Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Scourge of Ages on April 11, 2012, 04:08:48 pm
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Flipping a coin gives a roughly 1/2 chance of landing on tails.
But the chance of flipping a coin twice, and getting tails both times is only 1/4, the chance of getting tails three times in a row is 1/8, and so on until the odds of flipping the same face x times is practically impossible. Even 10 times in a row is 1/512(?)
Right?
Hypothetically, if you've already flipped a coin and gotten 9 tails in a row, the chance of getting tails that 10th time is still 1/512. Is that mathematically accurate?
Logically though, when you flip the coin the 10th time, there's no luck magic saying that this coin has a 511/512 chance of coming up heads, it's still logically a 50% chance.
How could something like a coin toss have a 1/512 chance of landing one way, but only 50% chance to land the other? Where's the other 49.xx%?
Am I confusing math and logic here? Am I missing a vital concept or step?
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Hypothetically, if you've already flipped a coin and gotten 9 tails in a row, the chance of getting tails that 10th time is still 1/512. Is that mathematically accurate?
No, the odds of getting tails are 1/2. The 1/512 probability is assigned to the entire configuration of 10 tails.
Assuming order is relevant, any unique configuration of 10 coin flips has a 1/512 probability of being flipped - for example, H H T H T T T H H H is just as improbable as H H H H H H H H H H.
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I've thought about this, too. I, honestly, don't really know for sure, but I'm pretty sure the chances are 50/50 again at that flip. The 2^x power thing is probably measured at the beginning and doesn't hold true as you go through them.
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I've thought about this, too. I, honestly, don't really know for sure, but I'm pretty sure the chances are 50/50 again at that flip. The 2^x power thing is probably measured at the beginning and doesn't hold true as you go through them.
Correct: the 50/50 probability distribution describes the distribution of a single coin flip, whereas the 1/512 odds are for drawing one configuration (all heads) out of the 512 possible configurations of 10 coins*
*I have not actually checked the math to be sure there are 512 possible configurations but the general principle should be sound
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If all you're interested in is the number of configurations that give you a number of heads or tails X, the relevant equation is C(10,X)=10!/[X!(10-X)!], which will tell you the number of configurations that contain X heads/tails.
Plugging in 10 tails, you see C(10,10) = 10!/(10!(0!)) = 1, or 1 configuration of ten coins that will give you ten tails.
(shouldn't there be 1024 unique combinations overall or am i dumb)
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Indeed, 10 coin flips will yeild 210 = 1024 possible unique outcomes.
1 coin: H, T = 2 outcomes
2 coins: HH, HT, TH, TT = 22 = 4.
3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT = 23 = 8
so on and so forth.
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The doubt can be cleared by thinking that events that have already occurred have no influence on future events (assuming of course, independence).
While at the beginning you have a 1/1024 chance you'll get that particular configuration, at the moment in time you are going to toss your 10th coin, the chances are 1/2, assuming the previous outcomes were successes.
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i prefer the game of russian roulette. each time the trigger is pulled the probability changes. and its always in a finite set. for your standard game one of six game, the probability of getting shot can be between 1/6 and 1, depending on the number of times the trigger has been pulled. if youre first up, a good tactic is to pull 3 and pass, since the probability on the first 3 shots is more favorable than the second 3. many will just pull 2 and pass. being second is another matter. as youre not presented with the same odds. if the guy before you has pulled twice and then passed your odds of dieing are 1 in 4. again the odds are in your favor. pulling twice has the same odds as pulling 3 times with a fresh gun. under pulling can be just as dangerous for you as over pulling, because the gun might come back around. so you want the other guys next turn to have as low odds as possible. if the other gimp had originally pulled 3 times, then a one pull is acceptable as it still has fairly good odds, pulling twice steps into the realm of unfavorable odds, at 2/3 chance of death. the 5th pull is always a fair coin, and the 6th is certain death.
here are a couple basic starting strategies. the idea is to consume the pulls with the best odds of survival. leaving the other player with less favorable odds.
three pull pass strategy, pull order = (p1,p1,p1,p2,p1,p2)
p1 odds = 1/6 + 2/6 + 3/6 + 1/2(3/6) = 1.5
p2 odds = 1/3 + 1 = 1.333_
note that this strategy gives p1 lower odds, but no chance of getting the shot with the worst probability of 1.
two pull pass strategy (p1,p1,p2,p2,p1,p2)
p1 odds = 1/6 + 2/6 + 1/2(3/6) = 1
p2 odds = 1/4 + 2/4 + 1 = 1.75
this strat favors p1. and again ensures that p1 never gets the last chamber
of course when it comes to rr there are essentially 3 kinds of players. chicken ****s, risk takers, and the cautious.0.
a chicken **** will always pull the least number of times, once, before passing. risk takers will take chances to ensure the other player have less favorable odds, the often pull to or past 1/2 probability. cautious players will pull more than once, but will prefer to stay in favorable odds. the guy who pulls 3 times of the first round is clearly a risk taker, and the guy who pulls twice is merely cautious. the chicken**** will pull once and pass. the lines do get blurred as the the game progresses past the starting strategy. a risk taker will only pull once if they are at chamber 5. or if it is impossible to get to 1/2 odds without passing it. such as for the 3/5 pull, where the risk taker behaves the same as the cautious player. lets see what happens when we pit a chicken **** against a risk taker:
chicken **** vs risk taker (p1,p2,p2,p2,p1,p2)
p1 odds = 1/6 + 1/3(2/6) = 0.5
p2 odds = 1/5(2/10) + 2/5(4/10) + 3/5(6/10) + 1/2(5/10) + 1 = 2.7
in this case the chicken **** gets the best odds. but what if the risk taker goes first? the pull order is the same as the three pull pass strategy from above. you notice a huge difference in the resulting odds. of course in the three pull pass strategy, the risk taker did not pass beyond 1/2 odds, and in the above example he did in the 3/5 pull. what happens if he did not (goes cautious):
chicken **** vs cautious (or risk taker behaving as cautious) (p1,p2,p2,p1,p2,p1)
p1 odds = 1/6 + 1/3(2/6) + 1 = 1.5
p2 odds = 1/5(2/10) + 2/5(4/10) + 1/2(5/10) = 1.1
here it pays off not to be a panicking wuss. its not a huge difference in odds, and it does make sure the chicken **** gets the last chamber. flip the player order (risk taker replaced with cautious as a risk taker would never do this) and you get:
cautious vs chicken **** (p1,p1,p2,p1,p2,p1)
p1 odds = 1/6 + 2/6 + 1/3(2/6) + 1 = 1.8333_
p2 odds = 1/4 + 1/2(2/4) = 0.75
massive difference here. and being a pussy pays off.
what about cautious vs risk taker and vise versa? well thats covered in the first 2 starting strategies. three pull-pass being the same as risk taker vs cautious or chicken ****. two pull pass has the same pull order as the cautious vs risk taker game (could also be considered cautious vs cautious). turns out in both those scenarios the cautious player always wins out. now is a good time to point out that the overall odds for completing the full game sequence is only 1/6, so a partial sequence may yeild more favorable odds to one player or the other than indicated here. overall the odds of a 2 player game are always 50/50. somones gonna die, and so the odds of the gun are irrelevant. the risk taker's strategy focuses mostly on giving the other player less favorable odds by consuming the favorable pulls. the chicken **** strategy revolves around pulling the gun the least number of times. a risk taker may need to pull more than half of the total pulls, increasing the odds for the chicken ****. being cautious of course can counter either strategy, as you are on the assumption that the game is 50/50 anyway, and so will try to keep the game as fair as possible. getting the first turn may also play a role, as it lets you choose the starting strategy, and then play using their typical strategy. it pays for the first player to be a risk taker, and then fall back to cautious or even chicken ****, as the other player will assume you will continue to take insane risks.
you can screw things around changing the number of barrels, or the number of shots, or number of players. you can play with the order of the shots in a multi shot game, and with multiple players keep the gun in play until all rounds are discharged. so russian roulette gives you a lot of options. the deer hunter strategy is fun to consider: when deniro convinces the gooks to put 4 rounds in, and pulls twice. his odds were:
deniro = 4/6 + 5/6 = 1.333_
gooks = 1 + 1 + 1 + 1 = 4
it was obvious to see how that would play out. after to pulls he uses the remaining four rounds to kill the gooks and free himself and his nam buddies. how do you like them odds :D
tldr: russian roulette is awesome
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Russian Roulette (http://www.cracked.com/video_18385_worlds-dumbest-movie-villain.html)
Thanks guys, I think I got it, about the coin thing anyway.
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Kind of a late reply but here goes...
Your talking about the differences between permutations and combinations.
Permutations care about sequence, while combinations do not.
...or something like that. My Probability and Random Signals course wasn't very pleasant for me. :no: