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"brycec" i can explain it to you |
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http://www.cut-the-knot.org/hall.shtml you will see the numbers for yourself! ;) |
I was sure that this thread would be on page 10 the day after I started it... :)
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ok assume the host is completely honest hard to believe but he will offer you the choice to switch regardless there are 2/3 empty boxes and 1/3 prize boxes if you select the prize on your first try the host has the discretion to reveal EITHER empty box therefore when you switch you will lose if you select an EMPTY box, there is only ONE other box that the host can possibly reveal - there is ALWAYS going to be two empty boxes when you switch on EITHER empty box you will get the prize 2/3 = 66% chance of winning |
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I forgot to mention I dont believe in statistics , I know the statistics would tell you should change |
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just click the radio buttons and play the game it will show you |
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if it isnt something you are considering, then your answer will be incorrect this is why when i play blackjack and poker i whip the shit out of your average layman |
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nope not true , it is 50% I compare this to asci code back in the day 10101010101 1 = you win 0 = you lose you claim there is 66% chance it will become 1 if you change but there are only two options 1 or 0 and thus it could be either way since it is ONE case and not hundreds of them if you would play this prize game 100 times you should indeed switch if you play the prize game once you have no reason too |
your idea of having 66% chance is based on playing more than once
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LOL whatever..it's a proven fact....it has been subject to a lot of studies before
it is easy to prove and you still won't believe it..not much I can do.... |
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even in an independant testing 66% is accurate you are considering "luck" you have to understand representative odds sure i can hit on 20 and pull a miracle ace out of the deck to hit 21 it did work that one time but that doesnt mean odds are 100% of pulling that ace |
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think I never been in this discussion before ? ? think again :Graucho :Graucho 95% of the people always dissagree with me and say I should learn math 5% doenst have a clue :1orglaugh :1orglaugh |
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here's a quote from the other thread: when picking randomly out of 3 boxes, you ALWAYS have a 1:3 chance of winning = which means you are going to be wrong 2:3. Eliminate 1 of the wrong answers AFTER you have taken, with 66% probability, the other wrong answer - and you have your explanation. |
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how can you have numbers for ONE single experiment ? you are giving numbers on this experiment done more than once |
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I wouldnt change and thats the end of it :) |
either way me and math dont get along
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they tried to teach me this in school and even then I strongly dissagreed with that , but appearantly all people consider this to be the truth and I think its crap dunno if I translated it correctly |
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it's easy... if you chose not to switch.. you have 1/3 chance of winning.. if you switch, you only LOSE if you chose the curtain with the prize behind it initially, thus you have a 1/3 chance of losing, and a 2/3 chance of winning... |
I ran the following c-program off this page .
It iterates 100,000 times through the random possibilites and here's the result. Code:
$ ./a.out |
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this is the kind of logic you are using lets say i picked 6 lottery numbers your logic is saying that there are two outcomes = win or lose therefore i can either do one or the other win or lose 50/50 i think get what you are saying - in one shot you could get lucky and beat the odds - but its the *illusion* of 50/50 the actual mathematical probabilityis 66% it is irrefutable |
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ok so you wouldn't change huh? Let's say the prize is behind door B On your first guess...you pick door A.... Host opens the door C...he ask you if you want to stay or switch you STAY with A...you lose (you should've switch) Let's say it's behind door C On your first guess, you pick A again...Hosts open door B (knowing it's behind door C) he ask you if you want to stay or switch.... You stay with A, you LOSE.... in BOTH CASE you would've win by switching! Only time you would win is if it's behind door A (your first pick) sooooo 1/3 of winning by staying... 2/3 of winning by switching! Repeat the process by assuming you take B as your first pick...same thing will happen! How can you fail to see this! |
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here ya go! it's no fucking philosophical debate LOL it's a proven FACT....you have MORE chance of winning by switching! thanks equinox... |
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well , I still dont think that is accurate let us say A / B / C / I chose C / >> B / gets eliminated so now it is down to A / and C / you are now saying because it wasn't B / and I choose C / automatically the best option and most winning opportunity will always be A / ? that just doenst make sense to me you will have a 50 % chance of it being true and a 50% chance of it not being true luck is an important factor in such a guessing game and should always be considered when calculating your odds I cant understand how in ONCE case I would be better of chosing A / because it isn't B / and I choose C / initially it would be if you dont count in luck , but who is the fool who does'nt count in luck in a guessing game ? ? |
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In my opinion these kinds of judgements are legitimate, but if you believe that the experiment is IDEAL and the results only depend on theoretical statistics, then you should definitely switch. there's no doubt about it... |
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No problem. Wasn't sure whether anyone had ran the program before, but I just did for kicks. And although it defies my logic, it truely seems to work out that way. Here's the code, and it's no cheating involved. **edit** GFY's gay censorship fucks up the code... |
With 300k tries:
$ ./a.out Switch won 100000 (66%), lost 49742 (33%) Stay won 50083 (33%), lost 100175 (66%) |
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there is math thats it gambling is all math |
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:Graucho it's simple LOL |
500k iterations:
$ ./a.out Switch won 167024 (66%), lost 82927 (33%) Stay won 83561 (33%), lost 166488 (66%) Any more questions? :1orglaugh |
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it's pure statistics!... |
1 Miiiiiiiiiiiiiiiiiiiiiilllliiiiiiiiiiiiiiiiion iterations:
http://pauldavidson.blogs.com/wfme/images/dr.evil.jpg $ ./a.out Switch won 333294 (66%), lost 165767 (33%) Stay won 167252 (33%), lost 333687 (66%) |
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no its not , there is not math involved in playing once on the lottery the 66% chance of winning when changing is based on many many tries not on ONE try on one try it will always be 50% chance of winning |
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we can MIT hack the city while the 50/50s lose their money on sucker bets |
10 Million iterations, and still no 50/50 to be found.
$ ./a.out Switch won 3334441 (66%), lost 1665671 (33%) Stay won 1668679 (33%), lost 3331209 (66%) |
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this is where it all comes down too , you have 50% chance of your 66% rule being in effect and 50% chance of it not being in effect on your ONE try thus it makes no difference to change or not on ONE try thats the math behind my reasoning and I know it is correct :2 cents: |
This is not goddamn math, it's stats.
If some statistical value is true for 10,000,000 tries, it is also valid for one try of the same experiment, Fake Nick. Get over it, please. I know it's hard to grasp, just get over it. |
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with that logic Quote:
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Please tell me this is some sort of sick joke and you don't actually believe this :ugone2far |
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When you roll a regular dice, you have 1/6 chances of getting a 6 and you replying: no, I have more chance, because 6 is my lucky number |
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you agree that your chance increases when you do multiple tries..then you also agree that your chance is greater when doing it once... it is the theoretical CHANCE of winning... of course you can see a better scale of results when you do this a million times |
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not at all your 6 numbers are 6 times 1 out of 42 you have a lucky number , there are many many more possiblities here there are two , on this once case the 66% will win or the 33% will win even thoug that the statistics will always say 66% will win , in one case they are equal |
i think youre just not understanding the theories behind the mathematics
if i was going to argue with a heart surgeon about how to operate on a patient i would be arguing out of total ignorance even if i had some kind of opinion i dont know what the fuck im talking about |
Should I run 100,000,000 iterations of the same program to prove the point?
Why are people so stubborn? Probability theory is not that hard, really. Especially not when it's been proven, black on white, how the outcome will be - the probabilities are set. You can toss a coin 10 times, and the probability distribution may just be 30% heads, 70% tails. But with a significant number of tries, you will get closer and closer to the 50/50 distribution for the experiment. |
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thing is...it doesn't take a brain surgeon to see that one LOL when we proved it with countless graphs and examples.... |
And "luck" certainly ain't no variable in probability theory. At least not the last time I checked. :1orglaugh
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if you play this game once you have absolutely no reason to change, the statistics never help you when playing once
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thats like saying theres a 5% chance for it to snow today and a 95% chance it wont however it will either snow or it wont so 50% 50% sure it might snow but the chance of it snowing is 5% thats the math thats the point |
Let me explain it to you:
If the host says "STC is the Greatest" you have 2 choices. If you choes to say 'Yes' you will have 100% chance of dying of natural causes while choosing 'No' leaves you with the obvious alternative. Not responding is stupid like imaginary numbers. |
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that is taking it to the extreme, if you play once statistics never help and I can prove that with stats :winkwink: |
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