

Go to page [Previous] 1~2~3~4~5~6~7~8 [Next]  
magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060327 15:27, Steve Martin wrote: You have to offset my expected gain, against my expected loss. Have a pound, win £2 = £1 gain. ..But then what happens? With your calculation, there is no way you can describe the 50% chance of me ending up with half what I started off with as a GAIN! If that was my gain, what was my loss? 

Andrei Veteran user Romania 353 Posts 
Gains should not be put in the same pot of stew as losses. That's what screws up the logic.
Andrei 

Steve Martin Inner circle 1119 Posts 
John,
The expected gain (as I said) is: (100 + (50))/2 The 100 is a positive gain. The 50 is a negative gain (i.e. a loss). There is no problem is describing them both as gains  it is just that one of them is a negative gain.
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 
If I place £1 bet that pays 1/2 my stake every time I win, I am going to win 50p every time I win.
...AND I am going to lose 50p every time I win. So my expectation of ever making a profit has to be zero. So I have a £1 coin... My PROFIT or LOSS if I swap... You have a £2 coin, I GET £2 and LOSE my original £1. (+£1 PROFIT) You have a 50p coin, I GET 50p and LOSE 50p and LOSE my original £1. (£1 LOSS) As each of the above is equally likely, my EXPECTED gain in pence is (100 + (5050100)) /2 = 0p ...Comments...? 

Steve Martin Inner circle 1119 Posts 
"You have a £2 coin, I GET £2 and LOSE my original £1. (+£1 PROFIT)"
Yes. "You have a 50p coin, I GET 50p and LOSE 50p and LOSE my £1. (£1 LOSS)" No. Why are you saying "and LOSE 50p" in the second statement? All you are getting is 50p, and all you are losing is £1. That's a net loss of 50p.
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 
If you bet £1 on the toss of my doubleheaded coin and I give you 50p on heads but take your stake,  you are in a winwin situation, every time you play you win 50p, but what is your expectation of making a profit?


Daegs Inner circle USA 4283 Posts 
Why do I feel like I have to type my post 3 times before someone address them lol...
Would you all agree that if you pay $1 to win $1.50, or win $.50, you are getting even odds? I give him a dollar and half the time I get $1.50 back, and half the time I get $.50 back... So over time I'll even out and I'm basically paying $1 to get $1. So can we agree that that case is even money? So if that is even money, then how can the same bet only earning $.50 MORE on the win($2 payed instead of $1.50) is not ABOVE even money and a good bet? The only way you can prove that a $1> $.50 or $2.00 scenario is equal, would be to prove that either it is the same as a $1> $.50/$1.50 scenario or prove that the $1> $.50/$1.50 scenario is actually a losing proposition.... 

Jonathan Townsend Eternal Order Ossining, NY 27017 Posts 
Does it help to imagine this as done with three boxes. One has a quarter, one a half dollar and one a silver dollar. Just before we start, we take two of the three boxes and give them to the host. The host enters the room and greets the guest with two boxes and an offer. The guest accepts one box as a gift, then the host makes the announcement about the possible relative values of the contents of the boxes and offers an opportunity to switch boxes.
Does this help vanish the paradox for you?
...to all the coins I've dropped here


Daegs Inner circle USA 4283 Posts 
I agree with tomas, the paradox isn't in that its really a 50/50... its the fact that you can state in multiple ways how its NOT 50/50 with completely true statements and logic...


Steve Martin Inner circle 1119 Posts 
Daegs  yes, you and I are in agreement. Your post at 9.22pm on 27th is fine. It agrees with what I have said, and what Tomas said on page 1 of this thread.
In the original problem, as posed, it can be shown that to swap IS a good bet because the expected gain is 0.25X (where X is what you originally hold). (John is trying to convince us, in a very strange way, that it is an even bet... I simply do not agree, mathematically, with what he is saying as he is counting the loss of 50p twice in his calculation.) As we have already said several times, the apparent paradox is that although swapping can legitimately be shown to be a good bet, we instinctively know that to do so is nonsensical since we could have chosen either envelope in the first place.
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 
Daegs…
I haven’t been ignoring your post, I’ve been try to come up with a convincing argument that a ( – X/2 profit) is a win of ½ and a loss of ½ at the same time with a net return of 0… (Still thinking about it.) Your scenario with three envelopes is fair enough, and your argument seems sound. But, in that scenario, I know that the difference between the envelopes is $.50, and I know that I hold $1 so switching will indeed be a even bet. In the two envelope scenario, even though it appears to be more favourable, it isn’t (as we all know), the difference is that the only time that I can gain anything, is when I hold the lower value envelope (and I am going to gain the value of the lower envelope), and the only time I lose, is when I hold the higher. (and I am going to lose the value of the lower envelope.) – An even bet. But this doesn’t wash with the “lose half or double your envelope “paradox”” .. and it makes it very difficult to relate your expected win or loss the contents of your envelope “X”. 

Andrei Veteran user Romania 353 Posts 
Steve Martin, it's not just 'instinctively' that we can make the clear, definite statement, that a swap brings in no extra profits, in the long run.
The question here is why does our APPARENTLY sound logic lead us to the conclusion that a swap IS profitable. I have the answer, but, while it's not sophisticated, it is difficult to word. English is my second language. I will do my best to word it properly and then I will post it. Andrei 

Steve Martin Inner circle 1119 Posts 
"The question here is why does our apparently sound logic lead us to the conclusion that a swap is profitable."
Absolutely. As I said, I have given an analysis of it in plain English, rather than using mathematical language. The paradox is a paradox even if you only count our instinctive knowledge (along with the expected mathematical gain). As I (and others) have said before, there is a deeper mathematical treatment possible to address the issue. None of us has gone anywhere near it. There are many web pages that address it. I look forward to hearing your thoughts. Posted: Mar 28, 2006 8:10am  John, You say: "I’ve been try to come up with a convincing argument that a ( – X/2 profit) is a win of ½ and a loss of ½ at the same time with a net return of 0" How can a negative profit possibly be the same as a breakeven situation? If you can prove that, please come and do my accounts for me.
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 
Hi Steve….
The problem that I am having is disproving Tomas’s original argument that your expected return by switching is (0.5 * 2X) + (0.5 * X/2) = 1.25X . So it is favourable to switch. There is clearly a mistake in the calculation. The premise is that 50% of the time, when you switch for the X/2 envelope, you will show a gain of 0.25X Now I could just argue that Tomas made a mistake i.e. his calculation should have shown that switching for the X/2 envelope should be represented as a potential loss and so should have been (0.5 * 2X) + (0.5 * –X/2) = 0.75X …. Which only creates another paradox in that now I have proved that it is unfavourable to switch! So which calculation is right? (The answer is obviously neither! Or possibly a bit of both!) So I think that there is an argument (somewhere) for saying that switching X for X/2 (50% of the time) results in a gain of 0.25X and a loss of –0.25X at the same time! (Showing a net gain of 0.) (If this can be demonstrated, it eliminates the paradox.) … but I’m finding it rather hard to justify!! 

Steve Martin Inner circle 1119 Posts 
There is nothing wrong (and no mistake) with the maths of the equation:
(0.5 * 2X) + (0.5 * X/2) = 1.25X which implies that it is favourable to switch. Solving the paradox is not about disproving that this equation holds.
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 
And just to be clear. The above is the expected value of the other envelope calculated according to the definition of expected value.
If you want the expected value of your _gain_ it is expressed as: 0.5 * (2XX) + 0.5 * (X/2X) = 0.25X which is positive. This of course is not a correct reasoning either (but the expression is absolutely correct IF there is a 50/50 chance of 2X and X/2 in the other envelope) so the puzzle is to find what's wrong in the reasoning. There was some good explanations in the old Two Envelopes thread. /Tomas 

Andrei Veteran user Romania 353 Posts 
The reasoning derails because even if you STICK with your choices, all the time, you will still have a 1.25X profit, because X is not a constant, but a probabilisticallydetermined variable.
Totally counterintuitive. I'm working on a writtenout explanation. I might have to use some advanced maths for it, contrary to my initial beliefs. Andrei EDIT: Or, better said, that you will always have an X profit, even if you switch. Calculating the average for X/2+2*X is mathematically unfounded in this particular case, because the X is always changing. 

TomasB Inner circle Sweden 1143 Posts 
Andrei...you open your envelope and find X in it. Let's say you open and find 100 dollars. Put that instead of X in the expressions for the expected value of the gain or the expected value of the other envelope. Thanks John for suggesting that you actually check what X is. It makes the problem so much better.
/Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060328 11:10, TomasB wrote: ... I think you mean "It makes the problem much more difficult!" 

Andrei Veteran user Romania 353 Posts 
I'll try to keep this as mathsfree as possible. Keep in mind that this is a simplified version of the whole thing, but studying it carefully MIGHT help you realize that there is really no paradox to begin with, because the premise, and deductions, are poorly made.
A guy walks into a room and sees two envelopes. He knows how the game works, and he thinks that if he switches all the time, he'll beat it. So he'll switch all the time. For simplicity's sake, I will assume that the envelopes contain either 50 and 100, or 100 and 200. You will see why in a second. Further, we must assume that the guy's choice will be perfectly random. He will not bias one or the other of his sides (left, right). Let us also assume that he is given 50 picks from the 50/100 combination and 50 picks from the 100/200 combination. Again, this would not be the case in real life, but to simulate real life we'd need a number of variables equal to double the number of picks, which would be hard to put succintly. Chances are, from his first 50 picks, he will pick the 50 envelope 25 times, and the 100 envelope 25 times. From his next 50 picks, he will pick the 100 envelope 25 times, and the 200 envelope 25 times. Proportionally, we have: 50  25 picks 100  50 picks 200  25 picks But, let's not forget that he SWITCHED THEM ALL. 50 turns into 100. 200 also turns into 100. Half of his 100s turn to 50s and half turn to 200s. so 50 > 100  25 picks 100 > 50  25 picks 100 > 200  25 picks 200 > 100  25 picks By adding them all up, we come right back down to our initial distribution, the only difference is that they've been switched around more, so the guy wasted more time, but ended up with the same money. If you're thinking "Well, that's all pokey. But where's the error in the 1.25X statement?". The short answer is that if you assume that X will always be the same, and calculate your average this way, then you are assuming that the guy will always pick the envelope with the same amount. You're turning a random choice into a choice which biases a specific result. (the reason for that, and the diabolic nature of this little puzzle, is that you're biasing it with the intention of making it easier to calculate) I hope I was somewhat clear. Let me know if there's anything I should expand on. Andrei 

The Magic Cafe Forum Index » » Puzzle me this... » » Two envelopes (0 Likes)  
Go to page [Previous] 1~2~3~4~5~6~7~8 [Next] 
[ Top of Page ] 
All content & postings Copyright © 20012020 Steve Brooks. All Rights Reserved. This page was created in 0.24 seconds requiring 5 database queries. 
The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement < 