A countable infinity of prisoners are placed in a room so that they can all see each other, but are not allowed to communicate in any way and cannot see their own heads. The warden places on the head of each prisoner a red hat or a black hat. The prisoners will each guess the color of their own hat. They will all be released if at most finitely many of them guess incorrectly, and they will all be killed otherwise. The prisoners know all of this, and may collude beforehand. The prisoners are all distinguishable—think of them as being numbered 1,2,3,.… Again, once the warden has placed the hats, the prisoners receive no information other than the color of their fellow prisoners’ hats. Prove that there is a strategy that guarantees a win for the prisoners.
Pbafvqre gur sbyybjvat eryngvba orgjrra vasvavgr frdhraprf bs pbybhef: “qvssrerag va bayl svavgryl znal cynprf”. Guvf vf na rdhvinyrapr eryngvba, naq jura gur ungf ner cynprq, nyy cevfbaref xabj juvpu rdhvinyrapr pynff gurl ner va. Guvf pynff vf pbhagnoyr, naq gur cevfbaref unir nterrq orsberunaq, gunaxf gb gur nkvbz bs pubvpr, ba n cnegvphyne rahzrengvba sbe rnpu rdhvinyrapr pynff.
Sbe rnpu cevfbare, rknpgyl gjb zrzoref bs gur pynff ner pbafvfgrag jvgu jung ur frrf. Ur thrffrf juvpurire nygreangvir pbzrf svefg va gur rahzrengvba. Bayl svavgryl znal bs gurz pna or jebat, orpnhfr gur pbzcyrgr frdhraprf vzcyvrq ol gurve jebat nafjref ner nyy qvssrerag, naq gur gehr frdhrapr unf bayl svavgryl znal naprfgbef va gur rahzrengvba.
Gur fnzr cebbs nccyvrf gb nal pbhagnoyr ahzore bs pbybhef. Vg’f abg pyrne gb zr lrg ubj zhpu ynetre gur ahzore bs cevfbaref naq gur ahzore bs pbybhef pna or. V fhfcrpg gung Bfpne_Phaavatunz wrfgf nobhg rkgraqvat gb erny-inyhrq pbybhef.
This was my solution, and it does work for arbitrary cardinalities of colors and prisoners, as long as you’re okay with the prisoners remembering arbitrary amounts of information. :)
Even harder version, to which this problem was a hint (and which I haven’t solved yet, so please continue to ROT13 solutions):
There are countably many boxes 1,2,3,..., into each of which Alice places an arbitrary real number. Bob then opens finitely many boxes, looking at the real numbers they contain as he goes, and then names a single real number and opens a single unopened box. Bob wins if that box contains the number he named. Bob may condition his choice of boxes to open on what numbers he has already seen, and at each time step, he may choose the next box to open by random choice out of finitely many boxes that he identifies at that time step. Show that Bob has a strategy such that no matter how Alice chooses her real numbers, Bob wins—correctly predicts a real number—with very high probability.
So, to repeat, it can indeed be proved that there is a rule—the “Hardin-Taylor rule”, I shall call it — that will, for any arbitrarily chosen function f, correctly predict most values of f on the basis of its past be- havior; that is, for most t the rule will correctly predict f(t) on the basis of f’s values at all s< t.
Oh dear, I suppose that rules out other “cheats” then: such as prisoner n guessing after n seconds. At any point in time, only finitely many have guessed, so only finitely many have guessed wrong. Hence the prisoners can never be executed. (Though they can never be released either.)
OK, I also got a “non-cheat” solution: unfortunately, it is non-constructive and uses the Nkvbz bs Pubvpr, so it still feels like a bit of a cheat. Is there a solution which doesn’t rely on that (or is it possible to show there is no solution in such a case?)
The warden places on the head of each prisoner a red hat or a black hat.
Heh. This is the easy version of the puzzle then. Hard version: The hats have arbitrary colours (suppose they are specified by infinite precision RGB values).
The participants have to base their decision on a non-measurable “events”, right? The guessing procedure ends up as contorted as the slices in Tarski and Banach’s sphere?
Do the prisoners all have to guess simultaneously? If they guess one by one, and hear their predecessors’ guesses, is it in the order 1, 2, 3..., or can guessing order be a part of their prearranged strategy?
Still confused. Richard Kenneway’s solution relies on the true sequence being at a finite place in some ordering. Doesn’t Cantor’s diagonal argument prevent you from having a countable ordering of all the sequences?
Doesn’t Cantor’s diagonal argument prevent you from having a countable ordering of all the sequences?
There’s no need for an enumeration of all the sequences, bayl na rahzrengvba bs gur rdhvinyrapr pynff gung gur cevfbaref frr gung gurl’er va jura gur ungf ner cynprq. Naq Bfpne_Phaavatunz’f fbyhgvba qbrfa’g arrq rira gung—gur rdhvinyrapr pynffrf pna or nal genafsvavgr fvmr jungrire.
I’m now wondering whether for the case of two colours, there is a computable algorithm. A prisoner would apply the algorithm by feeding it an infinite tape listing all the colours of the hats, with a blank for his own, and the algorithm would in a finite time say what guess to make.
It seems unlikely. In a finite time it’s impossible to get any idea whatsoever what equivalence class they’re in, so the solution, if there is one, would need to be very different.
I’m now wondering whether for the case of two colours, there is a computable algorithm. A prisoner would apply the algorithm by feeding it an infinite tape listing all the colours of the hats, with a blank for his own, and the algorithm would in a finite time say what guess to make.
Idea for a proof: we could assume the warden chooses colours randomly for each prisoner, iid with probability a half. Then there might be a probabilistic proof that the puzzle is impossible. This proof would fail to rule out the true solution because gur frgf vaibyirq jbhyqa’g or zrnfhenoyr va gung pnfr, ohg gurl jbhyq or zrnfhenoyr va rirel pbzchgnoyr pnfr.
Proof that there is no sequences of algorithms A1, A2, …, assigned to each prisoner, giving a winning strategy (assuming a computable warden given indices for the Ak):
Gur jneqra fvzhyngrf N-bar ba mreb mreb mreb… hagvy vg bhgchgf bar be mreb nsgre ernqvat x ovgf bs gur vachg. Gur jneqra gura cynprf gur bgure pbybe ung ba cevfbare 1, jub jvyy sbyybj N-bar naq thrff vapbeerpgyl; gur jneqra rafherf guvf ol cynpvat mreb ba cevfbaref gjb guebhtu x. Gur jneqra ercrngf guvf jvgu N-x+1 naq cevfbare x+1, fb gung x+1 jvyy thrff vapbeerpgyl. Naq fb ba. Fvapr rnpu Nx unygf nsgre ernqvat svavgryl znal ovgf sebz vgf benpyr, gur jneqra pna sbepr na vapbeerpg nafjre jvgu bayl svavgryl znal ovgf. Guvf jnl, gur jneqra pna sbepr vasvavgryl znal vapbeerpg nafjref. (Guvf eryngvivmrf gb nal benpyrf lbh pner gb tvir gb gur cevfbaref, nf ybat nf gur jneqra unf npprff gb gur pbhagnoyr wbva bs gubfr benpyrf.)
Do the prisoners guess all at the same time, or in order (and thus after hearing finitely many other guesses)? If the second, is the guessing order known beforehand, or will it be at the whim of the guards?
With guessing in order, I observe that for every finite subset there are either an even or an odd number of red hats, and prisoner 1 can indicate which it is by his guess; then everyone in that subset can count the red hats and figure out which colour his own must be to make the total number even or odd. Let the size of the finite subset go to infinity.
This is a good idea, and solves the similar problem with finitely many prisoners getting at most one guess incorrect. But...
Let the size of the finite subset go to infinity.
I don’t see how this immediately gives a solution; any finite set of hats has either an even or an odd number of red hats, but an infinite set of hats may have an infinite number of red hats and an infinite number of black hats, and infinities are neither odd nor even.
Either this problem is not completely stated, there are important things left out, or the answers are gibberish. In any case, I have no idea what is going on even after reading the answers. I can collude with prisoners ahead of time but I can’t communicate with them. This presumably means I can’t smile or wink or grab or stare meaningfully or longingly at them. I can’t tell them what my guess is.
So I find myself in a room full of people, some with red hats and some with black hats, in what is essentially a still picture, for all intents and purposes equivalent to freezing the room full of other people with their red and black hats and I can then examine this fixed scene. How does this even conceivably impact what I would guess the color is of the hat on my head?
There are important things left out. Specifically, this is a math problem and putting it in terms of prisoners or humans is intended to be ignored. You should of course just know that when someone is a human in a math puzzle, what they really mean is a hypercomputer or oracle that can return correct answers for uncomputable algorithms. Here’s another example:
There is a prisoner in a cell, the same as the prisoners in the question above. Each day the warden hands him an infinitely complex algorithm, and he is released if and only if he can correctly tell the warden whether it halts or not. On what day is he released?
The problem is correct as stated, and solutions above by RichardKennaway and Oscar_Cunningham are correct. I think you may have missed that the prisoners are all distinguishable, a.k.a. they are numbered 1,2,3,.… Or you are confused about the win condition; we don’t have to guarantee that any particular prisoner guesses correctly, just that only finitely many guess incorrectly.
Sub-puzzle: prove definitively that if the prisoners are not distinguishable, then there is no winning strategy.
In the winning strategy, do fewer than half the prisoners guess wrong? Do more prisoners guess correctly than incorrectly? I’m trying to get a handle on whether it is worth my while to try to penetrate the jargon in the “correct solutions.”
What do you mean by half the prisoners? Let’s start there.
How about I choose a prisoner at random from among all the prisoners in the problem. What is the probability that the prisoner I have chosen has correctly stated the color of the hat on his head? In particular, is that probability more than, less than, or equal to 0.5?
While we are in the neighborhood, if there is a prisoner who is more likely to get the answer correctly than not, if you could tell me what ihis step by step process of forming his answer is, in detail similar to “if he is prisoner n, he guesses his hat color is the opposite of that of prisoner n^2+1” or some such recipe that a Turing machine or a non-mathematician human could follow.
How about I choose a prisoner at random from among all the prisoners in the problem. What is the probability that the prisoner I have chosen has correctly stated the color of the hat on his head?
So what do you mean to choose a prisoner at random when one has infinitely many prisoners?
Whatever mwengler’s answer is, your answer is going to have to be “in that case the set you asked about isn’t measurable, and so I can’t assign it a probability”.
Maybe this way then. I set down somewhere in the universe in a location that I don’t reveal to you ahead of time and then identify the 100 prisoners that are closest to my position. If the 100th and 101st furthest prisoners from me are exactly equally distant form me I set down somewhere else in the universe and I keep moving until I find a location where I can identify the 100 closest prisoners to my current position.
Of that 100 prisoners, I count the number of prisoners who identified their hat color correctly.
My question is what is the probability that I have counted 50 or fewer correct answers? Is it greater than, less than, or equal to the probability that I have counted 51 or more correct answers?
Thanks to you and JoshuaZ for trying to help me here.
Maybe this way then. I set down somewhere in the universe in a location that I don’t reveal to you ahead of time and then identify the 100 prisoners that are closest to my position. If the 100th and 101st furthest prisoners from me are exactly equally distant form me I set down somewhere else in the universe and I keep moving until I find a location where I can identify the 100 closest prisoners to my current position.
So how have you set up the prisoners in the universe in advance and how do you decide on the location you set down?
Is it safe to say that this problem, this result, has no applicability to any similar problem involving a merely finite amount of prisoners, say a mere googol of them?
Yes. But I do think that thinking critically about the assumptions you are making, in particular that you can meaningfully talk about what it means to pick a random individual in a uniform fashion, is worthwhile for understanding a fair bit of probability and related issues which are relevant in broader in contexts.
As to critically understanding what it means to pick a random individual in a uniform fashion, yes it is worth understanding what one means, as a tremendous amount of mischief is done in the name of randomness. With a finite number of prisoners, I would actually not need to pick prisoners randomly in order to gather statistics. If I would simply count up all the prisoners who got the hat color correct and all that got it wrong, and I do the experiment say 100 times and plot a histogram of the results and then decide whether the results deviate from a normal distribution with mean 50% by enough to make it practically interesting.
But since you say this result has no applicability to a finite population of prisoners, I am assuming there is nothing here that would push the result away from 50:50?
Is the mathematician’s world really so insular,that someone from outside asking some questions about how a problem relates to concepts he understands gets downvoted?
Or are you pretending that unless you skate with ease over three different kinds of infinities, their differences and similarities, and the paradoxical results of probability problems with infinity in the numerator and denominator, that you are just a time wasting intruder on an otherwise valuable conversation?
Or did my questions, which I’d love to know the answer to, come across as a veiled negative comment?
Your questions were easily answered by looking up the definitions of the terms “finitely many” and “countably infinite”.
Are you aware that having a mathematician tell you a question is easily answered without actually answering it is actually the punch line to a joke? Closest I can find on the web is the 2nd one on this page.
You asked why you were downvoted. I told you why; you asked a question that showed you hadn’t made even a cursory attempt to understand the terms in the question.
The answers, in case you still haven’t put in the minimal effort required, are
Yes, a finite portion of an infinite set is infinitely less than half.
Yes, all but a finite number of an infinite set is infinitely more than a finite number.
This is not fancy jargon. These are terms anyone who has taken highschool calculus would know.
Puzzle:
A countable infinity of prisoners are placed in a room so that they can all see each other, but are not allowed to communicate in any way and cannot see their own heads. The warden places on the head of each prisoner a red hat or a black hat. The prisoners will each guess the color of their own hat. They will all be released if at most finitely many of them guess incorrectly, and they will all be killed otherwise. The prisoners know all of this, and may collude beforehand. The prisoners are all distinguishable—think of them as being numbered 1,2,3,.… Again, once the warden has placed the hats, the prisoners receive no information other than the color of their fellow prisoners’ hats. Prove that there is a strategy that guarantees a win for the prisoners.
(On my honor, this is possible.)
Pbafvqre gur sbyybjvat eryngvba orgjrra vasvavgr frdhraprf bs pbybhef: “qvssrerag va bayl svavgryl znal cynprf”. Guvf vf na rdhvinyrapr eryngvba, naq jura gur ungf ner cynprq, nyy cevfbaref xabj juvpu rdhvinyrapr pynff gurl ner va. Guvf pynff vf pbhagnoyr, naq gur cevfbaref unir nterrq orsberunaq, gunaxf gb gur nkvbz bs pubvpr, ba n cnegvphyne rahzrengvba sbe rnpu rdhvinyrapr pynff.
Sbe rnpu cevfbare, rknpgyl gjb zrzoref bs gur pynff ner pbafvfgrag jvgu jung ur frrf. Ur thrffrf juvpurire nygreangvir pbzrf svefg va gur rahzrengvba. Bayl svavgryl znal bs gurz pna or jebat, orpnhfr gur pbzcyrgr frdhraprf vzcyvrq ol gurve jebat nafjref ner nyy qvssrerag, naq gur gehr frdhrapr unf bayl svavgryl znal naprfgbef va gur rahzrengvba.
Gur fnzr cebbs nccyvrf gb nal pbhagnoyr ahzore bs pbybhef. Vg’f abg pyrne gb zr lrg ubj zhpu ynetre gur ahzore bs cevfbaref naq gur ahzore bs pbybhef pna or. V fhfcrpg gung Bfpne_Phaavatunz wrfgf nobhg rkgraqvat gb erny-inyhrq pbybhef.
Correct! But it can be simplified (note that this is also a spoiler for the hard version):
Sbe rnpu rdhvinyrapr pynff cvpx n ercerfragngvir zrzore. Gura unir rirelbar thrff nf vs gurl jrer va gung frdhrapr. V guvax guvf jbexf sbe neovgenel pneqvanyvgvrf bs cevfbaref naq pbybhef.
EDIT: I remembered to ROT13 it.
This was my solution, and it does work for arbitrary cardinalities of colors and prisoners, as long as you’re okay with the prisoners remembering arbitrary amounts of information. :)
Even harder version, to which this problem was a hint (and which I haven’t solved yet, so please continue to ROT13 solutions):
There are countably many boxes 1,2,3,..., into each of which Alice places an arbitrary real number. Bob then opens finitely many boxes, looking at the real numbers they contain as he goes, and then names a single real number and opens a single unopened box. Bob wins if that box contains the number he named. Bob may condition his choice of boxes to open on what numbers he has already seen, and at each time step, he may choose the next box to open by random choice out of finitely many boxes that he identifies at that time step. Show that Bob has a strategy such that no matter how Alice chooses her real numbers, Bob wins—correctly predicts a real number—with very high probability.
I don’t believe you.
This sounds related to this “proof of induction” by Alexander George. Sample quote:
Okay. I’m Alice. I placed random numbers into all boxes.
Your turn.
Yes! How simple!
I suspect an April Fool:
Cevfbare a+1 gnxrf gur ung sebz cevfbare a naq chgf vg ba uvf bja urnq. Gura nyy cevfbaref (ncneg sebz cevfbare 1) thrff gur pbybe pbeerpgyl!
No April Fool here.
Oh dear, I suppose that rules out other “cheats” then: such as prisoner n guessing after n seconds. At any point in time, only finitely many have guessed, so only finitely many have guessed wrong. Hence the prisoners can never be executed. (Though they can never be released either.)
OK, I also got a “non-cheat” solution: unfortunately, it is non-constructive and uses the Nkvbz bs Pubvpr, so it still feels like a bit of a cheat. Is there a solution which doesn’t rely on that (or is it possible to show there is no solution in such a case?)
Heh. This is the easy version of the puzzle then. Hard version: The hats have arbitrary colours (suppose they are specified by infinite precision RGB values).
I think it still works if the infinity is uncountable.
Is this a case where you can prove a solution exists but you can’t say what it is? (Because you have to invoke the Axiom of Choice or some such?)
The participants have to base their decision on a non-measurable “events”, right? The guessing procedure ends up as contorted as the slices in Tarski and Banach’s sphere?
Right.
Do the prisoners all have to guess simultaneously? If they guess one by one, and hear their predecessors’ guesses, is it in the order 1, 2, 3..., or can guessing order be a part of their prearranged strategy?
Is the sequence of hats pbzchgnoyr? I don’t see how to do this if it’s not.
Not necessarily.
EDIT: To be fair I should say that the “algorithm” they follow in the solution isn’t computable either.
Yes, thank you for saying this.
Still confused. Richard Kenneway’s solution relies on the true sequence being at a finite place in some ordering. Doesn’t Cantor’s diagonal argument prevent you from having a countable ordering of all the sequences?
There’s no need for an enumeration of all the sequences, bayl na rahzrengvba bs gur rdhvinyrapr pynff gung gur cevfbaref frr gung gurl’er va jura gur ungf ner cynprq. Naq Bfpne_Phaavatunz’f fbyhgvba qbrfa’g arrq rira gung—gur rdhvinyrapr pynffrf pna or nal genafsvavgr fvmr jungrire.
I’m now wondering whether for the case of two colours, there is a computable algorithm. A prisoner would apply the algorithm by feeding it an infinite tape listing all the colours of the hats, with a blank for his own, and the algorithm would in a finite time say what guess to make.
It seems unlikely. In a finite time it’s impossible to get any idea whatsoever what equivalence class they’re in, so the solution, if there is one, would need to be very different.
Idea for a proof: we could assume the warden chooses colours randomly for each prisoner, iid with probability a half. Then there might be a probabilistic proof that the puzzle is impossible. This proof would fail to rule out the true solution because gur frgf vaibyirq jbhyqa’g or zrnfhenoyr va gung pnfr, ohg gurl jbhyq or zrnfhenoyr va rirel pbzchgnoyr pnfr.
Proof that there is no sequences of algorithms A1, A2, …, assigned to each prisoner, giving a winning strategy (assuming a computable warden given indices for the Ak):
Gur jneqra fvzhyngrf N-bar ba mreb mreb mreb… hagvy vg bhgchgf bar be mreb nsgre ernqvat x ovgf bs gur vachg. Gur jneqra gura cynprf gur bgure pbybe ung ba cevfbare 1, jub jvyy sbyybj N-bar naq thrff vapbeerpgyl; gur jneqra rafherf guvf ol cynpvat mreb ba cevfbaref gjb guebhtu x. Gur jneqra ercrngf guvf jvgu N-x+1 naq cevfbare x+1, fb gung x+1 jvyy thrff vapbeerpgyl. Naq fb ba. Fvapr rnpu Nx unygf nsgre ernqvat svavgryl znal ovgf sebz vgf benpyr, gur jneqra pna sbepr na vapbeerpg nafjre jvgu bayl svavgryl znal ovgf. Guvf jnl, gur jneqra pna sbepr vasvavgryl znal vapbeerpg nafjref. (Guvf eryngvivmrf gb nal benpyrf lbh pner gb tvir gb gur cevfbaref, nf ybat nf gur jneqra unf npprff gb gur pbhagnoyr wbva bs gubfr benpyrf.)
Vf guvf cbffvoyr jvgubhg gur nkvbz bs pubvpr?
Do the prisoners guess all at the same time, or in order (and thus after hearing finitely many other guesses)? If the second, is the guessing order known beforehand, or will it be at the whim of the guards?
All at the same time. Solving the case with guessing in order is a good intermediate step.
With guessing in order, I observe that for every finite subset there are either an even or an odd number of red hats, and prisoner 1 can indicate which it is by his guess; then everyone in that subset can count the red hats and figure out which colour his own must be to make the total number even or odd. Let the size of the finite subset go to infinity.
This is a good idea, and solves the similar problem with finitely many prisoners getting at most one guess incorrect. But...
I don’t see how this immediately gives a solution; any finite set of hats has either an even or an odd number of red hats, but an infinite set of hats may have an infinite number of red hats and an infinite number of black hats, and infinities are neither odd nor even.
Guessing in order with the other prisoners hearing these guesses would violate the stricture that the prisoners cannot communicate with each other.
Either this problem is not completely stated, there are important things left out, or the answers are gibberish. In any case, I have no idea what is going on even after reading the answers. I can collude with prisoners ahead of time but I can’t communicate with them. This presumably means I can’t smile or wink or grab or stare meaningfully or longingly at them. I can’t tell them what my guess is.
So I find myself in a room full of people, some with red hats and some with black hats, in what is essentially a still picture, for all intents and purposes equivalent to freezing the room full of other people with their red and black hats and I can then examine this fixed scene. How does this even conceivably impact what I would guess the color is of the hat on my head?
Fix!
There are important things left out. Specifically, this is a math problem and putting it in terms of prisoners or humans is intended to be ignored. You should of course just know that when someone is a human in a math puzzle, what they really mean is a hypercomputer or oracle that can return correct answers for uncomputable algorithms. Here’s another example:
There is a prisoner in a cell, the same as the prisoners in the question above. Each day the warden hands him an infinitely complex algorithm, and he is released if and only if he can correctly tell the warden whether it halts or not. On what day is he released?
Gur svefg qnl.
The problem is correct as stated, and solutions above by RichardKennaway and Oscar_Cunningham are correct. I think you may have missed that the prisoners are all distinguishable, a.k.a. they are numbered 1,2,3,.… Or you are confused about the win condition; we don’t have to guarantee that any particular prisoner guesses correctly, just that only finitely many guess incorrectly.
Sub-puzzle: prove definitively that if the prisoners are not distinguishable, then there is no winning strategy.
In the winning strategy, do fewer than half the prisoners guess wrong? Do more prisoners guess correctly than incorrectly? I’m trying to get a handle on whether it is worth my while to try to penetrate the jargon in the “correct solutions.”
What do you mean by half the prisoners? Let’s start there.
How about I choose a prisoner at random from among all the prisoners in the problem. What is the probability that the prisoner I have chosen has correctly stated the color of the hat on his head? In particular, is that probability more than, less than, or equal to 0.5?
While we are in the neighborhood, if there is a prisoner who is more likely to get the answer correctly than not, if you could tell me what ihis step by step process of forming his answer is, in detail similar to “if he is prisoner n, he guesses his hat color is the opposite of that of prisoner n^2+1” or some such recipe that a Turing machine or a non-mathematician human could follow.
Thanks in advance
So what do you mean to choose a prisoner at random when one has infinitely many prisoners?
Whatever mwengler’s answer is, your answer is going to have to be “in that case the set you asked about isn’t measurable, and so I can’t assign it a probability”.
Maybe this way then. I set down somewhere in the universe in a location that I don’t reveal to you ahead of time and then identify the 100 prisoners that are closest to my position. If the 100th and 101st furthest prisoners from me are exactly equally distant form me I set down somewhere else in the universe and I keep moving until I find a location where I can identify the 100 closest prisoners to my current position.
Of that 100 prisoners, I count the number of prisoners who identified their hat color correctly.
My question is what is the probability that I have counted 50 or fewer correct answers? Is it greater than, less than, or equal to the probability that I have counted 51 or more correct answers?
Thanks to you and JoshuaZ for trying to help me here.
So how have you set up the prisoners in the universe in advance and how do you decide on the location you set down?
Is it safe to say that this problem, this result, has no applicability to any similar problem involving a merely finite amount of prisoners, say a mere googol of them?
Yes. But I do think that thinking critically about the assumptions you are making, in particular that you can meaningfully talk about what it means to pick a random individual in a uniform fashion, is worthwhile for understanding a fair bit of probability and related issues which are relevant in broader in contexts.
As to critically understanding what it means to pick a random individual in a uniform fashion, yes it is worth understanding what one means, as a tremendous amount of mischief is done in the name of randomness. With a finite number of prisoners, I would actually not need to pick prisoners randomly in order to gather statistics. If I would simply count up all the prisoners who got the hat color correct and all that got it wrong, and I do the experiment say 100 times and plot a histogram of the results and then decide whether the results deviate from a normal distribution with mean 50% by enough to make it practically interesting.
But since you say this result has no applicability to a finite population of prisoners, I am assuming there is nothing here that would push the result away from 50:50?
Is the mathematician’s world really so insular,that someone from outside asking some questions about how a problem relates to concepts he understands gets downvoted?
Or are you pretending that unless you skate with ease over three different kinds of infinities, their differences and similarities, and the paradoxical results of probability problems with infinity in the numerator and denominator, that you are just a time wasting intruder on an otherwise valuable conversation?
Or did my questions, which I’d love to know the answer to, come across as a veiled negative comment?
Your questions were easily answered by looking up the definitions of the terms “finitely many” and “countably infinite”.
Are you aware that having a mathematician tell you a question is easily answered without actually answering it is actually the punch line to a joke? Closest I can find on the web is the 2nd one on this page.
You asked why you were downvoted. I told you why; you asked a question that showed you hadn’t made even a cursory attempt to understand the terms in the question.
The answers, in case you still haven’t put in the minimal effort required, are
Yes, a finite portion of an infinite set is infinitely less than half.
Yes, all but a finite number of an infinite set is infinitely more than a finite number.
This is not fancy jargon. These are terms anyone who has taken highschool calculus would know.
Infinity is weird.