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B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
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August 2016
bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Jul 12, 2022 9:55:58 GMT
Koalas sleep 18-22 hours a day. Male lions sleep 20 hours a day. ...And humans are supposed to be the intelligent species? <definitely going back to bed very soon.>
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mousestalker
Inactive Moderator
ღ The Untitled
Just here for the cosplay
Staff Mini-Profile Theme: Mousestalker
Games: Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, Baldur's Gate, Neverwinter Nights, Jade Empire, Mass Effect Andromeda, SWTOR
Posts: 12,116 Likes: 30,354
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Post by mousestalker on Jul 12, 2022 15:53:55 GMT
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Another Crazy Finn
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rewindbutton
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Post by rewindbutton on Jul 13, 2022 11:07:33 GMT
James Webb is going to revolutionize the way we look at space. edit: the pic's back.
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save yourself
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, Mass Effect Andromeda
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Post by q5tyhj on Jul 17, 2022 0:44:27 GMT
James Webb is going to revolutionize the way we look at space. I forget where I saw it, but someone had done a side-by-side comparison between the JWST's deep field image and a Hubble deep field image- looking at the two together, its really amazing how much more detail the JWST has. (and with those deep field images we're talking about galaxies that are billions and billions of light years away) And maybe the most exciting of the new JWST images was the least interesting to look at, i.e. the one that was a spectrographic analysis of some distant gas giant; the ability to look at and analyze exoplanet atmospheric composition for Earth-like planets in their stars "habitable zone" (the area where liquid surface water is possible) is something we've never had before, and could very plausibly lead to our first unambiguous evidence of alien life. And very possibly in the not too distant future. Which, needless to say, would be freakin' awesome.
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B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
3,752
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bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Jul 27, 2022 14:35:26 GMT
Do you know what to do, if:
You're one of ten prisoners who are considered for execution, ...or pardon and release. All ten will either be executed together, or released together. All ten prisoners are numbered, have a number from 1 to 10.
Each is, one at a time, required to go into a room. Inside that room, there is a row of 10 fixed boxes that cannot be moved. They can be opened to look inside, but they cannot be kept opened or otherwise marked in any way. In the bottom of each box, there is a number written. The boxes contain one of all the numbers 1 to 10, randomly distributed.
Each prisoner have 8 attempts to find their own number, in one of the boxes. If he finds it, he leaves the room through another door and cannot speak to the other prisoners again. If he fails, if he has opened 8 boxes but still hasn't found his number, all ten prisoners will be immediately executed.
Only if all ten prisoners succeed in finding their own number in eight attempts or less, will everybody be pardoned and released.
The prisoners can talk with each other before the test begins.
No. I didn't come up with it myself. The origin of the puzzle is some University Math paper. I posted a video by Dr Derek Muller on the same subject in another thread, though the problem was formulated differently.
And I would definitely assume that people, who are able to come up with a solution by themselves, are extremely rare, because this is pretty far out. Which is why it helps to know what to do in advance. Not because you might happen to be in exactly this situation (well, there are RPGs...), but because there might be other problems that might benefit from a creative solution that may be similar in some sense.
If each prisoner acts alone, the probability of that everyone will be executed is quite high, 89.263%. Almost 90%. Or seen from the other angle, just a 10.737% chance of survival.
And you shouldn't be happy with that and only resort to prayers, because, if all prisoners cooperate on a common strategy, they can influence their chances.
The first step in developing a collective strategy, is that all prisoners identify the boxes in exactly the same way. That is, each box gets a label, and all prisoners label the boxes in the same way. This can be easily done. They agree on using two walls, like the left and the opposite, to the wall they enter through. They will look at the row of boxes, facing these walls, or one of them, and then label the boxes with numbers in reading order, left to right, 1,2,3,4,5,6,7,8,9,10.
This way, the prisoners can make sure they will all refer to each box in exactly the same way, when they enter the room.
If they then, for example, all agree to search the exact same 8 boxes, they will lower their survival chances from 10.7% to zero, to accomplish a guaranteed collective suicide, because exactly two of them will always fail to find their numbers in this way.
That's not what we want, of course, but we can clearly see that a common strategy can affect the outcome.
What we want is something different. We want each prisoner to search a personal group of boxes, maximum eight, and hope that everybody finds his number.
The crux is of course that each prisoner's personal chance of finding his number is always going to be 80%, no matter what we are doing. We can't do anything about that. But that was also the case with the suicide strategy. -Yet, we could still guarantee that two would always fail! Just not who would fail.
This is a collective game. If one fails, everybody fails. If all succeeds, everybody succeeds. So can there be a way, that we can couple together everybody's chance of failure, 20%, in a way that guarantees that if anyone fails to find their number, that will be, and only be, if EVERYBODY else also fails to find their number? This is mind-boggling of course, but as each only have a 20% chance of failure, and nothing can change that, wouldn't that then also automatically result in that there would be an 80% chance of everybody succeeding to find their number, together, at the same time?
For this collective mass failure, everybody would have to, at the same time, leave the correct box to the last two, the boxes not searched because of the 8 limit. And if that didn't happen, everybody would succeed. Can this be done?
- YES! Every box now has two numbers. The hidden one inside, and the labels the prisoners have agreed to designate them with. This fact can be used to link the boxes together in some order.
This is the way to do it: Every prisoner first goes to the box that is labeled with the prisoners own number. Let's say it's prisoner no 3. He first goes to the box that all the prisoners have agreed to label #3. He opens the lid. Inside the box he reads 8. He now goes to the box labeled #8. He reads 4. He goes to #4. He reads 2. He goes to #2. He reads 1. He goes to #1. He reads 7. He goes to #7. He reads 3, and he has found his own number. He has only opened 6 boxes, so he has success.
And actually, he can now tentatively also celebrate already, because if everybody carefully follows the rules, he can be sure they will now all succeed. They are guaranteed to. Why?
Look at his numbers: 8-4-2-1-7-3. 3 (inside #7) points back to the box he started with. These six numbers and their labels form a closed loop. Every prisoner with number 1,2,3,4,7,8 will search the same loop and will also find their own number in exactly six tries.
The other four numbers, 5,6,9,10 will also form closed loops/loop. We don't know exactly how, but it's impossible that any of the other prisoners will have to search more than 4 boxes. So everybody wins together.
To more clearly see what's going on here, let's see what must be happening if he is to fail. Remember, he has only a 80% personal chance to succeed. There is always the 20% personal chance of failure. So, if failure, the #7 box would not have contained his number, 3, and would not have pointed back to the box he started with, but something else, like: 8-4-2-1-7-10-6-9 So, eight boxes opened, and he has failed and everybody will be executed. But examine what is going on here. There can be two cases: 8-4-2-1-7-10-6-9-3, a nine long loop before it returns by pointing back to #3, and the last box will be #5 and it will always contain its own number, in this case 5. Or, the only other possibility is that all numbers are connected into only one loop, 10-long. Like: 8-4-2-1-7-10-6-9-5-3. The last box, points back to the start and contains the searched for number.
So we have almost accomplished what we wanted. By starting on their own number, at least nine prisoners will always fail together (prisoner 5 will succeed), if anyone fails. Alternatively, all ten prisoners will fail together, if anyone fails. We have coupled together the prisoners' individual 20% chances of failure, into a collective probability of failure that should be close to 20% (21.1%), and an almost 80% (78.9%) chance of collective survival.
Now, if he had found a 3 inside the #3 box, he would have found his own number, but he couldn't have been sure everybody would succeed. Why not? Because it leaves 9 numbers. There is still a possibility of that the other nine numbers are linked together in a 9-long loop, in which case everybody else will fail. If he's curious, he can start on another number and open up to seven other boxes. If the loop closes in seven or less, he can be certain of survival.
Nine prisoners will fail if nine numbers link together in a 9-long loop. Ten (all) prisoners will fail if all ten numbers link together in a single 10-long loop. If, however, the longest loop is 8 or shorter, everybody will always succeed together.
The probability of there being a 10-long loop is: <number of possible 10-long loops> = 10! <number of unique 10-long loops>10!/10 <ways of arraigning 10 numbers in 10 boxes> = 10! Probability of there being a 10-long loop = 10!/10/10! = 1/10. The probability of there being a 9-long loop is: 10*9!/9/(10*9!) = 1/9.
Probability of there being either a 9-long loop or a 10-long loop = 1/9 + 1/10 = 19/90 = 0.211... or 21.1% which translates into a 78.9% success rate. Which is certainly much to favor over 10.7%
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Post by themikefest on Jul 30, 2022 12:21:14 GMT
wrong thread. my bad
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Post by Lavochkin on Aug 1, 2022 1:19:13 GMT
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The world just opened up...l'm now hearing sounds that I haven't heard in quite some time!
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Mass Effect Andromeda, Anthem
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Post by Energizer Bunny 211 on Aug 1, 2022 11:23:36 GMT
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B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
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bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Aug 2, 2022 0:14:47 GMT
Yeah, I'm not sure if the reasons are exactly scientific, but yeah, like Chinese writing, Latin or English, crosses all borders and languishes.
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B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
3,752
August 2016
bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Aug 2, 2022 14:14:27 GMT
And this is Derek Muller, btw. Probably my favorite mad scientist.
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Beerfish
N7
Little Pumpkin
Games: Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights, Mass Effect Andromeda, Anthem, Mass Effect Legendary Edition
Origin: Beerfish
XBL Gamertag: Beerfish77
Posts: 15,177 Likes: 36,346
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https://bsn.boards.net/user/314/personal
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Post by Beerfish on Aug 2, 2022 16:08:51 GMT
That And this is Derek Muller, btw. Probably my favorite mad scientist. That is wild and surprising to me.
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The world just opened up...l'm now hearing sounds that I haven't heard in quite some time!
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Jan 15, 2017 18:43:23 GMT
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Mass Effect Andromeda, Anthem
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Post by Energizer Bunny 211 on Aug 2, 2022 16:15:48 GMT
That And this is Derek Muller, btw. Probably my favorite mad scientist. That is wild and surprising to me. For a second I thought it was going to be a cat tripping a woman in a 'slinky' dress when she is walking down the stairs. Slinkies are like people.....Mostly useless and often boring until/unless you push them down the stairs Then it becomes the "Science of Falling Down the Stairs" and "The Art of NOT Breaking Your Neck While Falling Down The Stairs" (Both are on the Best-Seller's List)
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Now Available As A Combo Meal!
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dragontartare
Add a cookie for just $1.99 (plus tax)!
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, Neverwinter Nights, Mass Effect Andromeda, Mass Effect Legendary Edition
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Post by dragontartare on Aug 2, 2022 20:33:43 GMT
And this is Derek Muller, btw. Probably my favorite mad scientist. That is the coolest thing I've seen in a long time.
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B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
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Post by B. Hieronymus Da on Aug 3, 2022 0:17:15 GMT
And this is Derek Muller, btw. Probably my favorite mad scientist. That is the coolest thing I've seen in a long time. The coolest thing I've encountered in a long time, is the thing I write about in the long TL:DR post above. I'm a bit disappointed that minds haven't been blown. Derek Muller did a 100 prisoners and 100 boxes, but only open 50 -version, in a video I posted somewhere, YouTube thread, I think.
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Add a cookie for just $1.99 (plus tax)!
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, Neverwinter Nights, Mass Effect Andromeda, Mass Effect Legendary Edition
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Post by dragontartare on Aug 3, 2022 0:27:28 GMT
That is the coolest thing I've seen in a long time. The coolest thing I've encountered in a long time, is the thing I write about in the long TL:DR post above. I'm a bit disappointed that minds haven't been blown. Derek Muller did a 100 prisoners and 100 boxes, but only open 50 -version, in a video I posted somewhere, YouTube thread, I think. It is cool, but I had already seen that video prior to you posting it, so my mind already blew up
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Beerfish
N7
Little Pumpkin
Games: Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights, Mass Effect Andromeda, Anthem, Mass Effect Legendary Edition
Origin: Beerfish
XBL Gamertag: Beerfish77
Posts: 15,177 Likes: 36,346
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Little Pumpkin
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Sept 30, 2024 16:15:40 GMT
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beerfish
https://bsn.boards.net/user/314/personal
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights, Mass Effect Andromeda, Anthem, Mass Effect Legendary Edition
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Post by Beerfish on Aug 5, 2022 19:40:48 GMT
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Post by q5tyhj on Aug 5, 2022 22:04:31 GMT
agreed, pretty hilarious
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Post by AnDromedary on Aug 6, 2022 17:44:18 GMT
This is great! Kinda shows to some extent why science shod stay away from twitter as much as possible. The journal system is messed up as hell but at least its peer reviewed.
While this was a harmless joke, the "argument from authority" makes twitter and other social media very dangerous to scientific integrity when it comes to more subtle things.
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Post by skekSil on Aug 7, 2022 19:11:54 GMT
This is great! Kinda shows to some extent why science shod stay away from twitter as much as possible. The journal system is messed up as hell but at least its peer reviewed. The Bogdanoff affair would like to have a word with you.
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mousestalker
Inactive Moderator
ღ The Untitled
Just here for the cosplay
Staff Mini-Profile Theme: Mousestalker
Games: Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, Baldur's Gate, Neverwinter Nights, Jade Empire, Mass Effect Andromeda, SWTOR
Posts: 12,116 Likes: 30,354
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Post by mousestalker on Aug 9, 2022 17:57:51 GMT
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Unapologetic Western Chauvinist. Barefoot. Great Toenails
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Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Aug 12, 2022 0:33:10 GMT
Do you know what to do, if: You're one of ten prisoners who are considered for execution, ...or pardon and release. All ten will either be executed together, or released together. All ten prisoners are numbered, have a number from 1 to 10. Each is, one at a time, required to go into a room. Inside that room, there is a row of 10 fixed boxes that cannot be moved. They can be opened to look inside, but they cannot be kept opened or otherwise marked in any way. In the bottom of each box, there is a number written. The boxes contain one of all the numbers 1 to 10, randomly distributed. Each prisoner have 8 attempts to find their own number, in one of the boxes. If he finds it, he leaves the room through another door and cannot speak to the other prisoners again. If he fails, if he has opened 8 boxes but still hasn't found his number, all ten prisoners will be immediately executed. Only if all ten prisoners succeed in finding their own number in eight attempts or less, will everybody be pardoned and released. The prisoners can talk with each other before the test begins. No. I didn't come up with it myself. The origin of the puzzle is some University Math paper. I posted a video by Dr Derek Muller on the same subject in another thread, though the problem was formulated differently. And I would definitely assume that people, who are able to come up with a solution by themselves, are extremely rare, because this is pretty far out. Which is why it helps to know what to do in advance. Not because you might happen to be in exactly this situation (well, there are RPGs...), but because there might be other problems that might benefit from a creative solution that may be similar in some sense. If each prisoner acts alone, the probability of that everyone will be executed is quite high, 89.263%. Almost 90%. Or seen from the other angle, just a 10.737% chance of survival. And you shouldn't be happy with that and only resort to prayers, because, if all prisoners cooperate on a common strategy, they can influence their chances. The first step in developing a collective strategy, is that all prisoners identify the boxes in exactly the same way. That is, each box gets a label, and all prisoners label the boxes in the same way. This can be easily done. They agree on using two walls, like the left and the opposite, to the wall they enter through. They will look at the row of boxes, facing these walls, or one of them, and then label the boxes with numbers in reading order, left to right, 1,2,3,4,5,6,7,8,9,10. This way, the prisoners can make sure they will all refer to each box in exactly the same way, when they enter the room. If they then, for example, all agree to search the exact same 8 boxes, they will lower their survival chances from 10.7% to zero, to accomplish a guaranteed collective suicide, because exactly two of them will always fail to find their numbers in this way. That's not what we want, of course, but we can clearly see that a common strategy can affect the outcome. What we want is something different. We want each prisoner to search a personal group of boxes, maximum eight, and hope that everybody finds his number. The crux is of course that each prisoner's personal chance of finding his number is always going to be 80%, no matter what we are doing. We can't do anything about that. But that was also the case with the suicide strategy. -Yet, we could still guarantee that two would always fail! Just not who would fail. This is a collective game. If one fails, everybody fails. If all succeeds, everybody succeeds. So can there be a way, that we can couple together everybody's chance of failure, 20%, in a way that guarantees that if anyone fails to find their number, that will be, and only be, if EVERYBODY else also fails to find their number? This is mind-boggling of course, but as each only have a 20% chance of failure, and nothing can change that, wouldn't that then also automatically result in that there would be an 80% chance of everybody succeeding to find their number, together, at the same time? For this collective mass failure, everybody would have to, at the same time, leave the correct box to the last two, the boxes not searched because of the 8 limit. And if that didn't happen, everybody would succeed. Can this be done? - YES! Every box now has two numbers. The hidden one inside, and the labels the prisoners have agreed to designate them with. This fact can be used to link the boxes together in some order. This is the way to do it:Every prisoner first goes to the box that is labeled with the prisoners own number. Let's say it's prisoner no 3. He first goes to the box that all the prisoners have agreed to label #3. He opens the lid. Inside the box he reads 8. He now goes to the box labeled #8. He reads 4. He goes to #4. He reads 2. He goes to #2. He reads 1. He goes to #1. He reads 7. He goes to #7. He reads 3, and he has found his own number. He has only opened 6 boxes, so he has success. And actually, he can now tentatively also celebrate already, because if everybody carefully follows the rules, he can be sure they will now all succeed. They are guaranteed to. Why? Look at his numbers: 8-4-2-1-7-3. 3 (inside #7) points back to the box he started with. These six numbers and their labels form a closed loop. Every prisoner with number 1,2,3,4,7,8 will search the same loop and will also find their own number in exactly six tries. The other four numbers, 5,6,9,10 will also form closed loops/loop. We don't know exactly how, but it's impossible that any of the other prisoners will have to search more than 4 boxes. So everybody wins together. To more clearly see what's going on here, let's see what must be happening if he is to fail. Remember, he has only a 80% personal chance to succeed. There is always the 20% personal chance of failure. So, if failure, the #7 box would not have contained his number, 3, and would not have pointed back to the box he started with, but something else, like: 8-4-2-1-7-10-6-9 So, eight boxes opened, and he has failed and everybody will be executed. But examine what is going on here. There can be two cases: 8-4-2-1-7-10-6-9-3, a nine long loop before it returns by pointing back to #3, and the last box will be #5 and it will always contain its own number, in this case 5. Or, the only other possibility is that all numbers are connected into only one loop, 10-long. Like: 8-4-2-1-7-10-6-9-5-3. The last box, points back to the start and contains the searched for number. So we have almost accomplished what we wanted. By starting on their own number, at least nine prisoners will always fail together (prisoner 5 will succeed), if anyone fails. Alternatively, all ten prisoners will fail together, if anyone fails. We have coupled together the prisoners' individual 20% chances of failure, into a collective probability of failure that should be close to 20% (21.1%), and an almost 80% (78.9%) chance of collective survival. Now, if he had found a 3 inside the #3 box, he would have found his own number, but he couldn't have been sure everybody would succeed. Why not? Because it leaves 9 numbers. There is still a possibility of that the other nine numbers are linked together in a 9-long loop, in which case everybody else will fail. If he's curious, he can start on another number and open up to seven other boxes. If the loop closes in seven or less, he can be certain of survival. Nine prisoners will fail if nine numbers link together in a 9-long loop. Ten (all) prisoners will fail if all ten numbers link together in a single 10-long loop. If, however, the longest loop is 8 or shorter, everybody will always succeed together.The probability of there being a 10-long loop is: <number of possible 10-long loops> = 10! <number of unique 10-long loops>10!/10 <ways of arraigning 10 numbers in 10 boxes> = 10! Probability of there being a 10-long loop = 10!/10/10! = 1/10. The probability of there being a 9-long loop is: 10*9!/9/(10*9!) = 1/9. Probability of there being either a 9-long loop or a 10-long loop = 1/9 + 1/10 = 19/90 = 0.211... or 21.1% which translates into a 78.9% success rate. Which is certainly much to favor over 10.7% I examined another case of this extremely counter-intuitive phenomenon: 100 prisoners get 95 attempts each, to find their number in 100 boxes. Normally, their chances to make it is only 0.59%, while, with the common strategy, it improves to 94.9%. Quite remarkably, and, considering the initial picture, crazy. So I came across another of these highly counter-intuitive puzzles about probability. This is something I actually got presented with already in high school. Don't know why. Our math teacher must have been fascinated by this. Anyway, it's the 3-door choice puzzle, which you may have seen before. You're presented with three hidden choices, one of them is good, the others are bad. So you obviously have 1/3 chance, or 33.33% chance of picking the right door. So you pick one. Then, one of the other doors is opened to show one of the poor, wrong choices. Now there are only two doors to choose between. What is the chance now, of you having picked the right door? 50%? One out of two, right? If you play Roulette - lets ignore there are 0 and 00 numbers -, and red has come up, what is the chance it will be red next time? -50%. What if red has come up 5 times in a row? What is the chance for red next time? For black? Most people have learned that it's still 50%. So why are the two remaining doors not equal, at 50%? (Because they're not) This puzzle typically includes that you get an opportunity to change your mind and choose the other door instead. Should you? Yes you should, because the probability that the right door is the other, is 2/3, while the chances for your already chosen door is only 1/3. Why is this? Why is it not 50/50? The explanation for this is - at face value - quite absurd: Because you have already chosen it once, and then the probability was only 1/3, and it hasn't changed!If you find this hard to wrap your head around, there's a trick that makes it more visible what is going on here: Imagine instead that there were 30 doors, instead of three. Your chances to pick the right door is 1/30 or 3.33%. The probability for the right door being one of the other 29 doors is of course 29/30 or 96.67%. Now also imagine that 28 'wrong' doors of those 29 are opened, leaving only one other door closed. So two doors. Is it 50/50 now? Of course not. The probability of that your first choice was right is 1/30. The probability of that the right door was among the other 29 doors was 29/30. Now 28 of them are opened, which means that the single, still closed door inherits all the 29/30. And the chances that you picked the right door the first time, is of course still only 1/30. It does not get promoted to 50%.
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inherit
11346
0
Sept 30, 2024 17:41:50 GMT
1,439
skekSil
1,200
November 2019
skeksil
Mass Effect Trilogy, Neverwinter Nights, Mass Effect Andromeda, Anthem, Mass Effect Legendary Edition
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Post by skekSil on Aug 12, 2022 19:31:50 GMT
So I came across another of these highly counter-intuitive puzzles about probability. This is something I actually got presented with already in high school. Don't know why. Our math teacher must have been fascinated by this. Anyway, it's the 3-door choice puzzle, which you may have seen before. This puzzle is better known as The Monty_Hall_problem. And, yeah, Theory of Probability can be quite unintuitive. That's why casinos are a thing.
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inherit
802
0
Sept 29, 2024 23:40:32 GMT
5,540
B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
3,752
August 2016
bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Aug 14, 2022 22:45:40 GMT
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inherit
Mr. Rump
46
0
Sept 29, 2024 2:16:59 GMT
8,995
Lavochkin
6,793
August 2016
lavochkin
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Jade Empire, Mass Effect Andromeda
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Post by Lavochkin on Sept 1, 2022 16:16:55 GMT
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inherit
802
0
Sept 29, 2024 23:40:32 GMT
5,540
B. Hieronymus Da
Unapologetic Western Chauvinist. Barefoot. Great Toenails
3,752
August 2016
bevesthda
Mass Effect Trilogy, Dragon Age: Origins, Dragon Age 2, Dragon Age Inquisition, KOTOR, Baldur's Gate, Neverwinter Nights
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Post by B. Hieronymus Da on Sept 3, 2022 16:40:25 GMT
This should require some popcorn.
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