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Server Time: 11/20/2009 8:37:09 PM PACIFIC |
 Better than Money Hall, stdioh, 8. May 2003 13:27 | ||
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| Now, Montey Hall is really easy to work through, but I've got one for you that is a real noodle baker. It is one where doing the math will give you one result and applying logic will give you another result and it takes a great deal of insight to really understand the solution. Here is the question. I randomly seed two boxes with money by the following algorithm. I flip a fair coin until I get a head and then record the number of times I've flipped the coin as n. I now put 3^n dollars in one box and 3^(n+1) dollars in the other. I hand you one of the boxes, but I don't tell you what n is. You open the box and can either choose to keep the money or exchange boxes. Obviously if there is $3 in the box then you should switch because the other box has $9. What should you do if there is $9 in the box? Switching will either get you $27 (+18) or $3 (-6). What should you do in general? I will post the solution to this tomorrow or some time after. | ||
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| One hint., stdioh, 8. May 2003 13:28 | ||
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| Just one hint. If you limit the size of n to be a maximum of some finite number, say 50, then the solution is different from when n is unlimited, as it is in this case. | ||
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| Wren., stdioh, 8. May 2003 13:30 | ||
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| Ok Wreny...don't ruin this one for everybody else just because you know the answer :) | ||
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| Re: Wren., Wren, 8. May 2003 13:45 | ||
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| I know what you have TOLD me is the answer, but that doesn't mean I understand or necessarily believe it :O) To be frank, I don't think anyone's gonna come up with the uw.general solution, but it'll be fun reading all the varying thoughts and opinions anyway. As a side note - shouldn't this go in NotQuitePoker? | ||
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| Re: Wren., stdioh, 8. May 2003 13:50 | ||
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| Good point...it very much is not quite poker. Too late. C'est domage. | ||
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| Re: Better than Money Hall, shorn, 8. May 2003 13:57 | ||
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| OK, maube I am really naive, but it appears to me that you would definitely choose to switch boxes. Here is why: Since it is a "fair" coin, the likelihood that the next flip is a "head" is 50%. So, the expected value of the second box is ((.5)*($27))+((.5)*(6)) or $16.50 Since box #1 only has $9, I think you go for the switch. | ||
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| Re: Better than Money Hall, ice, 8. May 2003 16:17 | ||
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| i think you should switch on the general premise that you are risking loosing six, to gain 18 (3 to1) if somebody would pay out 3 to one on a legit coin toss with me, i'd do it all day long ;) *********** ¡¢£ßøx | ||
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| Re: Better than Money Hall, Mark, 9. May 2003 09:22 | ||
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| shorn, i think you were very close but i would like to change your calculations a little. You did not consider the coins probability correctly. I want to show a quick list of coin flip probablilities. i think it must factor into the equation. This is the probabilities of the number of coin flips neccessary to get heads 3^1 50% 1:1 3^2 25% 3:1 3^3 12.5% 7:1 and so on > Since it is a "fair" coin, the likelihood that the next flip is a "head" is 50%. > So, the expected value of the second box is > > ((.5)*($27))+((.5)*(6)) or $16.50 > > Since box #1 only has $9, I think you go for the switch. 1st if you have $9 in your box either you picked box A and N = 2 or you picked box B and N = 1 CASE 1 ssume you have Box A, N=2 If you swith you will get Box b with $27, +$18 profit. you have to factor in the probability that N will reach 2, which is 25%. CASE 2 now assume you have box b, n=1 if you switch you will get Box A with $3, -$6 loss. the probablility of N = 1 is 50% so the EV of switching is case 1 * probability + case 2 *probability (+18) *.25 + (-6)*.5 = 4.5 - 3 = +1.5 so it is clearly advantagous to switch for $9. i don't know if this will always produce a positive expectation with large #s. | ||
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| Re: Better than Money Hall, shorn, 9. May 2003 09:34 | ||
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| Mark- Thanks. But I am not sure that I agree with the coin flip probabilities. With a fair coin, each flip is an independent trial, so you really have no information as to what the "next flip" is or was just because there is $9 in the first box. In other words, just because you flipped the coin twice and got heads twice, the likelihood that the next flip is a head is still 50%. So, to me (and again, I am no calculus whiz), you can't link the coin flip probabilities together to assign a lower % than 50% to either of the boxes. Therefore you should have a 50% chance of the next box containing either $27 or $6. Anyway, I appreciate your thought process, and it would work if we were talking about a finite set of outcomes (say, if you draw two cards from a deck and they are both hearts, what are the chances that the next card drawn is also a heart). However, in this case, any flip can give you the previous result, so linking doesn't apply. Anyone else disagree? Anyway, my 2 cents. Steve | ||
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| Re: Better than Money Hall, Mark, 9. May 2003 09:43 | ||
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| > Thanks. But I am not sure that I agree with the coin flip probabilities. With a fair coin, > each flip is an independent trial, so you really have no information as to what the "next flip" > is or was just because there is $9 in the first box. In other words, just because you flipped > the coin twice and got heads twice, the likelihood that the next flip is a head is still 50%. > So, to me (and again, I am no calculus whiz), you can't link the coin flip probabilities > together to assign a lower % than 50% to either of the boxes. Therefore you should have a 50% > chance of the next box containing either $27 or $6. > > Anyway, I appreciate your thought process, and it would work if we were talking about a finite > set of outcomes (say, if you draw two cards from a deck and they are both hearts, what are the > chances that the next card drawn is also a heart). However, in this case, any flip can give > you the previous result, so linking doesn't apply. > > Anyone else disagree? > > Anyway, my 2 cents. > > Steve i see your point, but do you think the odds of flipping 4 heads in a row are 50%? each flip is independant of one another, i agree. but because we are talking about the odds of sucessive flips, it becomes important to factor in the probabilities. mark | ||
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| Re: Better than Money Hall, Mark, 9. May 2003 09:46 | ||
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| maybe i'm wrong, let me think about it. | ||
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| Re: Better than Money Hall, shorn, 9. May 2003 09:51 | ||
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| > i see your point, but do you think the odds of flipping 4 heads in a row are 50%? > > each flip is independant of one another, i agree. but because we are talking about the odds of > sucessive flips, it becomes important to factor in the probabilities. > > mark Mark- No. The odds of flipping 4 heads in a row is not 50%. But what you have just done is change the problem. If it had read "what are the odds of flipping a head if you flipped a coin 4 times?" Now, all 4 flips are related to each other because you have 4 shots at flipping a head. As I read the problem laid out, the only question is "What is the likelihood of the box you have opened being n or n+1?" Since you have no other information, it seems (IMHO) to me that you cannot make assumptions about how many flips it took to get to $9 (since the coin is fair). So, you have to view the chance that the other box has $6 or $27 as a 50/50 shot. I think we both have good arguments here. I woud like to challenge others to join us and maybe collectively a number of other posters can change my mind or yours. I am not bullheaded enough to think that there is no way I am wrong! Steve | ||
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| Re: Better than Money Hall, Snorbolus, 9. May 2003 10:54 | ||
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| Shorn and Mark, You can link the coin flip probabilities because we are not asking "what is the chance that the next result will be heads?" We are counting the number of flips untill a head is scored. This is very different. For example: if we know that the nubmer of flips before a head was scored were either 3 or 4, then the quesiton is "did we get (a) TTTH or (b) TTTTH?" None of the other possible outcomes match the criteria. The probability of result (a) (3 tails in a row) is 0.5^3 (ie 0.5X0.5X0.5) - the probability that there were fewer than 3 tails in a row is 1-0.5^3 The probability of result (b) (4 tails in a row) is 0.5^4 - the probability of fewer than 4 tails in a row (either 0, 1, 2 or 3) is 1-0.5^4 Snorbolus > Thanks. But I am not sure that I agree with the coin flip probabilities. With a fair coin, > each flip is an independent trial, so you really have no information as to what the "next flip" > is or was just because there is $9 in the first box. In other words, just because you flipped > the coin twice and got heads twice, the likelihood that the next flip is a head is still 50%........ | ||
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| Re: Better than Money Hall, shorn, 9. May 2003 11:17 | ||
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| In this case, we don't know how many flips it took to get a heads, so you can't link. The number it took to get heads is irrelevant as the question is stated. Thewre isn't enough information to determine it. Now, I agree with your other post that as the amount of $$ in the box we open increases, then the general likelihood that we hold the n+1 box is higher. HOWEVER, with the box that we opened containing $9 (or 3^2), I still think it is a 50/50 shot between the other box having $27 (3^3) or $3 (3^1). | ||
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| Re: Better than Money Hall, Mark, 9. May 2003 09:48 | ||
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| I screwed up earlier, this is my new answer 1st if you have $9 in your box either you picked box A and N = 2 or you picked box B and N = 1 CASE 1 ssume you have Box A, N=2 If you swith you will get Box b with $27, +$18 profit. CASE 2 now assume you have box b, n=1 if you switch you will get Box A with $3, -$6 loss. > so the EV of switching is case 1 * probability + case 2 *probability (+18) *.5 + (-6)*.5 = 9 - 3 = +6 so it is clearly advantagous to switch for $9. mark | ||
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| Re: Better than Money Hall, shorn, 9. May 2003 09:55 | ||
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| Now I think we are on the same page. I just did the math slightly differently. | ||
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| Re: Better than Money Hall, Schuster, 8. May 2003 17:23 | ||
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| Pretty sure I will get this wrong the first try, but here's a shot. Don't read below if you want to figure it out for yourself. - - - - - - - - - - The EV of the box you are given will always be 3^m (I set the given box integer as m to avoid possible confusion). So, your only decision then is the other box being higher EV or not. The EV of the other box is going to be the odds of it being the higher amount times the higher amount plus the odds of it being the lower amount times the lower amount. The probability of any given n being selected goes as 2^(-n). IE, the odds of n being 3 are 1/8. Odds of the other box being m + 1 are 1 / 2^(m+1), while the odds of the other box being m - 1 are 1 / 2^(m-1). This is the part I'm not sure about, but I think you have to normalize each probability, making sure that the total probability of the other one being m+1 or m-1 is just 1, as compared with the total probability of staying also being 1. It's 4 times more likely that the box you have is the box with more money than the box you have is the one with less money, and when normalized, you get 0.8 and 0.2. So your EV for switching would be (4/5) * 3^(m-1) + (1/5) * 3^(m+1). Going back to earlier, your EV for staying would simply be 3^m. Compare the two results and you get that your EV is higher if you stay. EV for staying is 1, where EV for switching is 13/15. Of course, the situation where m is 1 has a special solution due to the boundary conditions. n will always be at least one, so there can be no solution where n = 0. I think that works out. In case this is wrong (and probably is) here's a few other results I came up with. At least you can tell me if I'm on the right track with one of these others. =) Switch if m is 1. Otherwise, stay. I ended up with EV of 1 compared to 5 / 2^(m+1). Your EV is 25% greater if you switch. Long jumbled calculations and I probably made an error somewhere. You're right on one thing, this really is baking my noodle! Lee | ||
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| Re: Better than Money Hall, Snorbolus, 8. May 2003 21:31 | ||
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| Stdioh, Yes this one is tough. I am having trouble with the maths. See below if you are interested in my workings so far. The thing about the Monty Hall problem is that, as you pointed out, the maths is very easy. Nevertheless, I found it very difficult to accept the correct answer once I had it. It is as if I find it annoying that things should be that way. Very strange. Working for difficult box of money problem: Box contains $3^x It is either boxA and n=x or boxB and n=x-1 Expectation if we switch: If we have boxA we give up $3^x and gain $3^x+1 Therefore we gain 3^x+1 - 3^x = 2X3^x (I think??) Probability we have boxA is pn=or>x which is 0.5^x If we have boxB we give up $3^x and gain $3^x-1 Therefore we gain 3^x-1 - 3^x = -2X3^x (I think??) Probability we have boxB is pn<x which is 1- 0.5^x Now we need to sum (p we have boxA X gain if we switch) and (p we have boxB X gain if we switch) to decide weather or not to switch. However (even if I haven't made a mistake so far), I am dammed if I can solve that for x. Snorbolus | ||
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| Re: Better than Money Hall, stdioh, 9. May 2003 08:50 | ||
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| Heh...I like what I am hearing so far. Let me give a couple of hints. First, you are handed the box of the two randomly, so before you look, there is a 50% chance that you have the big one and a 50% chance that you have the small one. If your strategy is to always chage boxes nomatter what is inside, then you always change and it doesn't matter what is inside, so you'll always end up with a 50% chance of having the big box. Hint #2: Use a telescoping series. | ||
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| Re: Better than Money Hall, Snorbolus, 9. May 2003 09:56 | ||
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| > Let me give a couple of hints. First, you are handed the box of the two randomly, so > before you look, there is a 50% chance that you have the big one and a 50% chance > that you have the small one. If your strategy is to always chage boxes nomatter what > is inside, then you always change and it doesn't matter what is inside, so you'll > always end up with a 50% chance of having the big box....... That is true but I had assumed that the object was to make decisions that would net the most money over repeated trials. In this case, my feeling is that the decision of wether to switch or not will depend on the size of M (where M= amount of money in the box that you are given). We know that M will always be 3^x and that x is either n or n+1 (where n is the number of coin tosses). Now: as M increases so does x (x is proportional to M) as x increases the probability that x < n decreases exponentially. That is to say it becomes more likely that we already have 3^n+1 box. So if we switch we will be giving up the larger box more frequently. as x increases the difference between M when x=n and M when x=n+1 increases exponentially. So those times that we have the smaller box and don't switch we will be giving up more as M (and x) increase. So: We need to find an expression for the cost of switching when we have the 3^n+1 box multiplied by the probability of having that box; plus the gain from switching when we have the 3^n box multiplied by the probability of having that box. Then we solve that equation for all values of M, graph the result and see at what value of M the expectation of switching dips below 0. Here is the rub. I haven't yet managed to generate the graph. My maths isn't up to snuff. Am I at least on the right track though? Snorbolus | ||
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| Re: Better than Money Hall, stdioh, 12. May 2003 08:47 | ||
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| Well, I'll clarify this a little. If we flip the coin up to 20 times, then the correct strategy will be to always switch unless we have a box that contains 21^3 ... since the value of n is unbounded we end up with an infinite telescoping sequence. Thus you need to look very closely at the math that generates your EV. You need to look more closely at the true relative probabilities. Let us assume that we flip our coin an infinite number of times, but we only count to the first head. Then we could have TTTH... or we could have TTHT... or TTHH... and all three of these yield the number 3 or 4. Thus, chance of getting a 3 is 1/8 and chance of getting a 4 is 1/16, but we see that there are exactly 3 infinite sequence classes which yiled 3 or 4 at 4 places and thus there is only a 1/3 chance that we have a 4. That said, we *still* have odds to switch boxes for the case of getting 9. Just letting you know that it is not 50/50 odds of having the big one. Now, work one level deeper and you should get the solution. | ||
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| Re: Better than Money Hall, Schuster, 9. May 2003 11:25 | ||
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| > First, you are handed the box of the two randomly, so > before you look, there is a 50% chance that you have the big one and a 50% chance > that you have the small one. Gah, this goes against my intuition a lot. If I have a box with 3^10 bucks in it, I'd be more inclined to think that this box was the big box rather than the small box. > Hint #2: Use a telescoping series. I'm not aware of these. Is it tough to explain or could you shed a little light? =) Lee | ||
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| Re: Better than Money Hall, stdioh, 12. May 2003 08:49 | ||
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| Well, if I want to figure out what the answer is to: 1+2+3+4+5+6+7+8+...+100 - ( 3+4+5+6+...+99 ) then I don't have to work everything out...I can cancel terms and end up with 1+2+100 which is easy to compute as 103. Likewise, if I want to take 1+2+3+... and subtract from it 3+4+5+... where the terms move up indefinitely and infinitely, then I can cancel all the big terms and end up with 3. Apply this method to this problem. | ||
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| Re: Better than Money Hall - Some longwinded math for y'all, Wren, 9. May 2003 11:34 | ||
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| Stdio told me not to "spoil" this problem for everyone else, but to be frank I still don't understand or necessarily accept the "correct" answer. So I'm gonna post my thought process here, and see if anyone can find any flaws: ------------------------------------------------------------------------------------------- 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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| My answer, Player X, 12. May 2003 01:29 | ||
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| Well I didn't take too much time figuring this out, so there's a good possibility of me making a mistake. X is the amount of money in the box X/3 is one possibility of the amount of money being in the other box 3X is the other possibility. It is four times as likely that there'll be X/3 in the box as 3x therefore... your expected value of switching is (13/15)X, meaning you should keep your box. That of course all excludes a situation like only $3 being in the box. X Writer www.pokerev.com | ||
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| Re: My answer, stdioh, 12. May 2003 08:52 | ||
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| It is not 4 times as likely. | ||
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| Re: My answer, Player X, 12. May 2003 09:59 | ||
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| lol...duh.. This shows how unclear headed you can be when fatigued. Guess I quit playing poker at about the right time last night. -X Writer www.pokerev.com | ||
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| Re: Better than Money Hall, scythide, 12. May 2003 05:04 | ||
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| Let me take a shot at this mathematically. The chance that n will be a specific number is (1/2)^n, so the chances of n being 2, giving you $9 or $27 is 1/4th. When you see $9, n can basically be one of two values: 1 or 2. The chance n=1 is 1/2, the chance n=2 is 1/4. Since n has to be either 1 or 2 in this situation, the relative probabilities are 2/3 and 1/3. There is a 2/3rds chance that the $9 is actually the higher amount and a 1/3rd chance that $27 is the higher amount. 2/3rds of the time you switch, you get -$6, and 1/3rd of the time you switch you get +$18, summing these up you get +$2 if you switch. Generalizing, 2/3rds of the time you lose 3^n-3^(n-1), and 1/3rds of the time you gain 3^(n+1)-3^n. Putting these into an equation, we get: 1/3[3^(n+1)-3^n] - 2/3[3^n-3^(n-1)] then simplify by factoring [note: 3^n-3^(n-1) factors to 3^(n-1) * (3-1) = 2 * 3^(n-1)] This leaves us with the result of EV of switching = 2/3 * 3^(n-1) This is consistent with our example for getting $9 in the box, n in this case would be 2. Checking with n=3 (for getting $27), we see an EV of +$6 (2/3rds chance of losing $18, 1/3rd chance of gaining $54). Using this equation, we will always switch because EV is always positive, and it only gets more positive as n increases. I'm not sure if my math is right or not, but I'm assuming it is. | ||
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| Solution, stdioh, 12. May 2003 09:30 | ||
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| ok...sorry I took so long to post it. Here is the solution for the 3^(n+1) dollars in the box problem. Let us look at the strategy for always switching and the strategy for never switching unless we have 3. 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. Probability (size of n)for the size of the box we get is: 50% (1), 25%(2), 12.5%(3), 6.25%(4), 3.125%(5), 1.5625%(6)... Let us break this into two problems, one where we start with the small box and one where we start with the big box. Assume that we always start with the small box, we then have the following EV if we don't switch, except for when we have 3: (3*.0)+(9*.75)+(27*.125)+(81*.0625)+(243*.03125)+(729*.015625)+... Now add the cases where we start with the big box: (9*.5)+(27*.25)+(81*.125)+... and then divide by 2, we get: (9*5/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Now we will calculate the same infinite sequence for the strategy of always switching, starting with the small box: (3*.0)+(9*.5)+(27*.25)+(81*.125)+(243*.0625)+(729*.03125)+... Now add in for the big box: (3*.5)+(9*.25)+(27*.125)+(81*.0625)+... and then divide by 2, we get: (3*1/4)+(9*3/8)+(27*3/16)+(81*3/32)+(243*3/64)+(729*3/128)+... Now we subtract the two totals: (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)+...} and we get a result of: (-3/4) + (9/4) and we see that the rest of the sequences starting with (27*3/16) are equal infinite sequences, so we can cancel them. If we only switch when we have the small box, vs. always switching, we see that we have an advantage of 9/4-3/4=6/4=3/2=1.5. That is, this strategy has an expectation of making $1.50 more than the strategy of always switching and the lion's share of this is based on the fact that the majority of the time you will be getting small boxes. It is trivial to show that the strategy of switching only when you have the smallest box is superior to the strategy of never switching at all...which we can show is logically equivalent to the strategy of always switching. So we see that *very counterintuitively* the correct solution is the one most obvious: switch if your box has 3 (thus knowing that you will get 9) and don't switch if you box has more than three. And yes, the math doesn't seem to make sense, but it is all spelled out above. Where this changes is when there is an upper bound to the number of coin flips possible...that changes everything and cascades down, since then you would end up with a huge number multiplied by a tiny fraction at the end of both sequences and that would change things drastically. If there was a bound on n then the correct strategy would be to always switch unless you had the biggest possible box, but I will leave that proof as an escersise for anybody who cares to complete it. Hope that you all enjoyed this problem as much as I did. | ||
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| Re: Solution, stdioh, 13. May 2003 13:03 | ||
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| For anybody who hasn't looked at the remainder of the thread in "not quite poker" I've posted a general solution for this problem. Yes, the solution I've given shows that switching except for when you see 3 is better than always switching. The general solution shows that for any *FINITE* (and that is very important) number you chose, switching when you see that number or less will be better the higher the number, so long as it is finite. | ||
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