * (0.5 * X/2 + 0.5 * 2X) + 1/3 * (1 * X/2 + 0 * 2X) = X and therefore there is no advantage in swapping.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

Steve Martin Inner circle 1119 Posts 
I suggest this is a good point on which to end this topic. Spinoff topics, please start a new thread.
Thanks to everyone who contributed... it was a good one.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 536 Posts 
For every possible value of the greater envelope, there is a corresponding "lesser" envelope. (And vice versa) the chances of picking either are 50/50.
Give me any value of the greater envelope and I will tell you the contents of it's corresponding pair. (But not infinity!) 

Josh the Superfluous Inner circle The man of 1882 Posts 
But this time it actually was better to switch. Oh well I guess half a million dollars is pretty good. But I tried to give you a chance. I thought for sure you'd get the hint when I offered you the chance to switch. Oh well.... (under breath) loser.
What do you want in a site? "Honesty, integrity and decency." Mike Doogan
"I hate it, I hate my ironic lovechild. I didn't even have anything to do with it" Josh #2 

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