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File: stay or switch.swf-(1.12 MB, 550x400, Other)
[_] Random mode Anon 3046117 Marked for deletion (old).
>> [_] Anon 3046125 inb4 50 replies
>> [_] Anon 3046140 I know this guy who literally believes he has bad luck. We play Magic: the Gathering together and he's often explained that dice will roll differently for him. Not if he touches the dice, just if the outcome is something he cares about. He said the average of the dice will always be predictably the same, but there will be higher variance for him, in a way that somehow leads to a bad outcome. I tried to explain that variance is measurable and predictable also, and he politely said that was my opinion and he disagreed with my opinion. The punchline (which only Magic players will understand) is that he always builds three or four color decks but he doesn't prioritize color fixing. He thought Brilliant Spectrum, the converge draw spell that has you draw up to four cards then discard two, was a good first-pickable card.
>> [_] Anon 3046147 Always switch.
>> [_] Anon 3046149 I remember reading about this. Apparently switching everytime is the better option. Reason being, they aren't eliminating at random. >># Tell him to work on his hypergeometric distribution game.
>> [_] Anon 3046153 >># As a huge FNM dweeb he's a moron. Especially with the current meta, unless your running all one mechanic like Slivers duo/mono decks will win hands down due to the extra color costs most late turn decks need. Even in a perfect game I've beat better decks simply because they couldn't combo well late-mid game.
>> [_] Anon 3046164 >># More often than not, a game will be decided due to consistent drops early on. I've beaten control blues with an aggro-combo gruul in as much as three to six rounds simply because I got fatties out fast. It should go without saying: minimum decksize, progressive cost and balanced color ratios, maximize key cards.
>> [_] Anon 3046183 >># I completely agree. I was talking about Battle for Zendikar booster draft and I play draft most often, but I'm sure the same holds in other formats. A lot can be done to increase the consistency of your deck. He builds decks with high variance, and instead of learning from his mistakes, blames luck. (My post wasn't closely related to the flash, but I think about this guy whenever questions of probability come up.)
>> [_] Anon 3046188 If your first pick was a car, there are two goats, they show one at random. If you switch, you'll land on the other goat. If your first pick was a goat, there is only one other goat, the have to open that door. Then if you switch, you'll get the car. The important point is that your first pick has only a 1/3 chance of being a car, so the second scenario is more likely. If you make the numbers more extreme, the right answer is more obvious. Imagine there are 100 doors and 99 goats. After your first pick, they can open 98 doors and show goats (they can always do this, regardless of which door you choose.) Do you want to stick with your one-in-a-hundred first pick, or switch to the door that will be correct if your first pick was wrong?
>> [_] Anon 3046194 >># >># >># >># >>>/tg/
>> [_] Anon 3046198 I like the story about the mathematician who belligerently denied the Monty Hall problem. He finally conceded after people showed him through the results of computerized simulations that that the increased likelihood of winning after switching empirically exists.
>> [_] Anon 3046201 >># >>>/fag/
>> [_] Anon 3046206 >># Currently at 9 wins in a row for switching. Rigged.
File: stay or switch.swf-(1.12 MB, Game)
[_] Stay or switch, /f/? Anon 1853435 Marked for deletion (old).
>> [_] Anon 1853447 fuck yeah i won a car
>> [_] Anon 1853448 wow this game is fucking retarded, why would it matter if you stay or switch
>> [_] Anon 1853452 >># Always switch. That way you have 1/2 instead of 1/3 chance. There were massive threads about this on /b/ for fucking ever...
>> [_] Anon 1853454 actually it does matter, i cant explain it, a friend studying math explained it to me sometime
>> [_] Anon 1853456 >># The correct choice is to Stay. It is a well known probability example, and the removal of a door does not change a 1/3 chance to a 1/2 chance.
>> [_] Anon 1853457 >># If you choose to stay you also have 1/2 of a chance to have picked the correct one /b/ is retarded
>> [_] Anon 1853462 love that car winning melody SOOOOOOOOOOO SATISFYING
>> [_] Anon 1853463 http://en.wikipedia.org/wiki/Monty_Hall_ problem let the doors be named x y and z. let x be the door you choose first. if x's chances of getting the car is equal to 1/3, then the chances of either of the other two doors having the car is 2/3. (Y + Z = 2/3) you are then shown which one of the other two doors actually has the goat - let's say it's door y. thus, you know that door y now has a 0/3 chance of having the car because it doesn't have it at all. Y + Z still = 2/3, however, because there is a 1/3 chance that X has the car and there are 3 doors. if we know that Y = 0, then that means 0 + Z = 2/3. thus, by simple algebra, Z has a 2/3 chance of having the car. this is one of the few demonstrable paradoxes that people get consistently wrong because life is not always intuitively correct
>> [_] Anon 1853469 >># All i know is I am now the proud owner of 3 cars and a goat.
>> [_] Anon 1853472 >># >># Switch is the proper choice. It's really quite silly.
>> [_] Anon 1853475 >># >># Reading the Wikipedia article did make me reconsider my thoughts on it, but it blows my mind too much
>> [_] Anon 1853480 >># Math has some things like that. Like how .999 repeating is the same as 1. That one seems more reasonable, but still trips people up sometimes.
>> [_] Anon 1853485 >># The Universe truly is a strange place to live in
>> [_] Anon 1853488 >># the answer's correct, but the reasoning's wrong. if you switch, you get a 2/3 chance of getting the car. >># >># common misconceptions. there's nothing compelling you to not choose the goat door that's revealed - it's just that it now has a 0/3 chance of having the car. there are still three doors to choose from.
>> [_] Anon 1853490 here's another explanation that doesn't involve algebra: let's call the first goat "goat A" and the second goat "goat B." at the start you can choose from any door. let's say you choose door number 1 for simplicity's sake, but this applies to every door. we also have to assume for the sake of the problem that the host will only reveal a goat - he'll never reveal the car. if you choose door number 1 and it actually has a car, then the host will reveal either goat A or B. if you choose to switch, then you get the goat. if you choose door number 1 and it actually has goat A, then the host will reveal goat B. if you choose to switch, then you'll get the car. if you choose door number 1 and it actually has goat B, then the host will reveal goat A. if you choose to switch, then you'll get the car. thus, switching has a 2/3 chance of getting a car.
>> [_] Anon 1853623 >># FUCK I finally get it. You aren't choosing between two doors, you are choosing between two unknown doors and a known goat. That's what makes the switch door more likely to be a car. Because the known door will never be a car.
>> [_] Anon 1853624 the math on this one always makes my head hurt.
>> [_] Anon 1853632 >># Right? It seems like all you've done is eliminate a door and so are left with a 50/50 choice, except that the door eliminated is based upon what door you chose. So if you pick goat A, they reveal goat B 100% of the time, and visa versa, and if you pick the car then they reveal either goat with a 50% chance. But that means that you still are picking between unknown 1, unknown 2 and a goat.
>> [_] Anon 1853633 Why won't this work right? I havn't stayed once, and I've played 25 games, won 15, so I have 60% Shouldn't it be 66%? If I play to 100 should it balance out?
>> [_] Anon 1853635 Another way of looking at this problem is this: you have a 2/3 chance of initially picking a goat, in which case the host effectively shows you precisely where the car is by eliminating the other goat. The only time the host does not show you the car is if you have picked a car to start with, which only happens 1/3 of the time.
>> [_] Anon 1853643 >># Probability is never truly clean or exact, if the sample is truly random (or close enough to it). There will be runs, even though switching gives you a car 2/3 of the time, where you will get 10 goats in a row.
>> [_] Anon 1853645 >># Exactly. Staying means you only have the same 1/3 chance of getting the car you had when you made the first choice, while switching converts that to only a 1/3 chance of losing the car you had.
>> [_] Anon 1853649 played 20 games won 10 times ...
>> [_] Anon 1853650 >># Random number generator
>> [_] Anon 1853660 fuck yeah i won a goat
>> [_] Anon 1853662 As long as a goat is revealed (which it always is), you have a 50/50 shot whether you switch or stay. FUCK EVERYTHING!
>> [_] Anon 1853678 I think that the 1,2/3rd people forget that since one goat is ALWAYS shown it means that one of the doors is now eliminated for a chance at a car. Door X, Y, Z. X has a goat Y and Z are the only 2 unknown doors left Ergo 1/2
>> [_] Anon 1853683 This is the best analogy I've seen for those who still don't understand why it's advantageous to switch: Imagine now that the game has 100 doors. You pick a door, and the host reveals that behind 98 other doors, there are all goats. Now, do you assume that your first guess, one door randomly picked out of 100, is correct, or do you switch to the one other door that hasn't been opened yet?
>> [_] Anon 1853685 dicks
>> [_] Anon 1853688 >># I am enjoying this. Imagine it with 30,000 doors, one for each person in a major stadium, and you pick one, and all but one other door is revealed to be empty. Do you stick with your original door? BALLS OF STEEL.
>> [_] Anon 1853689 >># No, not at all. It's all about the initial choice before a goat is revealed. If you select a goat initially, the only door that can be revealed will be the other goat, so when you switch you will get the car. The probability of selecting a goat initially is 2/3.
>> [_] Anon 1853690 >># i see your math there, but if you think about wouldn't z or y have a 1/2 probablity since one of them is always goat then that door '1/3' is not part of the equation either i.e goat=x car=y door=z (2x+y/3z) - (x/z)= x+y/2z or 1goat+1car/2doors
>> [_] Anon 1853694 >># I get it now.