Ad Astra - Scipio and Hannibal - Vol. 8 Ch. 51 - A Contest of Probabilities
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I'm afraid I don't follow the logic in it. Certainly picking the bigger group especially after having removed portions would in theory give you a higher chance, yet the moment Scipio had been asked if he could pick a different choice this meant that essentially it was already a new round so to speak. There was only 2 cups, and hence it's back to 50/50 no? The removal of the cup in the 2/3rd is not done after Scipio made his second choice, beforehand, so regardless of which he chooses, it's still 50/50.
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@DexerMang
The idea is that the cup you chose has a 1/3 chance of holding a grape while there is a 2/3 chance the grape is in one of the other two cups, so if you switch you have 2/3 odds of winning instead of 1/3 if you stay by your choice.
Google the Monty hall problem for a more accurate explanation.
The idea is that the cup you chose has a 1/3 chance of holding a grape while there is a 2/3 chance the grape is in one of the other two cups, so if you switch you have 2/3 odds of winning instead of 1/3 if you stay by your choice.
Google the Monty hall problem for a more accurate explanation.
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I've heard of this mathematical schtick before, and never really bought it until just now when I googled it and looked over the Wikipedia entry. It's actually a modern thing, called the "Monty Hall Problem", originally posed in terms of prizes behind door #1, door #2 and door #3.
The reason it works that way is that the host's choice of doors/cups to reveal isn't random--Archimedes isn't going to pick the cup with the grape, he's only going to reveal an empty cup. So the deal is, you have a 1 in 3 chance of your initial choice being correct. If your initial choice was correct, then switching after Archimedes' offer will make you lose. But you have a 2 in 3 chance of your initial choice being wrong, an empty cup. And since Archimedes showed the other empty cup, the remaining one has the grape--so if you picked wrong, switching after Archimedes revealed the other wrong choice gets you the win.
It was a pretty cute maneuver by the mangaka--using a modern mathematical problem, but a simple enough one that someone certainly could have solved it back in the day, and able to be applied to a variation on the old "shell game", which probably existed in antiquity.
The reason it works that way is that the host's choice of doors/cups to reveal isn't random--Archimedes isn't going to pick the cup with the grape, he's only going to reveal an empty cup. So the deal is, you have a 1 in 3 chance of your initial choice being correct. If your initial choice was correct, then switching after Archimedes' offer will make you lose. But you have a 2 in 3 chance of your initial choice being wrong, an empty cup. And since Archimedes showed the other empty cup, the remaining one has the grape--so if you picked wrong, switching after Archimedes revealed the other wrong choice gets you the win.
It was a pretty cute maneuver by the mangaka--using a modern mathematical problem, but a simple enough one that someone certainly could have solved it back in the day, and able to be applied to a variation on the old "shell game", which probably existed in antiquity.
@norctune
@ Purplelibraryguy
But isn't that 2/3 advantage merely illusory and irrelevant? I mean say you repeat that scene 100 times during the 2nd choice after Archimedes takes away 1 cup. If you randomized where the grape is, there's no functional way picking the 2/3rd choice would net you 2/3 wins over 100 attempts no? Because even if you pick out of the 2/3 group, it's already turned into 1/2 at this point, 1 being the '2/3' and the other 1 being the '1/3' you abandoned. There's simply no other factors in there in play. It sounds like sophistry, and heck I'm almost certain that even if you pulled this move of Archimedes from scratch 100 times over, it would still end up being close to 50/50 the higher the sample size, edging further and further away from 2/3. Because in the end you are only allowed to choose a single cup, no matter how you dance around the grouping it doesn't change the fact that Scipio is only gamble on a single outcome. It would only be of value if the rules were changed to benefit this '2/3' no? Like if it was 3/4 instead and you can open up two cups instead of one if you switched.
Or am I wrong even on that? Cause I'd really like to see some youtube video showing how this would happen.
@ Purplelibraryguy
But isn't that 2/3 advantage merely illusory and irrelevant? I mean say you repeat that scene 100 times during the 2nd choice after Archimedes takes away 1 cup. If you randomized where the grape is, there's no functional way picking the 2/3rd choice would net you 2/3 wins over 100 attempts no? Because even if you pick out of the 2/3 group, it's already turned into 1/2 at this point, 1 being the '2/3' and the other 1 being the '1/3' you abandoned. There's simply no other factors in there in play. It sounds like sophistry, and heck I'm almost certain that even if you pulled this move of Archimedes from scratch 100 times over, it would still end up being close to 50/50 the higher the sample size, edging further and further away from 2/3. Because in the end you are only allowed to choose a single cup, no matter how you dance around the grouping it doesn't change the fact that Scipio is only gamble on a single outcome. It would only be of value if the rules were changed to benefit this '2/3' no? Like if it was 3/4 instead and you can open up two cups instead of one if you switched.
Or am I wrong even on that? Cause I'd really like to see some youtube video showing how this would happen.
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@DexerMang
One grape hidden among three cups: [1] [2] [3], you pick cup [1]
If the grape is in cup [1], you lose if you change.
If the grape is in cup [2], GM opens cup[3], you win if you change.
If the grape is in cup [3], GM opens cup[2], you win if you change.
That's 2/3 odds for winning if you change. In the case where the grape is in cup [1] it doesn't matter if he opens cup [2] or [3] because the GM knows which cup has a grape and will only open a cup with no grape in it.
One grape hidden among three cups: [1] [2] [3], you pick cup [1]
If the grape is in cup [1], you lose if you change.
If the grape is in cup [2], GM opens cup[3], you win if you change.
If the grape is in cup [3], GM opens cup[2], you win if you change.
That's 2/3 odds for winning if you change. In the case where the grape is in cup [1] it doesn't matter if he opens cup [2] or [3] because the GM knows which cup has a grape and will only open a cup with no grape in it.
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Man, Archimedes is a bro. Makes the shit that will happen later all the more depressing.
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Archimedes' wording is weird, or is that the point? "Among the two cups you didn't choose, I will lift the cup without the grape." "The cup", not "one cup". Isn't he directly admitting the other unpicked cup has the grape? Or is it just a slight translation error? Or am I putting too much weight on semantics?
@norctune Basically I'm saying the first choice and removal of a cup is all fluff. It's irrelevant, no matter what cup you choose at the start, it doesn't change the fact that a non-grape cup will be removed and you will be asked the same question again. At that point is when your decision truly is a decision, and thus is it not 50/50? Because no matter which cup you picked at the start, it would not have changed the fact that you would pick from 2 cups with 1 having a grape. That wasn't even a real choice at that point. I feel like I'm repeating myself. I admit I don't really grasp the logic of the theory but I'd really like to see an argument of an actual result. It's like the argument in the chapter was half baked or that scenario should have been more complicated. Even in the supposed 99 cups, so long as those 99 cups are taken out before Scipio gets to choose again, then ultimately he's still picking from 2 cups no matter how you try to word it. That 1 removed cup will never ever become a real variable because it's guaranteed to be removed. As for how that theory could become a practical application, I don't know but maybe someone can clue me in. From what I can tell though, it'd only make sense if switching gave Scipio 2 chances instead of removing 1 cup, cause then that unknown cup is still a variable. But removing it before a choice was made simply eliminates it from the equation yes? That or the challenge is done in two parts, so the 1st choice would actually matter in the final result, as such switching over will give Scipio 1 win 1 loss whereas sticking with his cup would be a gamble. That's the best I can think up.
@boldmonkey
I also thought that too, but perhaps it's just a mistake in translation.
@boldmonkey
I also thought that too, but perhaps it's just a mistake in translation.
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@DexerMang I see your point since I struggle to understand the logic behind this as well...
I think you could approach this from the negative point of view:
Your first pick is a 99/100 chance for a blind (using the 100 cup version), while it is 1/2 with your second guess (I know this aint the proper math but it shows why you switching makes sense even when thinkig of them as two sep choices..) you could argue that its still 50/50 and that would be true if you hadnt had a chance of 99/100 to save a blind from being removed while the gamemaster has a "98/99 chance" to keep the right cup in the game, which he does since he knows, hence 99/100 chance thats the right cup. its all in thanks to the gamerules and "the gamemaster knows" why you have to think of it as "one" choice afterall..
I think you could approach this from the negative point of view:
Your first pick is a 99/100 chance for a blind (using the 100 cup version), while it is 1/2 with your second guess (I know this aint the proper math but it shows why you switching makes sense even when thinkig of them as two sep choices..) you could argue that its still 50/50 and that would be true if you hadnt had a chance of 99/100 to save a blind from being removed while the gamemaster has a "98/99 chance" to keep the right cup in the game, which he does since he knows, hence 99/100 chance thats the right cup. its all in thanks to the gamerules and "the gamemaster knows" why you have to think of it as "one" choice afterall..
It's a famously counter-intuitive probability problem.
@DexerMang @TheInvisableKid
You choose one cup -- you now know that there's a 1/3 chance of the grape being under your cup, and 2/3 chance of the grape being under one of the other two cups. Or to think of it another way, there's a 1/3 chance you're in a universe where neither of the two other cups have a grape, and a 2/3 chance you're in a universe where one of the two cups you didn't choose has the grape.
Now Archimedes removes an empty cup that you didn't choose. There's a 1/3 chance that neither of the cups had grapes anyway (and so the remaining cup is empty) and a 2/3 chance that one of the cups had a grape (so Archimedes removed the empty cup and the remaining cup has a grape under it). For you, there is 2/3 chance there's a grape under the other cup, only because you saw Archimedes take both cups into account, and remove the one without a grape.
Now consider a soldier who walks into the room and knows only that a grape has been hidden under one of the two cups. For them, they could only guess that the probability is 1/2. But you know better -- you saw Archimedes remove a cup. The difference in information changes what a rational observer can determine about the situation.
And as for Archimedes, he knows better still -- the odds of the other cup containing a grape are 0%. He already knows the grape is under the first cup because he put it there.
See, probability is not a function of the physical state of the grape -- probability is a function of what you know about the grape. It's a calculation based on all the hints, clues, and information you have about the situation. And in this case, Archimedes removing one of the cups adds real information to the game. When you have more information you can make better decisions-- throwing out relevant information doesn't get to a more correct answer.
====
One more way to think about it: Instead of choosing one cup, imagine instead that you divide the cups into two groups: two cups on the left, and one cup on the right. You can either choose the one cup on the right, or both cups on the left. If your group has the grape under it, you win. Which would you choose?
@DexerMang @TheInvisableKid
You choose one cup -- you now know that there's a 1/3 chance of the grape being under your cup, and 2/3 chance of the grape being under one of the other two cups. Or to think of it another way, there's a 1/3 chance you're in a universe where neither of the two other cups have a grape, and a 2/3 chance you're in a universe where one of the two cups you didn't choose has the grape.
Now Archimedes removes an empty cup that you didn't choose. There's a 1/3 chance that neither of the cups had grapes anyway (and so the remaining cup is empty) and a 2/3 chance that one of the cups had a grape (so Archimedes removed the empty cup and the remaining cup has a grape under it). For you, there is 2/3 chance there's a grape under the other cup, only because you saw Archimedes take both cups into account, and remove the one without a grape.
Now consider a soldier who walks into the room and knows only that a grape has been hidden under one of the two cups. For them, they could only guess that the probability is 1/2. But you know better -- you saw Archimedes remove a cup. The difference in information changes what a rational observer can determine about the situation.
And as for Archimedes, he knows better still -- the odds of the other cup containing a grape are 0%. He already knows the grape is under the first cup because he put it there.
See, probability is not a function of the physical state of the grape -- probability is a function of what you know about the grape. It's a calculation based on all the hints, clues, and information you have about the situation. And in this case, Archimedes removing one of the cups adds real information to the game. When you have more information you can make better decisions-- throwing out relevant information doesn't get to a more correct answer.
====
One more way to think about it: Instead of choosing one cup, imagine instead that you divide the cups into two groups: two cups on the left, and one cup on the right. You can either choose the one cup on the right, or both cups on the left. If your group has the grape under it, you win. Which would you choose?
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@DexerMang I too have trouble "believing" it, even when the logic is laid out and practical demonstrations are made.
But since the evidence indicates it is true, then the problem lies with our human intuition. It never evolved to deal with things like sub-atomic physics, the real world difference between a million and a billion, and other way-out subjects. I'm actually surprised to find the shortcomings of our intuition coming up this soon, when faced with what at first seems to be a trivial puzzle: Three cups and a grape!
But since the evidence indicates it is true, then the problem lies with our human intuition. It never evolved to deal with things like sub-atomic physics, the real world difference between a million and a billion, and other way-out subjects. I'm actually surprised to find the shortcomings of our intuition coming up this soon, when faced with what at first seems to be a trivial puzzle: Three cups and a grape!
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@DexerMang
Let's say W = win and L = lose.
In the beginning you have:
[W] [LL]
[L] [WL]
[L] [LW]
You choose one of the three options with probability 1/3.
Then one of the L cups is removed.
One can think that the situation is
[W] [L]
[L] [W]
But that's wrong, because this way you don't consider the situation in the beginning.
The right answer is
[W] [L] (the position of the two L in this case doesn't count)
[L] [W ]
[L] [ W]
The odds that W is in the second group is 2/3.
The point is, the grape is put under a cup, and the subject make a choice, BEFORE one of the wrong cups is removed. This makes the history of the experiment relevant. The chanches are fixed in the beginning of the experiment.
Let's say W = win and L = lose.
In the beginning you have:
[W] [LL]
[L] [WL]
[L] [LW]
You choose one of the three options with probability 1/3.
Then one of the L cups is removed.
One can think that the situation is
[W] [L]
[L] [W]
But that's wrong, because this way you don't consider the situation in the beginning.
The right answer is
[W] [L] (the position of the two L in this case doesn't count)
[L] [W ]
[L] [ W]
The odds that W is in the second group is 2/3.
So the guy that won more in this whole mess has finally entered the fray. I mean Masinissa, OFC .. the one that ended his life at 96 as king and with a LOT of children 
( P.S Sorry for the 2 millenia old spoilers )
P.S Was not expecting to see a probabilities discussion in here
( P.S Sorry for the 2 millenia old spoilers )P.S Was not expecting to see a probabilities discussion in here
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You'll most likely pick an empty cup initially, by 2/3rds odds, right? If that happens, Archimedes then removes the remaining empty cup, because he never removes the cup hiding the grape, as that would prematurely reveal the grape, nor does he remove the cup you picked. So, you're left with only one remaining cup to change to: the grape cup.
You also have 1/3rds chance of picking the grape cup right away. In this case, changing cups means you get an empty cup, regardless of which empty cup Archimedes removes. But, if you don't change cups, you still only have a regular 1/3rd chance of your first cup being correct, and 2/3rds chance of it being incorrect.
You also have 1/3rds chance of picking the grape cup right away. In this case, changing cups means you get an empty cup, regardless of which empty cup Archimedes removes. But, if you don't change cups, you still only have a regular 1/3rd chance of your first cup being correct, and 2/3rds chance of it being incorrect.
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@lupus_in_fabula congratulations man, i think you just made the most clear, graphic, short, and easy to digest explanation to a logical problem that messes up our normal intuition, i may even link to your comment whenever i encounter this puzzle and have to explain it, the one i was using before listed all the choices and out comes (9), so it was a bit longer.
@rrolo1 is it really spoilers when the manga itself tells us that he goes and unifies Numidia and becomes king?
@rrolo1 is it really spoilers when the manga itself tells us that he goes and unifies Numidia and becomes king?
@Gnauga
I cannot agree with the remaining cup having a 2/3rd chance, because you're arbitrarily splitting up the cup you chose with the other two cups the moment you picked, yet it serves no purpose and only blinds you to the fact that a guaranteed empty cup was removed. and that you get to pick again anyways.
@lupus_in_fabula
Why is the history relevant? Please explain. No matter how I look at it, if you repeat the experiment, whether you pick the cup with the grape or not, an empty cup that isn't yours should you pick an empty one will be removed, leading you to choose again.
@boldmonkey
So how does that allow always choosing to change the cup during the second round more likely to win?
@TheInvisableKid
I honestly couldn't follow you on that. The gist of what I grasp is you're telling me the first choice 'matters' even though it doesn't matter, and that the mathematical theory is designed for accounting of that first irrelevant choice. If that's the case, then I understand somewhat but at the same time can't find it anything more than showing off, because it doesn't even apply to the match itself as I pointed out earlier. If the results were based on both his choices, then I can understand, but the first choice as far as I can tell has no meaning.
I just want to know, and I'd like an answer this time for real. If you repeat what Scipio did 100 times, 1000 times, 1000000 times, infinite times, will you ever closer to 2/3rd by doing such a thing? Or will you simply stay roughly 50/50 in probability?
The way I see it, and this might be very arrogant of me, is that if the theory can't even be applied to something as simple as Archimedes dice game with tangible results, then there's only a few possibilities. Either the theory is wrong (like missing variable) or the theory was applied to the wrong scenario. I explained in my previous posts why I believe it is the latter, because ultimately if we ignore the hidden victory condition of Scipio explaining his thought process (and honestly I half expected Scipio did it simply so he can appeal to Archimedes' mathematician heart) then the only objective of Scipio is to do his best and take the best choice possible to win, nothing more nothing less. Whatever mental mode or way of thinking shouldn't matter if it cannot actually increase his chance of victory in reality. The way I see it, if someone actually believed in this theory applies to this specific dice game (with the dealer always removing 1 empty cup that isn't the player's before asking again) then they are in for a big disappointment. It's practically the sophistry of math.
Of course I could be wrong, if so please explain. Not through theory, but actual concrete evidence. To begin with, it makes no sense that what I mentioned can't be used as a theory itself, to rebuke Scipio's theory. After all, mine is easily proven in real life. All you have to do is simulate X amount of games running Scipio's method, then X amount of games where you just don't care and randomly pick. The larger X is, the higher the chance that win rates will level out close to even for both approaches. Because the results isn't round 1 + round 2, it's simply whether or not you pick the right cup at round 2.
I cannot agree with the remaining cup having a 2/3rd chance, because you're arbitrarily splitting up the cup you chose with the other two cups the moment you picked, yet it serves no purpose and only blinds you to the fact that a guaranteed empty cup was removed. and that you get to pick again anyways.
@lupus_in_fabula
Why is the history relevant? Please explain. No matter how I look at it, if you repeat the experiment, whether you pick the cup with the grape or not, an empty cup that isn't yours should you pick an empty one will be removed, leading you to choose again.
@boldmonkey
So how does that allow always choosing to change the cup during the second round more likely to win?
@TheInvisableKid
I honestly couldn't follow you on that. The gist of what I grasp is you're telling me the first choice 'matters' even though it doesn't matter, and that the mathematical theory is designed for accounting of that first irrelevant choice. If that's the case, then I understand somewhat but at the same time can't find it anything more than showing off, because it doesn't even apply to the match itself as I pointed out earlier. If the results were based on both his choices, then I can understand, but the first choice as far as I can tell has no meaning.
I just want to know, and I'd like an answer this time for real. If you repeat what Scipio did 100 times, 1000 times, 1000000 times, infinite times, will you ever closer to 2/3rd by doing such a thing? Or will you simply stay roughly 50/50 in probability?
The way I see it, and this might be very arrogant of me, is that if the theory can't even be applied to something as simple as Archimedes dice game with tangible results, then there's only a few possibilities. Either the theory is wrong (like missing variable) or the theory was applied to the wrong scenario. I explained in my previous posts why I believe it is the latter, because ultimately if we ignore the hidden victory condition of Scipio explaining his thought process (and honestly I half expected Scipio did it simply so he can appeal to Archimedes' mathematician heart) then the only objective of Scipio is to do his best and take the best choice possible to win, nothing more nothing less. Whatever mental mode or way of thinking shouldn't matter if it cannot actually increase his chance of victory in reality. The way I see it, if someone actually believed in this theory applies to this specific dice game (with the dealer always removing 1 empty cup that isn't the player's before asking again) then they are in for a big disappointment. It's practically the sophistry of math.
Of course I could be wrong, if so please explain. Not through theory, but actual concrete evidence. To begin with, it makes no sense that what I mentioned can't be used as a theory itself, to rebuke Scipio's theory. After all, mine is easily proven in real life. All you have to do is simulate X amount of games running Scipio's method, then X amount of games where you just don't care and randomly pick. The larger X is, the higher the chance that win rates will level out close to even for both approaches. Because the results isn't round 1 + round 2, it's simply whether or not you pick the right cup at round 2.
@DexerMang
Look, this article has the statistics:
https://wizardofvegas.com/articles/monty-hall-problem/
This one has a version you can try yourself:
https://betterexplained.com/articles/understanding-the-monty-hall-problem/
Look, this article has the statistics:
https://wizardofvegas.com/articles/monty-hall-problem/
This one has a version you can try yourself:
https://betterexplained.com/articles/understanding-the-monty-hall-problem/
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@DexerMang
Because two out of three initial cup options lead to the same final outcome if you opt to change: randomly picking either of the two empty cups and electing to change leads to the same outcome, regardless of which of the empty cups you randomly picked.
Let's say the grape is under the first cup. Let's call that 1G (cup nr. 1, Grape), and the others 2E (cup 2, Empty) and 3E (cup 3, Empty). Remember, he only ever removes one empty cup.
1) If you pick 2E, which empty cup can he remove? Which cup is left for you to change to? Do you win or lose by changing?
2) If you pick 3E, which empty cup can he remove? Which cup is left for you to change to? Do you win or lose by changing?
3) If you pick 1G, which empty cups can he remove? Which cup is left for you to change to? Do you win or lose by changing?
I understood the problem by throwing out mathematic calculation entirely: I'm not schooled enough to define this problem elegantly on paper, much less in my head (look at this shit: https://en.wikipedia.org/wiki/Monty_Hall_problem#Direct_calculation ). Instead, I pictured the cups in my mind, with knowledge of what was under each imaginary cup, and what would happen if I was the game master with knowledge of where the grape was, and another person had to select a cup at random.
Because two out of three initial cup options lead to the same final outcome if you opt to change: randomly picking either of the two empty cups and electing to change leads to the same outcome, regardless of which of the empty cups you randomly picked.
Let's say the grape is under the first cup. Let's call that 1G (cup nr. 1, Grape), and the others 2E (cup 2, Empty) and 3E (cup 3, Empty). Remember, he only ever removes one empty cup.
1) If you pick 2E, which empty cup can he remove? Which cup is left for you to change to? Do you win or lose by changing?
2) If you pick 3E, which empty cup can he remove? Which cup is left for you to change to? Do you win or lose by changing?
3) If you pick 1G, which empty cups can he remove? Which cup is left for you to change to? Do you win or lose by changing?
I understood the problem by throwing out mathematic calculation entirely: I'm not schooled enough to define this problem elegantly on paper, much less in my head (look at this shit: https://en.wikipedia.org/wiki/Monty_Hall_problem#Direct_calculation ). Instead, I pictured the cups in my mind, with knowledge of what was under each imaginary cup, and what would happen if I was the game master with knowledge of where the grape was, and another person had to select a cup at random.
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