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On January 26 2016 01:04 Toadesstern wrote:I wasn't entirely sure what part of Germany / Austria / Switzerland that could be aside from having to go south. So just take it as as literal as possible since it's people living in the alps (mountains), right? At least it sounds funnier than Berlin dialect. All I can do as someone from around Frankfurt is turn "Ich" into "Isch" while changing every k into a g for every word in a sentence 
The first time I was in Berlin I just switched to English to buy my public transport ticket from the friendly lady at the ticket office since neither of us had any clue what the other was trying to say in their own version of German ;P
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What percentage of government revenue comes from seigniorage in western countries like US, Canada, Germany, etc?
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Zurich15365 Posts
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Anyone have any experience shipping a motorcycle cross country in the states?
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Been a long day and I may be a bit intoxicated, could someone explain to me again (sorry Professor whatever your name was) how the odds on winning coin flips works again.
I know each time it's flipped there's a 50-50 chance of guessing correctly but what are the odds of guessing right 5 out of 5 times and how does that work?
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That is called a Bernoulli-chain. The answer is the Binomial Distribution, dependent on the probability of your coin hitting Heads (p), the amount of times you throw the coin (n), and the amount of Heads that you want. (k)
The complete formula is:
B(n, k, p)= (n choose k)*p^k*p^(n-k)
where (n choose k) is the Binomial Coefficient n!/(k!*(n-k)!)
Or in your case: (10 choose 5)0.5^5*(1-0.5)^(10-5)
Lets dissect that formula.
p^k*p^(n-k) is the probability of any exact chain like "HHTTTHTTHH" that involves exactly k = (5) Heads (And thus (n-k) = (10-5) Tails.)
(n choose k) is the amount of different such chains that are possible, because it is the amount of different ways you can choose k (5) items out of a set of n (10), while ignoring the order in which they are pulled.
n!/(n-k)! = something like 10*9*8*7*6
Which you use because you have 10 choices for which place the first H is at, 9 choices where the second H is at (because one is already occupied), and you divide that by
k!= something like 1*2*3*4*5
because that is the amount of different such chains you can create which only differ in the order in which they are drawn, but for our purposes it does not matter if the first H is on position 3 or the last H.
This is a bit harder to explain in text form on a forum as opposed to in person when you are capable of drawing things on paper, rest assured that i am usually less opaque with my students.
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On February 02 2016 15:34 GreenHorizons wrote:
Been a long day and I may be a bit intoxicated, could someone explain to me again (sorry Professor whatever your name was) how the odds on winning coin flips works again.
I know each time it's flipped there's a 50-50 chance of guessing correctly but what are the odds of guessing right 5 out of 5 times and how does that work?
Let me give you a bit shorter answer than Simbertos (that is correct, but will be hard to follow if you don't know maths well). Not sure which level you are at, so choose answer accordingly. 
The principle here is that you multiply probabilities if you want all to happen:
First flip turning up heads is 50%, which is a probability of 0.5.
Second flip also has a 0.5 probability of turning up heads.
Probability of both turning up head? 0.5*0.5 = 0.25.
A way of understanding this is to imagine the situation where you do your rolls 1000 times. About 500 of them will turn up head first flip. And about 250 of those 500 will turn up head on the second roll as well. So 250 out of 1000 will turn up head and head, which is 0.25 = 25%.
Continuing on to 5 flips, you get with the same princinple
0.5*0.5*0.5*0.5*0.5 = 0.5^5 = 0.03125
Just above 3%.
If you start asking what the chances are to get 4 heads and 1 tail, it turns a bit more complicated (as the tail can appear in any of the five flips), and I refer to Simbertos post for that. 
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Oh sorry, i totally misread that (It was early in the morning) I read his question to be "guess correctly 5 out of 10 times". Your answer is much better to the actual question.
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On February 02 2016 18:56 Simberto wrote:
Oh sorry, i totally misread that (It was early in the morning) I read his question to be "guess correctly 5 out of 10 times". Your answer is much better to the actual question.
Does anyone post here that isn't drunk or sleep deprived? >_>
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what i know is that a coin is usually heavier on one side(it weights more due to the engravings and such) so those 50/50 odds to start with are skewed in favor of the lighter side.
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So if we assume that we don't really know the probability p_c of the coin, but we estimate that the probability distribution P(p_c) of the true coin-probability is a normal distribution around 0.51 with a width of 0.005, limited to the (0,1) interval, but we don't know which side it is biased towards, so it is actually a sum of two normal distributions around 0.49 and 0.51, how many times do we have flip the coin to have a 95% probability of having 95% certainty of whether the coin probability p_c is larger or smaller than 0.5?
You guys want to start a Tuesday Power Calculation tradition here in this thread? Could be fun! FUN!!
A bit like day9's Funday Monday, but it's not Monday, and it's not fun?
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United States43989 Posts
On February 02 2016 15:34 GreenHorizons wrote:
Been a long day and I may be a bit intoxicated, could someone explain to me again (sorry Professor whatever your name was) how the odds on winning coin flips works again.
I know each time it's flipped there's a 50-50 chance of guessing correctly but what are the odds of guessing right 5 out of 5 times and how does that work?
Two equally (for the purposes of our calculations) likely results means the chance halves with each consecutive correct guess.
1 flip can go H or T so you have 1/2
2 flips can go HH, HT, TH, TT so 1/4
3 flips can go HHH, HHT, HTT, TTT, HTH, THH, THT, TTH so 1/8.
and so forth
by 5 flips we're at 32 different possible outcomes of which just one is HHHHH.
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are we all forgetting about the possibility of the coin landing on the side in here?
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Patch 3.1.2 seems unnecessary to me. What do you guys think?
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United States43989 Posts
On February 03 2016 00:32 Toadesstern wrote:
are we all forgetting about the possibility of the coin landing on the side in here?
Or not landing at all! What if a sparrow catches it in a coconut?
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On February 03 2016 01:24 KwarK wrote:Show nested quote +On February 03 2016 00:32 Toadesstern wrote:
are we all forgetting about the possibility of the coin landing on the side in here? Or not landing at all! What if a sparrow catches it in a coconut?
That counts as heads, so says the Magna Carta.
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On February 03 2016 01:15 EsX_Raptor wrote:
Patch 3.1.2 seems unnecessary to me. What do you guys think?
You may not have needed it, but I had enough trouble with the idle connection timeout to be glad it was finally fixed.
(I would agree that OAuth2 endpoints in OpenID-Connect flows wasn't that important)
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On February 03 2016 01:24 KwarK wrote:Show nested quote +On February 03 2016 00:32 Toadesstern wrote:
are we all forgetting about the possibility of the coin landing on the side in here? Or not landing at all! What if a sparrow catches it in a coconut?
Is it an european sparrow or an american sparrow ?
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context! it's about the US election so unless someone imports large amounts of european sparrows to screw with a cointoss it's obviously going to be an american one.
actually, it probably was about the "what are the chances to get 50 states right" and not an actual cointoss.
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United States43989 Posts
Nope, the Bernie vs Clinton tie was settled by a series of coin tosses in which Bernie got very unlucky. Literal cointosses. GH suspects foul play, not fowl play.
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EPT
