can you ask the same god more than one question?
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Conundrums, Puzzles, Logic Problems (1 Viewer)
- Thread starter fatguyinalittlecoat
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Anonymous Internet User
Footballguy
FragileFred
Footballguy
I would say yes based on how it is written. I copied it directly off this website: http://philosophy.hku.hk/think/logic/hardest.php
Aabye
Footballguy
Forcing the integers to be different just ensures that the answer must always be either "higher" or "lower". Using only integers just makes the solution easier to spell out. I think the logic is the same if you allow Al to choose any real numbers, but Bob's strategy is more complicated.And remember, Bob only has to increase his expectation above 50%. So if he can get an expectation of 50.00000001%, he has a winning strategy.
Anonymous Internet User
Footballguy
yeah, but a strategy usually involves interaction either by asking questions or by using previous knowledge. every trial here is independent of the others, and he cannot gather any information. he's only allowed to guess between two options of equal probability. is there something i'm missing here?
put it in spoiler text darnit
Anonymous Internet User
Footballguy
Aabye
Footballguy
I messed around with this problem for a long time and I think eventually figured it out. There's a FBG thread about it, and I think my solution is in there. I'll abstain, but be warned: it's a major pain in the ###.If people want a link to the thread with the (proposed) solution, I can try to dig it up and spoiler tag it.But that sort of ruins the fun of trying to solve the thing.
Mr. Know-It-All
Footballguy
The answer here is obvious. You guys are thinking waaaaaaay too hard. Set the ropes on fire and then use your stop watch to get the time right.bostonfred said:
i know. i want the answer in spoiler text!!!!
Aabye
Footballguy
Yeah, but the intuition is quite common. I mean, all you have is a random number. That doesn't tell you anything about the other number. So it doesn't look like there's anything to work with.But yes, there is a winning strategy. In fact, Bob can tell Al exactly what his strategy is and Al still can't prevent Bob from winning (although he can reduce Bob's advantage to a miniscule fraction of a percent).
Anonymous Internet User
Footballguy
If he did, They never would have felt a sense of relief, and they all would have died the next dayAre you sure the logician didn't say that there is at least 1 dragon with BLUE eyes?they all have red eyes. that was stated in the second or third sentence of the original problem. and yes, i've already conceded that my answer doesn't work. welcome.This doesn't make any sense at all.For starters, where was it stated that a dragon must have either red or blue eyes? Why couldn't his eyes be any other color?mytagid = Math.floor( Math.random() * 100 );document.write("actually, the dragons know that there are at least 98 other dragons on the island with red eyes. they can see them. look at it as if there are two dragons on the island. they both have red eyes. when the logician appears and says "at least one of you MFers has red eyes" each dragon assumes it's the other dragon that has red eyes and assume that they themselves have blue eyes. if it is true that one of those dragons had blue eyes, then the other dragon would see those blue eyes and know he has red eyes, then he'd be dead the next day...
but, both dragons have red eyes, and they can see the other has red eyes. therefore, when neither of them dies on the first day, it is impossible that either of the dragons have blue eyes, because neither of them died. so they both die.
now, just extrapolate that out with 99 dragons.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***
");document.close();
Aabye
Footballguy
Ask and ye shall receivemytagid = Math.floor( Math.random() * 100 );document.write("
Bob randomly generates an integer prior to choosing one of Al's hands - call that number N. He then adds 1/2 to that number - call that number N+1/2. If the number in the revealed hand is greater than N+1/2, Bob guesses that the concealed number is lower. If the number in the revealed hand is smaller than N+1/2, Bob guesses that the concealed number is higher.
This yields 3 possibilities:
1. Both numbers are higher than N+1/2 - in this case, Bob's expectation is 50%, since either number will mean that Bob says "lower" and he'll pick the lower number 50% of the time.
2. Both numbers are lower than N+1/2 - in this case, Bob's expectation is 50%, since either number will mean that Bob says "higher" and he'll pick the higher number 50% of the time.
3. N+1/2 lies between the revealed number and the concealed number.
In this case, Bob will guess "lower" when the revealed number is the higher number and "higher" when the revealed number is the lower number. He'll be right 100% of the time.
Now, N+1/2 will always have the potential of lying between the two integers that Al picks. This is the case even if Al picks two consecutive integers.
So the expectation will have to be better than 50%.
This is a neat little puzzle since it seems completely impossible to develop a winning strategy, but once the strategy is given, it is clearly a winning one.
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
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It didn't make any ####in' sense anyway...
FragileFred
Footballguy
mytagid = Math.floor( Math.random() * 100 );document.write("This solution raises an interesting question. Is it even possible to produce a random integer of any value? That is, you can produce a random integer between 1 and 6 by rolling a die. But how do you produce a random integer between negative infinity and infinity?
Assuming Bob can in fact do such a thing, does it actually help him? Let's say Al writes 11 and 12. Bob chooses a random integer, of which there are an infinite number. He has a one out of infinity chance of choosing 11 and thereby avoiding a coin flip scenario. So he has increased his chances by one divided by infinity. But what is one divided by infinity???*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***
");document.close();
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Aabye
Footballguy
For some reason I can't embed a 3rd spoiler tag, but this is in reference to Fragile Fred's post:
mytagid = Math.floor( Math.random() * 100 );document.write("
I'm not sure I understand the first question exactly. You can produce an integer of any arbitrary size, however. I think that's all you need. As long as Bob's random number generation is as good as Al's, he can win.
It is quite true that there are infinitely many choices. But the chance of choosing any one of those numbers is nonzero. By choosing consecutive integers, the incredible minuteness of the advantage that Bob can gain is evident. But the fact that 11 can be randomly generated, yields an expectation of 100%, and all other numbers yield an expectation of exactly 50% means that the total expectation will have to be above 50%.
I think the fact that Al has to generate two integers of some size, however arbitrary, means that Bob can generate a number that falls between them. And the fact that Bob can generate this number (even though it is extremely unlikely) is enough to give him an edge.
I'm by no means a math guy and my grasp of infinite sets is, um, limited, but I think this is enough to show that the expectation is >50% even though we can't put an actual number on it.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
mytagid = Math.floor( Math.random() * 100 );document.write("
I'm not sure I understand the first question exactly. You can produce an integer of any arbitrary size, however. I think that's all you need. As long as Bob's random number generation is as good as Al's, he can win.
It is quite true that there are infinitely many choices. But the chance of choosing any one of those numbers is nonzero. By choosing consecutive integers, the incredible minuteness of the advantage that Bob can gain is evident. But the fact that 11 can be randomly generated, yields an expectation of 100%, and all other numbers yield an expectation of exactly 50% means that the total expectation will have to be above 50%.
I think the fact that Al has to generate two integers of some size, however arbitrary, means that Bob can generate a number that falls between them. And the fact that Bob can generate this number (even though it is extremely unlikely) is enough to give him an edge.
I'm by no means a math guy and my grasp of infinite sets is, um, limited, but I think this is enough to show that the expectation is >50% even though we can't put an actual number on it.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
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I don't buy itmytagid = Math.floor( Math.random() * 100 );document.write("
scenario 3
let's call Al's number A1 and A2
Scenario 3 will happen this often:
|A1-A2|/infinity = 0
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
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Aabye
Footballguy
mytagid = Math.floor( Math.random() * 100 );document.write("If that were the case, then it would be impossible for Bob to generate a random number that fell between A1 and A2.
But that's clearly wrong. Here's a counter-example:
A1 = 0
A2 = 2
Bob's random number = 1
EDIT TO ADD:
A different way to look at it:
Let's start by looking Bob's N+1/2, then try to figure out whether it is possible for Al to choose two numbers that fall on either side of it (one larger and one smaller). This clearly is possible, so it can't have probability 0.
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***
");document.close();
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mytagid = Math.floor( Math.random() * 100 );document.write("
i am not seeing the solution as being an actual solution. you are using backwards thinking. if you pick 100.5 and the first number chosen is 10000 you will say lower. What are your odds that the next number will be lower than 10,000? 50/50.
if you pick 1000000.5 and the first number chosen is 2 you will say higher. What are the odds that the number will be higher than 2? 50/50
the fact that looking backwards at it when that number happens to encapsulate your number your guess was always right is not important. In fact you could pick your number after seeing the first number and the backwards analysis would be the same.
for example in the right hand is 723000. You then say 100 and then say lower. If the next number is lower than 100 when you chart your stats you will have been 100% correct here.
the number you pick first is irrelevant. you are just assigning a number value to whether or not u say higher or lower.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
i am not seeing the solution as being an actual solution. you are using backwards thinking. if you pick 100.5 and the first number chosen is 10000 you will say lower. What are your odds that the next number will be lower than 10,000? 50/50.
if you pick 1000000.5 and the first number chosen is 2 you will say higher. What are the odds that the number will be higher than 2? 50/50
the fact that looking backwards at it when that number happens to encapsulate your number your guess was always right is not important. In fact you could pick your number after seeing the first number and the backwards analysis would be the same.
for example in the right hand is 723000. You then say 100 and then say lower. If the next number is lower than 100 when you chart your stats you will have been 100% correct here.
the number you pick first is irrelevant. you are just assigning a number value to whether or not u say higher or lower.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
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Aabye
Footballguy
mytagid = Math.floor( Math.random() * 100 );document.write("The numbers chosen are random. But start by considering what numbers Al could have. It should be clear that Bob can generate a number that falls between them with some probability > 0.
Now, any time Bob generates such a number, he is guaranteed to win.
Just pick two random numbers for Al.
Make it simple and say 100 and 500. There is a nonzero probability that Bob generates a number such that 100<N+1/2<500. In this case, it doesn't matter which of Al's numbers he picks. If he picks 100, N+1/2>100 so Bob says "higher". If he picks 500, N+1/2<500 so Bob says "lower". He's guaranteed to get it right when he randomly hits a number that falls between the two numbers generated by Al. And when he doesn't randomly hit between Al's numbers, his expectation is still exactly 50%.
So, in a sense, it's all luck. Bob has to keep playing until he gets lucky and generates a number that falls between Al's two numbers. But when this does happen, Bob always wins.
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***
");document.close();
Aabye
Footballguy
One thing people seem to be sticking on is that Bob has a 0% chance of hitting a "magic" number. I think this should show that it's not the case:
mytagid = Math.floor( Math.random() * 100 );document.write("
Imagine a lottery in which a randomly generated number of lottery machines (those things with the balls labeled w/ digits 0-9 and the air blowing the balls around) are placed in order and random digits are drawn from each machine to form one long number.
What is the probability that any particular number is drawn?
It can't be 0%, since that would mean that every possible natural number has probability 0.
But clearly some number has to be drawn, so the probability of any particular number can't be 0.
The same reasoning applies to Bob's random number generation. It can lie between Al's numbers, so the probability of generating that number can't be zero.
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
mytagid = Math.floor( Math.random() * 100 );document.write("
Imagine a lottery in which a randomly generated number of lottery machines (those things with the balls labeled w/ digits 0-9 and the air blowing the balls around) are placed in order and random digits are drawn from each machine to form one long number.
What is the probability that any particular number is drawn?
It can't be 0%, since that would mean that every possible natural number has probability 0.
But clearly some number has to be drawn, so the probability of any particular number can't be 0.
The same reasoning applies to Bob's random number generation. It can lie between Al's numbers, so the probability of generating that number can't be zero.
*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();
FragileFred
Footballguy
Okay, so after some quick internet review of calculus, I have been reminded that 1/infinity only makes sense in terms of limits. The limit of 1/x as x approaches infinity = 0. I think this brings us back to my first question. If Bob could truly choose a random integer from an infinite number of integers, his strategy would yield a 0% increase over 50%. But that doesn't really make sense. The reason it doesn't, I think, is that Bob can only "randomly" choose an integer from a finite number of integers. His percentage increase over 50% approaches 0 as the number of integers he is choosing from increases, but it will never be zero. So his strategy works, ASSUMING Al's two integers fall in the range of integers Bob chooses from. But there in lies the problem. They could fall outside Bob's range, no matter how large it is.
Aabye
Footballguy
That sounds right. I think I said earlier that Bob should be able to match Al's ability to randomly generate integers. If not, then Al can make his integers fall outside of Bob's range. But if we give both an equal ability to generate random integers, which seems most reasonable since neither is constrained by limits like time, computational ability, or paper to write on, then Al's range won't be larger than Bob's range, so there shouldn't be a problem.
according to this math I could say that the times when both numbers are below the n+.5 he will be wrong more than 50% of the time.
for example if bob picks 10.5 and the next number picked is -1000 bob will say higher. well there are only 989.5 numbers that could come up that will make him right every other number below -1000 will make him wrong, if both numbers are lower.
understand?
for example if bob picks 10.5 and the next number picked is -1000 bob will say higher. well there are only 989.5 numbers that could come up that will make him right every other number below -1000 will make him wrong, if both numbers are lower.
understand?
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FragileFred
Footballguy
that would make him right.not that would make him right if both numbers are below N+.5and my math was horrible, lol 1010 not 989.5
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Aabye
Footballguy
No, I don't see why this would be entailed.As far as infinite series go, (the set of integers > 100) = (the set of integers > 200).I'm assuming that you're not arguing that this is incorrect, but that my solution somehow implies that (the set of integers > 100) > (the set of integers > 200), which is false. I don't know why it would imply that but if it does, that's very bad news for me.EDIT: I think I misread your criticism.
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your statement before states that he will win 50% of the time that both numbers are below N+.5this is in fact untrue.if n+.5 is 1.5 and the first number picked is -10000 you would say "higher". Well if both numbers are under n+.5 you are going to lose this guess way more than you win it.there are only 10001 ways to win and still have both numbers below n+.5. there are an infinite amount of ways to lose
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taking the opposite scenario if N+.5 is 1.5 and the first number is 10,000,000,0000 once again there are way more ways to lose here than win. ten trillion is a big number, but it is not infinity
edited to add I mean if both numbers are above n+.5
edited to add I mean if both numbers are above n+.5
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Aabye
Footballguy
Ah, I think I see.Look at it this way. Suppose n+5 is 1.5 and both of Al's numbers are lower. That means that Bob will say "higher", regardless of which number he picks.What is the probability that Bob picks the lower of Al's two numbers and (therefore) guesses correctly? Exactly 50%. He's just picking the right or left hand, either of which has a 50% chance of containing the lower number.
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it is a very fun discussion, but do you see now why looking backwards at it is very misleading?
FragileFred
Footballguy
I see what you are saying now, but I still think there is something wrong with it. It clashes with the following way of looking at the situation:There are two numbers less than n+.5 in Al's hands, one of which is less than the other. Surely the chances of the lower one being uncovered first is exactly 50%, which would make Bob 50% likely to be correct when he predicts the second one will be higher.
FragileFred
Footballguy
Aabye
Footballguy
I don't see how this is looking backwards.If Bob's N+1/2 is above both of Al's numbers, then Bob always says "higher". So it doesn't matter which number he chooses. He'll randomly choose the higher number 50% of the time and the lower number 50% of the time. It's a total wash.Exactly the same argument applies when Bob's N+1/2 is below both of Al's numbers. He'll always say "lower" and again he'll be right when he randomly chooses the higher number (50% of the time) and wrong when he chooses the lower number (50% of the time).But any time Bob's number falls between Al's higher and lower numbers, Bob always guesses correctly.It has nothing to do with looking back after you know the numbers. Those scenarios are just an exhaustive list of the three possibilities. The first two are exactly random, since Bob's guess will be the same regardless of which number he picks and he'll pick the higher/lower number with identical probability.Bob gets an edge because sometimes he randomly generates an N+1/2 that falls between Al's numbers and he guesses with 100% accuracy when that happens. The probability of generating an N+1/2 that falls between Al's numbers is nonzero, so Bob has to get a mathematical edge.
FragileFred
Footballguy
I think ELM is right about the backwards thing. There are two numbers in Al's hands, one low and one high. There is exactly a 50% chance of either being uncovered first and therefore exactly a 50% chance Bob will be correct when he guesses higher or lower. Surely, the act of Bob "thinking of a number" before choosing a hand does not affect those percentages.
ok one more time here before I go to bed.The 100% area you speak of is false. It is a made up figure. In fact the more I read this scenario it is fairly ridiculous as you predict that Bob could gain a HUGE edge. You predict that he should win almost 67% of the time if Al was truly randomly picking integers.I really dont know how else to explain it so I will offer you a bet. We can play this game and you can pick your N+.5 and then I will pick my two numbers. We will do it 200 times. For every time that both of my numbers were greater than yours and you guessed correctly I will give you a dollar. Every time you guessed wrong you will give me one dollar. Every time both of my numbers were above yours and you guessed right I will give you one dollar and everytime you guessed wrong you will give me one dollar. Every time one of my numbers is lower and one of my numbers is higher and you guessed right I will give you one dime. I concede that you will not guess wrong here so you will not need to give me any dimes.But according to your theory you should spilt the first two scenarios with me and that should be a wash. Then you will make a dime everytime the third scenario comes up.If this is not enough money to justify your time we can happily up the stakes as long as the ratios stay the same. Deal?
It took awile to "get" it, but th best way to explain this is to not try to explain ot organicallyhere goesIf there's 1 dragon he dies on day 1If 2 dragns then each is thinking "if I have blue eyes, that other poor sap is dead" When the other dragon lives they both know they're going to be dead on day 2if 3 dragons then each is thinking" if I have blue eyes then those other two poor saps are going to go through the two day process outlined above and be dead on day 2" When they survive all of the dragons know they're going to be dead on day 3if 4 dragons then each is thinking" if I have blue eyes then those other 3 poor saps are going to go through the3 day process outlined above and be dead on day 3" When they survive all of the dragons know they're going to be dead on day 4and so onI'll try to demonstrate the extra information given by the logician for 4 dragons :A knows the following :B,C,D have red eyesB,C,D know at least 1 has red eyesB,C,D know at least 2 have red eyesnow A does NOT know that B knows he(A) has red eyes, so A has the following information regarding B's knowledge:A knows that B knows :C,D have red eyesC,D know at least 1 has red eyessimilarly, going down the chainA knows that B knows that C knowsI think this breaks down when applied to 4 or more dragons the way the puzzle is worded. Every dragon sees 98 other dragons with red eyes. His eyes could be red or not red once you got to 4 or more dragons, because it wouldn't matter what he assumed about his own eyes, as he will always know that their is at least one set of red eys, as the logician said.
has red eyesAfter the revelation, the extra information is that :A knows that B knows that C knows THAT D KNOWS AT LEAST 1 HAS RED EYESAfter re-reading this, I don't think this will help as it becomes way too abstract. The new knowledge obtained is not about what each dragon know by itself, but what dragons know about the knowledge of other dragons. This is a nasty problem.Majorum
Aabye
Footballguy
That's what makes the solution so counter-intuitive. It just doesn't look like any strategy should ever be better (or worse) than 50/50.There is a 50% chance that either number is picked.When both numbers are higher than Bob's N+1/2, he always says "lower". So there is a 50% chance that he's right.When both numbers are lower than Bob's N+1/2, he always says "higher". So there is a 50% chance that he's right.Both of these cases yield an expectation of exactly 50%. It is the same scenario as if Bob just announces "higher" or "lower" before he picks Al's first number.But there is a nonzero probability that Bob's number falls between Al's numbers, regardless of the numbers that Al picks. The actual probability can't be given as a number - it's vanishingly minute but it can't be zero.So you have the following probability distribution:Any time Bob's number is higher or lower than both of Al's numbers, his expectation is exactly 50%. But there is a nonzero probability that Bob's number falls between Al's numbers and the expectation there is 100%, so the overall expectation approaches, but cannot be as low as, 50% overall. The nonzero probability of choosing a number between Al's guarantees that.The only point of generating the random number that falls between the integers is that Bob creates the possibility of of getting between Al's numbers.I guess another way to say it is that by employing this strategy, Bob's expectation can be seen as approaching 50% from the high end. But this limit can never be reached, since to do so would mean that the probability of Bob's choosing a number falling between Al's numbers had dropped to zero, which is strictly impossible.So it has to be a higher expectation than 50%.FragileFred said:
Aabye
Footballguy
Bob can win easily if Al is randomly picking integers.OK, imagine that three different random numbers are written on three pieces of paper and the pieces of paper are folded over. Al then randomly picks two pieces and Bob get the third. Bob reveals his number, then picks one of Al's numbers to reveal. He uses the same strategy outlined above for generating his guess. The only difference is that the number Bob gets is randomly picked for him.ELM said:
Look at the 3 possibilities:
Bob will randomly pick the highest number 1/3 of the time. When he does this, he'll guess "lower" when he sees either of the numbers in Al's hands. He'll be right exactly 50% of the time.
Bob will randomly pick the lowest number 1/3 of the time. When he does this, he'll guess "higher" when he sees either of the numbers in Al's hands. He'll be right exactly 50% of the time.
Bob will randomly pick the middle number 1/3 of the time. When he does this, he'll guess "higher" when he happens to pick Al's lower number and "lower" when he happens to pick Al's higher number. He'll be right exactly 100% of the time.
So if this is the game, it's clear that Bob has a big advantage. So far I think we're in agreement. Bob should play this game.
I'd be more than happy to play this game with you, using the dollars and dimes. I certainly have the edge.
But Al doesn't have to generate two random numbers. He can just generate one random number, then add 1 to it. This makes Bob's edge very small. Imagine that Al tells Bob beforehand that he's going to choose two positive two-digit numbers.
So we'll allow Al's numbers to be anything between 00 and 99 (so 100 possibilities).
Say Al picks 46. He then adds 1 to get his second number - 47. Now Bob generates a random number between 00 and 99. 99% of the time he'll pick some number other than 46. In that case, Bob's N+1/2 will be either higher than both of Al's numbers (between .5 and 45.5) or lower than both of Al's numbers (between 47.5 and 99.5) and he'll have exactly a 50% expectation. But 1% of the time, Bob will randomly pick 46, then his N+1/2 will be higher than Al's lower number and lower than Al's higher number. So he'll win 100% of the time here. So even using these very restricted conditions, Al can reduce Bob's expectation from 66.66666---% to 50.5%.
The point is that if Al knows Bob's strategy, he can reduce Bob's edge to a miniscule amount.
Since you know my strategy and can plan against it, you can reduce my edge to a paltry 50.00000000001% (actually much less than this).
If you want to play the first game (with three random numbers), then I'd certainly accept. I have a big mathematical edge.
If you want to play a game in which you get to pick your own random numbers, my edge reduces to a pathetic amount and there's no good reason to play the game. You could offer me a million dollars when my number fell between yours and it wouldn't be worth it to play, since you could just inform me that you'll be randomly generating a 100 digit integer, adding 1 to it to get your 2nd number, and I can play from now until I die and have a vanishingly small chance of ever collecting the money.
But yeah, if you wanted to play the first game with 3 randomly generated numbers I'd take that action.
sartre
Footballguy
FragileFred said:
This makes great sense. If there is no limit, the solution doesn't work. If you say, Al picks any two integers from -100 trillion to +100 trillion, then it bears some validity.But to say ANY two integers, well, Bob's strategy couldn't improve over 50/50.Let's say Al DOES use a random number generator. Bob could simply say (as someone else suggested) "if the first number is negative, I will guess higher, if the first number is non-negative, I will guess lower" and achieve the same advantage - which again is proportional to the size of the integer pool that Al is using. (The point of Bob generating a third random number is only to foil Al from using this against him if Al generates his numbers non-randomly.)
sartre
Footballguy
That's the way you framed the original riddle though. Do you see how if you say, Al picks two random integers from -1000 to 1000, it changes everything?As long as you put a limit on the range ELM can choose from, sure you'll have an edge. But if you say ANY integer, then that edge becomes zero.
Aabye
Footballguy
If Bob's edge is to become zero, the probability that Bob could generate a number that falls between Al's numbers must hit zero. But since Bob has a process that can always generate a number that falls between Al's integers (Bob can use decimal numbers but Al cannot), the probability cannot fall to strictly zero. The chance of Bob's number falling between Al's numbers is not, and cannot be, zero. So Bob's edge cannot be reduced to zero. Bob's edge approaches zero as a limit, but it cannot hit zero.It's easy to see this in the case of bounded ranges where Al has finite options. But even given infinite possibilities, Al cannot guarantee that Bob's number will not fall between his numbers, so he cannot reduce Bob's edge to zero.
******* spoilers below ********
when it comes to arguments about infinity. logic breaks down
It simply is not logical that Bob could come up witha strategy that will yeild him better than 50%
Infinity - 10 trillion is still infinity.
If you don't buy that, then lets's calculate probabilities
there are 4 scenarios
Both Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%
One is high and one is lower bob will Win always - 50%
Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%
Bob will win 75% of the time right
I'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.
********************************
wrap your head around that one...
when it comes to arguments about infinity. logic breaks down
It simply is not logical that Bob could come up witha strategy that will yeild him better than 50%
Infinity - 10 trillion is still infinity.
If you don't buy that, then lets's calculate probabilities
there are 4 scenarios
Both Al's numbers are greater than Bob's n+.5 = 25 percent of the time - Bob will guess lower and Win half of the time - 12.5%
One is high and one is lower bob will Win always - 50%
Both Al's numbers are less than Bob's n+.5 = 25 percent of the time - Bob will guess higher and Win half of the time = 25 percent of the time - win half 12.5%
Bob will win 75% of the time right
I'm sure somebody can use this to prove that .75 = .5 at which point all Rational Mathematics breaks down.
********************************
wrap your head around that one...
Aabye
Footballguy
If the potential pool is all of the integers then for any given integer, the set of number of integers larger than the given integer is infinite (strictly, this set is the size of the natural numbers) and the set of integers larger than the given integer is also infinite (strictly, this set is also the size of the natural numbers).Al doesn't have a finite integer pool. He's not limited on either side. In addition, Al can simply ensure that when he chooses his second integer, it is larger than zero if the original integer is larger than zero and smaller than zero if the original integer is smaller than zero. That would defeat this potential strategy for Bob.
Aabye
Footballguy
Logic doesn't break down when we start talking about infinite sets.Common sense breaks down.I certainly agree that you can take any finite number of elements out of an infinite set and the infinite set won't become finite. You can't thin an infinite set.If Al and Bob choose 3 different integers comletely at random, Bob will get the highest number 1/3 of the time, the middle number 1/3 of the time, and the lowest number 1/3 of the time - it's not 25%, 50%, and 25%. Finally, none of the solution relies on funky math. It's flies in the face of common sense, but given that we're talking about infinite sets, the fact that the solution doesn't pass the smell test doesn't mean anything. All sorts of facts about infinite sets fly in the face of common sense (e.g. that some infinite sets are strictly larger than other infinite sets - proved by Cantor a while ago).As I said, if Bob's edge is supposed to reduce to zero, then you'd have to show that his probability of generating a number that falls between Al's numbers has to hit zero. If someone could show that, the solution would be invalidated.I guess what I'm asking is, is this reasoning incorrect? If not, what's the problem with the solution?
