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 Better than Money Hall (Moved groups), Snorbolus, 11. May 2003 16:18 | ||
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| The end of last week was a bad time for my powers of rational thought. Please disregard my previous posts on this topic - they were largely nonsense. Shorn you are, of course, right that the chance of having each box is 0.5. Wren, thanks for posting your workings on this problem. I think that I can see where you have gone astray, but I couldn't have come up with my present answer without looking at yours first. Very interested to hear if you think I have it right. You should always switch. Here goes with the explanation: n= number of coin tosses. BoxA contains 3^n coins BoxB contains 3^n+1 coins There are 3^x coins in your box. When x>1 there are exactly two possibilities: Case 1 you have BoxA and x=n Case 2 you have BoxB and x=n+1 Now, as Wren pointed out, we know that the first x-2 coin tosses must have been tails. There is no possible case in which they were not, therefore we can exclude them from our analysis. The only real question is: was toss number x-1 a head or a tail? Because (and this is important) the outcome of toss x-1 fixes the outcome of toss x. Toss number x is either a head or it does not occur at all. There is no possible case in which toss number x is a tail (p=0). As with the first x-2 tosses we can exclude toss x from our analysis; because there is no variability in any of the possible outcomes. If toss number x-1 was a tail then x=n and we have BoxA If toss number x-1 was a head then x=n+1 and we have BoxB It is a fair coin, so the chance of any single toss being either heads or tails is 1/2 (p=0.5). Therefore: We are equally likely to have either box, in all cases where x>1 (ie whenever there are more than 3 coins in the box). What is the net result of switching in each case? Case 1, you have BoxA: When you switch you give up 3^n (3^x) coins and you take 3^n+1 (3^x+1) coins. Net result is 3^x+1 - 3^x = +2(3^x) Case 2, you have BoxB: When you switch you give up 3^n+1 (3^x) coins and you take 3^n (3^x-1) coins. Net result is 3^x-1 - 3^x = -2(3^x-2) The probability of having either box is 0.5 (when n>1) so the expectation of switching is: 0.5*2(3^x) - 0.5*2(3^x-2) = 3^x-3^x-2 This is a positive number. Therefore, by switching, you gain more those times that you have BoxA and switch to BoxB than you loose those times that you have BoxB and switch to BoxA. Thus you should always switch. I am not sure that I have this right. I have already corrected one mistake (using the edit feature) since I first posted this. Snorbolus | ||
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| Re: Better than.. No that still doesn't make sense?!, Snorbolus, 11. May 2003 19:55 | ||
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| Nope, Thought about it some more and my answer sitll doesn't make sense. I am having the same trouble as Wren. If there is a positive expectation to switching no matter what you see in the box why bother to look? But there can not be a positive expectation to switching blind - if there is why not switch back again for even more increase in expectation? It does not make any sense. This must have something to do with the relative frequency of the various values of n - but I can't see what ????!!!! This is all too difficult for me. I will just have to lie down and hope somebody tells me the answer soon. Snorbolus | ||
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| Re: Better than.. No that still doesn't make sense?!, Mark, 11. May 2003 22:35 | ||
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| I am also very troubled by this situation, I've put some of my thoughts in with your post. > Nope, > > Thought about it some more and my answer sitll doesn't make sense. I am having the > same trouble as Wren. If there is a positive expectation to switching no matter what > you see in the box why bother to look? But there can not be a positive expectation to > switching blind - if there is why not switch back again for even more increase in > expectation? There would not be an increased expectation by switching back unless you could look in the box. With out looking in the box you don't gain anymore information and therefore the expectation doesn't change. >It does not make any sense. I know > > This must have something to do with the relative frequency of the various values of > n - but I can't see what ????!!!! I agree, the only determining factor i can see deals with the possibility of arriving at the "N" you have and weighing it against the pros and cons of switching. > > This is all too difficult for me. I will just have to lie down and hope somebody > tells me the answer soon. > > Snorbolus All of the possible answers i come up with call for always switching, but is does not make sense, because you never have any clue as to which box you start with. As a general philosophy, i think that as the values of N go up, it becomes more likely that you have the higher value box. For example, you are more likely to only get 2 heads in a row that 32 heads in a row. So, the higher the value you have, the less likely it is that there is a higher value in the other box. i don't know. this one is tough mark | ||
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| Re: Better than.. No that still doesn't make sense?!, Player X, 12. May 2003 02:23 | ||
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| You can check out my post in the other forum, to see the answer i've come up with. Even if it were to turn out to be wrong, I think it at least answers some of the questions you both have. -X Writer www.pokerev.com | ||
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| Re: Better than.. No that still doesn't make sense?!, Snorbolus, 12. May 2003 06:41 | ||
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| Mark, Yes, I have changed my mind yet again about the relative chance of having each box. I now agree with you that when the box has more than $3 (3 coins) you are more likely to already have BoxB (the n+1 box). If we toss a fair coin 100 times the expected outcomes are: H 50 TH 25 TTH 12.5 TTTH 6.25 and so on.... So, if there are more than $3 (3 coins) in the box then there are twice as many ways for it to be BoxB than there are for it to be BoxA. p BoxA 0.33. pBoxB 0.66. But for the expectation of switching to be 0 (which it surely must be) the relative probabilities of having each box need to be 0.75 BoxB and 0.25 BoxA ????!!!!! That is to say, you need to be 3 times as likely to have BoxB already whenever you see more than 3 coins. > As a general philosophy, i think that as the values of N go up, it becomes more likely > that you have the higher value box....... | ||
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| Re: Better than.. No that still doesn't make sense?!, scythide, 12. May 2003 05:29 | ||
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| You don't switch back again because you're assuming that n can only take on one of two values, switching endlessly will only cycle between these two. However, since it is more likely that you did not get the highest value box to begin with (2/3rds chance you got the lower value box), you always switch the first time. After that, you either have the higher value box or you don't, you can't switch off for another higher valued box. | ||
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| Re: Better than.. No that still doesn't make sense?!, Wren, 12. May 2003 11:20 | ||
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| I'm in the same boat. This apparent contradiction is what makes the problem so perplexing; mathematics professors have apparently debated over it! So I don't feel so bad that I have a BMath and am still confused as hell about it. I have made some progress on understanding this though: I am now swayed by the "always stay if you don't have 3" argument. Stdio put it simply: If you DON'T look inside the boxes, the always stay and always switch strategies will have the same expected value, since 50% of the time you initially get the bigger box, and 50% of the time you initially get the smaller box. However, once you look inside the box, there are times that you KNOW you must switch - when you see 3 coins. To balance this, you MUST have a +EV to stay in all the other cases (when x > 3). I just want to know what is wrong with my math! I think you might be right about something being wrong with my 1/3 / 2/3 probabilities, but I'm not sure what it is. | ||
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| You should always switch, 3Kings, 12. May 2003 08:35 | ||
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| The chances of having the first head are: 2nd flip = .5 3rd flip = .25 4th flip = .125 nth flip = 1/2^(n-1) Therefore, comparing trials, you are twice as likely to get a head on the Nth flip as you are on the (N+1) flip. The odds are 2:1. When looking in the box and you see for example 27 coins, it is more likely that you are looking at the n+1 box rather than the n box. By switching you are twice as likely to get the box with only the n amount of coins in it. However, the one in three times you do have the n box you will get more than you lose the other times. You see the box with 27 coins. Case #1: n = 2, n+1=3. You switch from the 27 coins to the 9 coins. You lose 18 coins Case #2: n = 3, n+1 = 4. You switch from the 27 coins to the box with 81 coints. You win 54. Case #1 will happen twice as often as Case #2. Therefore, your expected value is [9 + 9 + 81]/3 = 99/3 = 33. Since 33 is more than 27, you will show a profit by switching. Similar cases will prove that this works for all n > 2. | ||
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| Re: Better than Money Hall (Moved groups), stdioh, 12. May 2003 09:54 | ||
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| I've posted the solution back in the original thread. It uses telescoping series'. Enjoy. | ||
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| Re: Better than Money Hall (Moved groups), Wren, 12. May 2003 11:14 | ||
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| I looked at your response, and the math is really frickin' ugly and I still don't really understand what you are doing. You might want to try to clean it up and post it again in this thread for all interested parties. Also - tell me what's wrong with my math, dammit! I don't think you've ever done this. I believe your result is right, but I can't help but be convinced by my own "proof". So couldja take a sec and look at it? (I'll post it again here). | ||
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| Re: Better than Money Hall (Moved groups), stdioh, 12. May 2003 12:16 | ||
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| Your proof is "good" and it is exactly the first thing I did in trying to figure out this problem a long time ago. The problem with the proof is a subtle one. You don't know exactly the chance of having the small box vs. having the big box. Yes, you can say that you have either 4 or 5 flips and given that you have at least 4 flips, the chance of having 5 is 1/3. What is wrong with this is that it does not take into account the chance of having 6 or more...now you know that you don't have more than 5, but the fact that there had been a chance for more than 5 skewed your relative incidence rates. Thus, you need to look at the entire sequence, multiply all your chances by your decisions for both cases, and average them. Thus, you deal with the infinite series and cancel terms. | ||
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| My "proof" for always switching - what's wrong with it?!??!?!, Wren, 12. May 2003 11:15 | ||
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| Hypothesis: You should always switch Proof: Case 1: You see 3 coins. You should obviously switch because the other box necessarily contains 9 coins. Case 2: You see 3^n coins, where n > 1. This leaves two possibilities: Case a: boxes contain 3^(n-1) and 3^n coins Case b: boxes contain 3^n and 3^(n+1) coins Probabilities: We know that in each case, the first (n-2) flips will be tails. Case a: flip n-1 must be a H. Probability: 0.5 Case b: flip n-1 must be a T AND flip n must be a H. Probability: 0.25 The other 25% here occurs when both flip n-1 and flip n are T. However, this is impossible since we already know that one of the boxes contains 3^n. Therefore, we must look at the relative probabilities: Case a: 0.5 / 0.5 + 0.25 = 2/3 Case b: 0.25 / 0.5 + 0.25 = 1/3 Now we can obtain the expected value of switching. Let x be the number of coins in the current box. EV(switch) = 2/3 * 1/3x + 1/3 * 3x = 2/9x + x, which is clearly > x. Since the EV of switching is greater than the EV of staying, you should always switch in the general case. ------------------------------------------------------------------------------------------- So there's my math "proving" that you should always switch, but I have trouble consolidating it with the conceptual idea that given one box at random (assuming you *don't* look inside of it), there is a 50% chance of it being the bigger box, therefore it really doesn't matter whether you switch or not - over N trials of switching (where N is sufficiently large), we should end up with the same number of coins as we would over N trials of not switching. However, by looking inside, there are times when we know with 100% certainty that we SHOULD switch: when we see 3 coins. This leads me to believe the opposite of what I just attempted to prove, that in the general case, it is better to stick with the current box than to switch. And THIS leads me to believe that there is something wrong with my math! No one has yet been able to find precisely what is wrong with my proof, so maybe someone here can :O) Whew. That was longwinded. | ||
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| Clarified solution, stdioh, 12. May 2003 12:33 | ||
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| ok. I'm going to edit my solution for clarity. First, probability that we have the bigger of the two boxes is 50% at the start, since there are exactly two boxes and we are chosing one at random. There is a 50% chance we flip a 1, a 25% chance we flip a 2, a 12.5%chance we flip a 3, etc. We will compare two strategies by applying them to all cases and then comparing the totals. Case 1: First, assume that we only switch if we see 3. Case 1:a: Assume that we initially pick up the smaller of the two boxes. We now have the following expected value, expressed as an infinite series: (3*0)+(9*3/4)+(27*1/8)+(81*1/16)+(243*1/32)+(729*1/64)+... Case 1:b: Assume that we initially pick up the larger of the two boxes. We now have the following expected value, expressed as an infinite series: (9*1/2)+(27*1/4)+(81*1/8)+... We now add cases 1a and 1b and divide by 2...we do this term by term and get an infinite series that tells us the expected value when there is a 50% chance of getting the big box and a 50% chance of getting the small box: (9*5/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Call this result 1. Case 2: We calculate the same sequence, but using our strategy of always switching: Case 2:a: we always get the small box and switch to the big box: (3*0)+(9*1/2)+(27*1/4)+(81*1/8)+(243*1/16)+(729*1/32)+... Case 2:b: we always get the big box and switch to the small box: (3*1/2)+(9*1/4)+(27*1/8)+(81*1/16)+... Now again we average these two cases to get: (3*1/4)+(9*3/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Call this result 2. Now we want to know which of the results is larger, representing a more positive EV strategy, so we subtract result 2 from result 1 and the difference will be the amount that result 1 is better than result 2. Result 1 minus result 2 equals: [(9*5/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+...] - [(3*1/4)+(9*3/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+...] Now how do we simplify an infinite series minus another infinite series? Notice that all the terms match up from term 2 of result 1 and term 3 of result 2. Thus, we can cancel them and telescope. And because they are infinite series (no bound on n) we can just dispose entirely of the right hand side. Thus we get: (9*5/8) - ((3*1/4)+(9*3/8)) = 9*2/8 - 3*1/4 = 9/4 - 3/4 = 6/4 =3/2 = 1.5 Thus method 1 is worth $1.50 more in EV than method 2. Thus it is better to only switch if you have $3 in the box you open than it is to always switch nomatter what, to the tune of $1.50 per trial. Now, if there were an upper bound, then the series wouldn't telescope so neatly and we would have a term left on the right hand side. This term would be badly bloated from the exponentiation, since ^3 greatly overbalances log-base2. Thus, the last term would compensate for the entire series and it would become worthwhile to always switch *unless* you got the biggest possible box. Since there is no biggest possible box in this case, it is improper to switch unless you have the smallest possible box (3). | ||
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| Re: Clarified solution, vanagon40, 12. May 2003 17:00 | ||
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| You still haven't sold me stdioh Assuming your math is correct, you proved that switching only on 3 is better than ALWAYS switching. But that same math shows that switching in 3 and 9 is better yet, and adding 27 is even better, but not as good as adding 81. It would appear that always switching is better, so long as there exists some very high number on which you will not switch, but the number is infinitely high, so switching is always better, and the circular reasoning continues. Math in next post | ||
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| Re: Clarified solution, vanagon40, 12. May 2003 17:09 | ||
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| We will compare two strategies by applying them to all cases and then comparing the totals. Case 1: First, assume that we only switch if we see 3 or 9. Case 1:a: Assume that we initially pick up the smaller of the two boxes. We now have the following expected value, expressed as an infinite series: (3*0)+(9*1/2)+(27*3/8)+(81*1/16)+(243*1/32)+(729*1/64)+... Case 1:b: Assume that we initially pick up the larger of the two boxes. We now have the following expected value, expressed as an infinite series: (3*1/2)+(27*1/4)+(81*1/8)+... We now add cases 1a and 1b and divide by 2...we do this term by term and get an infinite series that tells us the expected value when there is a 50% chance of getting the big box and a 50% chance of getting the small box: (12*1/4)+(27*5/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Call this result 1. Case 2: We calculate the same sequence, but using our strategy of always switching: Case 2:a: we always get the small box and switch to the big box: (3*0)+(9*1/2)+(27*1/4)+(81*1/8)+(243*1/16)+(729*1/32)+... Case 2:b: we always get the big box and switch to the small box: (3*1/2)+(9*1/4)+(27*1/8)+(81*1/16)+... Now again we average these two cases to get: (3*1/4)+(9*3/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Call this result 2. Now we want to know which of the results is larger, representing a more positive EV strategy, so we subtract result 2 from result 1 and the difference will be the amount that result 1 is better than result 2. Result 1 minus result 2 equals: [(12*1/4)+(27*5/16)+(81*3/32)+(243*3/64)+(729*3/128)+...] - [(3*1/4)+(9*3/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+...] Now how do we simplify an infinite series minus another infinite series? Notice that all the terms match up from term 2 of result 1 and term 3 of result 2. Thus, we can cancel them and telescope. And because they are infinite series (no bound on n) we can just dispose entirely of the right hand side. Thus we get: (9*1/4) + (27*2/16) -(9*3/8) = 9/4 +27/8 - 27/8 = 9/4 = 2.25 | ||
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| Re: Clarified solution, stdioh, 13. May 2003 11:46 | ||
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| First of all, there is a world of difference between picking a very high, in fact arbitrarily high, number on which to switch and in always switching. I can specify a 3D geometrical figure with a finite volume and in infinite surface area, so you could fill it with paint, but you could not paint the inside of it. This is the nature of math that approaches infinity, so don't think you can make statements like, "the bigger the better so the best is infinitely high." We have shown that allowing this bound to travel arbitrarily high is fundamentally incorrect. Now, all I said was that switching only the 3-box is better than always switching. This is true. Your proof shows that switching the 3 and the 9 is better than switching the 3 only. Have you actually tried it with 27? With 81? Even if you can show them, you have not proven anything. I would suggest an induction proof for solving this problem, however, I have already shown that it does not hold in the general case, so no - there must be some bound. Thus, one of two possibilities remain: 1) the return increases logarithmically as the bound for switching increases, but drops off in a piecewise fashion if there is no bound specified at all. 2) the return is not monotonically increasing and at some point starts decreasing - at the point of inflection is the optimal place to decide whether to accept a box or switch it. The idea of a third possibility...a function that moved around with respect to optimality with multiple ranges for optimal switching is ruled out by the fact that this function must be convex. I'll see if I have time to work out a general solution to what the optimal switching number bound is. | ||
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| Re: Clarified solution, vanagon40, 13. May 2003 12:26 | ||
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| > First of all, there is a world of difference between picking a very high, in fact arbitrarily > high, number on which to switch and in always switching. I can specify a 3D geometrical figure > with a finite volume and in infinite surface area, so you could fill it with paint, but you > could not paint the inside of it. This is the nature of math that approaches infinity, so don't > think you can make statements like, "the bigger the better so the best is infinitely high." We > have shown that allowing this bound to travel arbitrarily high is fundamentally incorrect. > > Now, all I said was that switching only the 3-box is better than always switching. This is > true. Your proof shows that switching the 3 and the 9 is better than switching the 3 only. Have > you actually tried it with 27? With 81? With 27 I get +3.375 and with 81 I get +5.0625. So I stick with my theory that the bigger the better, so the best is infinity. > Even if you can show them, you have not proven > anything. I would suggest an induction proof for solving this problem, however, I have already > shown that it does not hold in the general case, so no - there must be some bound. > There is no bound, and if fact, the positive expectation grows exponentially. > Thus, one of two possibilities remain: > 1) the return increases logarithmically as the bound for switching increases, but drops off in > a piecewise fashion if there is no bound specified at all. > > 2) the return is not monotonically increasing and at some point starts decreasing - at the > point of inflection is the optimal place to decide whether to accept a box or switch it. > > The idea of a third possibility...a function that moved around with respect to optimality with > multiple ranges for optimal switching is ruled out by the fact that this function must be > convex. > > I'll see if I have time to work out a general solution to what the optimal switching number > bound is. For better clarification, see my last post from yesterday. The upper bound is the box with an infinite amount of cash. Unless I open the box with an infinite amount of cash, I am better off switching because I know that I have a 2 out of 3 chance of having the "big" box. The other box has either 1/3 the amount or 3 times the amount. By switching, I lose 2/3 of the prize 2/3 of the time, but gain 3 times the prize 1/3 of the time, giving me a positive expectation of 11/9 of the amount in the box. This works whether the number is 9, 27, 81, or 99 trillion. Therefore, I must always switch unless there is an infinite amount of money in the box, and then I keep the box. But as there is a zero percent chance that the box has an infinite amount of money, I will always switch. You can probably outplay me and beat me at hold'em all afternoon holding 57o against my QJs, but I can hold my own in math. | ||
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| The final word, stdioh, 13. May 2003 12:59 | ||
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| I too can hold my own at math. I have a math degree from a university with some clout indeed. I've done *all* the gruntwork for the general case here. The A case works out as follows: Coefficiets follow this sequence: 1/2,1/4,1/8,1/16,... 0,3/4,1/8,1/16,... 0,1/2,3/8,1/16,... 0,1/2,1/4,3/16,... Thus we can reduce this to a generating function. We see that the following sum represents the first sequence: sum(n=1,infin,(3/2)n) ... we will call this sum A. Now we see that the second production resolves to A+2*3^2/2^2-3^1/2^1 = A+3 The third production resolves to A+2*3^3/2^3-3^1/2^1 = A+21/4 The fourth production resolves to A+2*3^4/2^4-3^1/2^1 In the general case, the nth production will resolve to A+ (2*3^n)/2^n -3/2. Now we shall work out the B side. Our coefficient sequences play out as follows: 0,1/2,1/4,1/8,1/16,... 0,1/2,1/4,1/8,1/16,... 1/2,0,1/4,1/8,1/16,... 1/2,1/4,0,1/8,1/16,... 1/2,1/4,1/8,0,1/16,... We see, trivially, that the first production is equal to 3A and that the second production is equivalent. Now we see that the third production is equal to: 3A + 3/2-3^2/2^1 = 3A-3 the fourth production is 3/2+3^2/2^2-3^3/2^2 = 3A-3 the fifth production is 3/2+3^2/2^2+3^3/2^3-3^4/2^3 = 3A-3 And of course we see that for all productions after that they are of the form: 3A - 3^n/2^n-1 + sum(n=1,infin,(3^n-1)/(2^n-1)). The proof that 3^n/2^n-1 - sum(n=1,infin,(3^n-1)/(2^n-1)) = 3 is trivial and I shall not include it here. Needless to Each leading term cancels out the next term, leaving only as much as the next term to be the new leading term...the result is always three, regardless of n. Thus our averages will proceed as follows: Never switch: (A+3A)/2 = 2A Switch on 3 only: (A+3 +3A)/2 = 2A+3/2 Switch on 3 and 9 only: (A+21/4 +3A -3)/2 = 2A + 21/8 -3/2 Switch on 3, 9, and 27 only: (A+69/8 + 3A -3)/2 = 2A + 69/16 -3/2 And now the real treasure, Swith on any box less than n for an arbitrarily large, finite n: 2A -3/2 + (3^n/2^(n-1)) - 3/2 = 2A + (3^n/2^(n-1)) -3. Thus the larger bound you chose for when to stop switching, the better you will do. Now, here is where I will make a very counterintuitive statement. Always switching is the *worst* thing you can do. Only switching when you have a box with $1,000,000,000,000,000,000 or less is a very good thing to do...a bigger number is even better. The point is that you have to have some finite bound for stopping the switch. Otherwise, the very very very large number at the end of the sequence does not get a chance to exist, and as a result you end up with the smallest possible expected value...that of never switching. You see, if you never switch vs. if you always switch are always equal, since you're getting one of the boxes at random anyway. By switching boxes up to some finite bound, you maximize your return. To put it another way, think of it in real money. The higher you set your limit the more potential money you can make, but the less likely you are to make a small amount of money. Thus if $1,000,000,000 was enough for you to not know what to do with, then you should set your bound accordingly and you'll have the highest EV that will pay you up to that amount. Setting it instead at $1,000,000,000,000 will give you a slightly higher EV, the your chance of winning the smaller amounts goes down in exchange for the slightly bigger chance at the more gigantic number. To make a long story short, there is no such number as "infinity". This is a rare case, where the larger number you limit yourself to, the better you do, but when you don't limit yourself at all, you do very poorly. I know that most of you won't be able to wrap your heads around it...I didn't really start to understand the true idea of infinity until about the third year of my math degree. Like I said before, I can give you a 3D figure that you can fill with paint, but cannot paint the inside of. This is another example of a theoretical construction of that type. In the real world there would be a limit to the amount of money in the bank and by knowing that number you could maximize your switching criterion. Thus in any lottery, where you know exactly what the most you an win it, there is always a precisely optimal move. Here the optimal move is undefined and you can make "more optimal moves" by adjusting your strategy to being as close as possible to infinity without "going over". If anybody has any questions about generating functions, infinite sums, or the likes, I'm game to field questions. Other than that, I hope that this closes discussion on this problem. This is a difinitive proof. And thanks to all who tried this. I hope it was a good mental workout. -<stdio.h> | ||
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| Re: The final word, shorn, 13. May 2003 13:06 | ||
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| You guys must be a blast at cocktail parties... :) | ||
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| Re: The final word, 4 POKER, 13. May 2003 13:30 | ||
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| "I got a mental workout just by reading it"! | ||
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| Re: Clarified solution, vanagon40, 12. May 2003 18:04 | ||
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| OK, after thinking about this all the way home from work, I am convinced that always switching is the correct answer. The most rational explanation for this solution is that no matter what amount of money is in the box, there is a 33% chance that the other box has three times that amount, and a 67% percent chance that the other box has one third the amount. Assuming the amount in the first box is X, by switching you will get (1/3 * 3X) + (2/3 * X/3) or 11/9 * X. The reason this flies in the face of logic is because the problem is premised on two contradicting theories. The first theory is that the person is able to look into the box and see a finite amount of money. The second theory is that it is possible for the box to contain an infinite amount of money. As stdioh noted, if a limit is placed on the number of coin flips (i.e., a cap on the amount in the box), then one should switch until one sees the maximum amount in the box. The solution is the same with the infinite series. One should switch until one opens the box and sees an infinite amount of money in the box. This is the reason that switching without looking does not work, because one might trade the box with the infinite amount of money. I am sticking with this solution until someone convinces me otherwise. | ||
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| Re: Clarified solution, 3Kings, 14. May 2003 20:57 | ||
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| I agree with you vanagon. Because you can see what is in the box, switching will always yeild a positive expectation. Series means nothing here because you aren't working with an infinite series, the box you look in will always have a finite number as well as the other two boxes. If the box you look in has 3^n, then the other box will always have 3^(n-1) or 3^(n+1). Mathmatics will prove (has proved) you will always have a positive expectation, by switching. | ||
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| Re: Clarified solution, stdioh, 15. May 2003 10:02 | ||
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| Sorry, but a mathematical proof is only as valid as the axioms upon which it is based. Since you are starting with false information, your proof is invalid. The false information is your assumption as to what the chance is, after looking, that you have the larger box. Assuming that you have a 1/3 chance of having the bigger of the two is fundamentally incorrect. You need the infinite series to approximate your exact probability of having the big box. | ||
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| Re: Clarified solution, 3Kings, 15. May 2003 12:24 | ||
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| on 15. May 2003 10:02 stdioh wrote: > Sorry, but a mathematical proof is only as valid as the axioms upon which it is based. Since you are > starting with false information, your proof is invalid. The false information is your assumption as > to what the chance is, after looking, that you have the larger box. Unless I missed something in the problem, if you are given a box either it is the one with more coins or less coins. That's it. Now comes the part where you have to decide which box it is. There is a 1 in 3 chance I have 3^n. I have a specific box not some arbitrary one. Then I only have two choices, not an infinite amount. I don't see where my false axiom is. Assuming that you have a 1/3 > chance of having the bigger of the two is fundamentally incorrect. You need the infinite series to > approximate your exact probability of having the big box. | ||
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| Re: Clarified solution, stdioh, 15. May 2003 14:52 | ||
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| You are assuming that you have a 1/3 chance, but that isn't true. Yes, there are two ways to put X in a box and one way to put 3X in the box, but there are a lot more ways to put <X in the box or >3X in the box and the frequency of those skew the relative frequency of the first two. You need to approach the problem wholistically. If you can't see it, then mull this over. If you always switch then you always get the box that I don't hand you first. If you never switch then you always get the box that I hand you first. If I am always handing you a random box first, then always switching and never switching *must* have the same expected value. Period. Thus, there must be a better method that always switching, because we know that switching only on 3 is better than never switching at all. | ||
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| Re: Clarified solution, 3Kings, 15. May 2003 12:29 | ||
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| As soon as I sent my last post, I figured something out. N will not be the same every time so you may pick the bigger box at n=3 and the smaller box at n =10 and come out way down. Now I will have to look at it a little more closely before I comment any more. | ||
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| Re: Clarified solution, stdioh, 15. May 2003 14:53 | ||
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| Fair enough. Indeed, it is important that you look at all possibilities multiplied by how often each possibility presents itself. | ||
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