Probability of a Probability

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JPC
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Main Question or Discussion Point

Hey, lets take for example the probabilities when u throw 100 times a coin (lets say it cant be corner)

ok , the probability is obviously 50 : 50

but , whats the probability that your results are same as the probability for this case
 

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radou
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The probability of what is '50 : 50' ? :wink:
 
JPC
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well

well if u throw 100 times a coin
the probability is : 50 heads, 50 tails
 
D H
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In other words, you want the probability of throwing exactly 50 heads (and 50 tails) in 100 tosses, any order of heads and tails.

Is this homework? If so, please show what you have done. If not, I (or someone else here) will show you how to compute such a probability.
 
ssd
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well

well if u throw 100 times a coin
the probability is : 50 heads, 50 tails
Not for any coin. Only for an unbiased coin P(Head)=P(Tail) in a single toss = 0.5, assuming that there is no other outcome of the toss.
Getting 50 heads and 50 tails is an event, it is not a probability.
Again, if P(H)=0.5 then "50 heads" is the "expected" number of heads in 100 throws....it is not a probability.
Now comming to your question: If you denote the no. of heads in 100 tosses by X, then X~Bin(100,0.5). Find P(X=50).
 
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JPC
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If you denote the no. of heads in 100 tosses by X, then X~Bin(100,0.5). Find P(X=50).
What does that mean , i don'k know all the conventions yet ?

and its not for homework
 
radou
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ssd was talking about random variables. You may want to google-up that term.
 
D H
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In particular, ssd was talking about a random variable with a binomial probability distribution, which applies only if the coin is unbiased and tosses are independent. Assuming an unbiased coin and indepdendent tosses, the answer to the original question is
[tex]\frac{100!}{50!\,50!}\;\frac 1 {2^{100}} \approx 0.0796[/tex]
 
ssd
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In particular, ssd was talking about a random variable with a binomial probability distribution, which applies only if the coin is unbiased and tosses are independent. Assuming an unbiased coin and indepdendent tosses, the answer to the original question is
[tex]\frac{100!}{50!\,50!}\;\frac 1 {2^{100}} \approx 0.0796[/tex]
Right you are. I did not mention the fact that the sum of Bernoullian trials will be Binomial if those trials are independent. Actually I really was focused on the statement of the problem.
 
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JPC
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hey , u didnt put any operator between the 50! and 50!
is it multiply ?

and , u didnt put any operator between the two fractions, is it also multiply ?

because , if it is , doesnt look very much like 0.0796

100! : 5050
50! : 1275
50!² : 1625625

5050 / (1625625 * 2^100) = 2.45...





and, also, in general terms, what is the logic between this equation ?
 
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I'm not that good at probability, but the vinculum affects the order of operations, i.e.
[tex]\frac{100!}{50!50!} \frac{1}{2^{100}} = ( 100! / 50!50! ) * ( 1 / 2^{100} ) = 0.0795892374[/tex]
 
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CRGreathouse
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because , if it is , doesnt look very much like 0.0796

100! : 5050
50! : 1275
50!² : 1625625

5050 / (1625625 * 2^100) = 2.45...
Wha...? 50! is a huge number with 65 digits, it's a lot bigger than 1275. 1275 = 50 + 49 + ... + 2 + 1, while 50! = 50 * 49 * ... * 2 * 1.
 
JPC
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oh damn , i made a mistake
the ! is a multiplication , not addition, not 0.5(n² + n)

i see my mistake
 
JPC
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but , how do u translate x! then ?

like whats the equation to find it
like if u want to find the sum of all the numbers from 0 to x its : 0.5(x + x²)
but what is it for the multiplication ?
 
CRGreathouse
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but , how do u translate x! then ?

like whats the equation to find it
like if u want to find the sum of all the numbers from 0 to x its : 0.5(x + x²)
but what is it for the multiplication ?
There's no polynomial that gives a value close (for any reasonable definition of "close") to x! for all positive x.

There are some approximation formulas; (x/e)^x ~ x! would be one of the simplest*, where e = 2.71828....

If you want the exact answer you're just going to have to multiply it out, though. For this problem that's pretty much all you can do. Notice, though, that 100!/50! = 100 * 99 * ... * 51 * 50 * 49 * ... * 2 * 1 / (50 * 49 * ... * 2 * 1) = 100 * 99 * ... * 52 * 51.

* I hope I got this right, it's from memory. There are much better ones if you need more accuracy; Gosper's reformulation of Stirling's estimate is pretty good.
 
JPC
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better solution (if x is smaller than 1 million)
learn visual Basics , and make a program to do it for you using 'do until' - 'loop'


But whats the logic in that equation ?
how can i make an equation like that for other cases ?
whats the clear syntax ?
 
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