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Why this is a reasonable approximation?

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    (Original post by gff)
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    Can you elaborate on the convergence of this power series you've posted? I don't see how it works out.
    I'm pretty sure if you substitute it into the equation Mx=e^x then LHS=RHS. I didn't check the convergence of it although I think comparison to a standard geometric series guarantees the convergence
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    (Original post by TheMagicMan)
    I'm pretty sure if you substitute it into the equation Mx=e^x then LHS=RHS. I didn't check the convergence of it although I think comparison to a standard geometric series guarantees the convergence
    I may be a bit ignorant, but to deduce the relationship I let x,M \to \infty, and these power series don't fit into that.
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    I interpret this as follows.

    \displaystyle f(x) = \log(x) \Rightarrow x = e^{f(x)} = \sum_{n=0}^{\infty} \frac{f^n(x)}{n!}.

    I well may be wrong, but don't see the method.
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    (Original post by hassi94)
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    (Original post by Slumpy)
    This is right.





    (Original post by gff)
    ...
    Any hints on how to approximate y in w=lnM+y? I assume y is a random variable.
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    (Original post by Dog4444)







    Any hints on how to approximate y in w=lnM+y? I assume y is a random variable.
    The chance of you having the biased coin is not 1/129. If you just picked a coin at random it would be, but if you picked up a coin, flipped it a thousand times, and got heads every time, the odds would be hugely in favour of you having picked the biased coin. Can you see this makes sense?

    (In the case in question there's 2/3 chance the coin you have is the biased one.)
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    (Original post by Slumpy)
    The chance of you having the biased coin is not 1/129. If you just picked a coin at random it would be, but if you picked up a coin, flipped it a thousand times, and got heads every time, the odds would be hugely in favour of you having picked the biased coin. Can you see this makes sense?

    (In the case in question there's 2/3 chance the coin you have is the biased one.)
    Nope.
    There are 129 coins. I pick one and flip it over 9000 times. But the chance that it's biased is still 1/129 no matter what?
    And when 2/3 comes from?
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    (Original post by Dog4444)
    Nope.
    There are 129 coins. I pick one and flip it over 9000 times. But the chance that it's biased is still 1/129 no matter what?
    Ok, try this:

    I have 2 coins, one is double headed, one is normal.
    If I've flipped a coin 3 times and got heads every time. The odds of this happening are:
    P(picked an unbiased coin)*P(heads each time)=1/2*(1/2)^3=1/16
    P(picked biased coin)*P(heads every time)=1/2*1^3=1/2

    The other 7/16's are situations in which you pick an unbiased coin, and don't get 3 heads.
    So, with the 3 heads options, the probability that the coin was biased was (1/2)/(1/2 + 1/16)=8/9
    Yes?
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    (Original post by Slumpy)
    So, with the 3 heads options, the probability that the coin was biased was (1/2)/(1/2 + 1/16)=8/9
    Yes?
    Can't follow this line. (1/2)/(1/2 + 1/16)=9/32.
    And still can't follow the logic. What does other 9/32 mean then (probability the coin is unbaised?) and what (32-2*9)/32 mean then?
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    (Original post by Dog4444)
    Can't follow this line. (1/2)/(1/2 + 1/16)=9/32.
    No, what I wrote there is correct.
    ( http://www.wolframalpha.com/input/?i...2F2+%2B+1%2F16 if you won't take it on faith:p:)
    Edit-to check; it's division, not multiplication.

    (Original post by Dog4444)
    But still can't follow the logic.
    Ok, consider an even simpler(if slightly odder) case. You have 2 coins. One is heads on both sides, the other is tails on both.
    You pick a coin and flip it once, getting a heads. Can you see the probability that you picked the double heads coin is 1?

    (Original post by Dog4444)
    What does other 9/32 mean then (probability the coin is unbaised?) and what (32-2*9)/32 mean then?
    Edit: Don't really know where you're getting this stuff from tbh.
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    (Original post by Slumpy)
    No, what I wrote there is correct.
    ( http://www.wolframalpha.com/input/?i...2F2+%2B+1%2F16 if you won't take it on faith:p:)
    Edit-to check; it's division, not multiplication.



    Ok, consider an even simpler(if slightly odder) case. You have 2 coins. One is heads on both sides, the other is tails on both.
    You pick a coin and flip it once, getting a heads. Can you see the probability that you picked the double heads coin is 1?



    Edit: Don't really know where you're getting this stuff from tbh.
    Yeah, now I see how crappy my logic is.

    But still can't follow why you divided instead of multiplying. Can you bring more maths in it like P(A|B) and stuff?
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    (Original post by Dog4444)
    Yeah, now I see how crappy my logic is.

    But still can't follow why you divided instead of multiplying. Can you bring more maths in it like P(A|B) and stuff?
    So, we have the 2 coins, coin X is HH, coin Y is normal.
    Event A is the event you pick coin X. Event B is the event you throw three heads in a row.

    P(A)=1/2, pretty clearly, but we want P(A|B)
    P(A|B)=P(AnB)/P(B)
    Now, we know that P(B)=(1/2)*P(B|A)+(1/2)*P(B|A'), yeah? (This is the law of total probability: http://en.wikipedia.org/wiki/Law_of_total_probability )

    so you get P(A|B)=2*P(AnB)/(P(B|A)+P(B|A')).

    P(B|A)=P(BnA)/P(A)=2P(BnA) and P(B|A')=P(BnA')/P(A')=2P(BnA') -this just comes from the definitions

    So we now have P(A|B)=P(AnB)/(P(AnB)+P(A'nB))
    Now, P(AnB)=P(A)=1/2, and P(A'nB)=1/2*(1/2)^3=1/16
    So we get P(A|B)=(1/2)/(1/2+1/16)
    Does this make sense?
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    (Original post by Dog4444)
    Any hints on how to approximate y in w=lnM+y? I assume y is a random variable.
    Well, have you read my post?
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    It is a matter of rearranging.

    Given that we write w = \log(M) + c, since c is in a sense arbitrary, we have

    c = w - \log(M) = (\log(M) + \log(w)) - \log(M) = \log(w) \approx \ \log(\log(M) + c)

    This gives a recursion, which I don't know how TheMagicMan has turned into power series.

    w = \log(M*(\log(M*(\log(M) + \{\bar c\})))

    etc..

    For example, for M = 10^{75}, you can approximate w by \log(10^{75}*\log(10^{75}*\log(1  0^{75})))
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    (Original post by Slumpy)
    So, we have the 2 coins, coin X is HH, coin Y is normal.
    Event A is the event you pick coin X. Event B is the event you throw three heads in a row.

    P(A)=1/2, pretty clearly, but we want P(A|B)
    P(A|B)=P(AnB)/P(B)
    Now, we know that P(B)=(1/2)*P(B|A)+(1/2)*P(B|A'), yeah? (This is the law of total probability: http://en.wikipedia.org/wiki/Law_of_total_probability )

    so you get P(A|B)=2*P(AnB)/(P(B|A)+P(B|A')).

    P(B|A)=P(BnA)/P(A)=2P(BnA) and P(B|A')=P(BnA')/P(A')=2P(BnA') -this just comes from the definitions

    So we now have P(A|B)=P(AnB)/(P(AnB)+P(A'nB))
    Now, P(AnB)=P(A)=1/2, and P(A'nB)=1/2*(1/2)^3=1/16
    So we get P(A|B)=(1/2)/(1/2+1/16)
    Does this make sense?
    Yes, thanks. Now I'm wondering how he managed to do everything in his head.

    (Original post by gff)
    Well, have you read my post?
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    It is a matter of rearranging.

    Given that we write w = \log(M) + c, since c is in a sense arbitrary, we have

    c = w - \log(M) = (\log(M) + \log(w)) - \log(M) = \log(w) \approx \ \log(\log(M) + c)

    This gives a recursion, which I don't know how TheMagicMan has turned into power series.

    w = \log(M*(\log(M*(\log(M) + \{\bar c\})))

    etc..

    For example, for M = 10^{75}, you can approximate w by \log(10^{75}*\log(10^{75}*\log(1  0^{75})))
    Nice, thanks.
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    (Original post by gff)
    I may be a bit ignorant, but to deduce the relationship I let x,M \to \infty, and these power series don't fit into that.
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    I interpret this as follows.

    \displaystyle f(x) = \log(x) \Rightarrow x = e^{f(x)} = \sum_{n=0}^{\infty} \frac{f^n(x)}{n!}.

    I well may be wrong, but don't see the method.
    Iw asn't using power series....I was just using log laws
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    (Original post by hassi94)
    I don't think so.

    I think we'd say:

    P(coin is unbiased) = 128/129
    P(coin is biased) = 1/129
    P(unbiased coin is heads 8 times) = (1/2)^8 = 1/256
    P(biased coin is heads 8 times) = 1

    So the probability of an unbiased coin doing what has been done is 1/258

    And the probability of a biased coin doing what is done is 2/258

    So the probability we have a biased coin is 2/3 and the probability we have a unbiased coin is 1/3

    Then finally the probability it will be heads again is (2/3)(1) + (1/3)(1/2) = 5/6

    I don't really know if this is correct, and sorry I couldn't use proper notation - I've never really done probability except a tiny bit in S1 so I've just done this out of logic alone.
    Sorry, I didn't understand the bold bit, how did the 6 change to an 8?

    Also could you explain the underlined if you can
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    is question 4, 1600 by any chance?

    6000 numbers
    remove multiples of 2 (half of them)
    6000-3000=3000
    remove multiples of 3 (half of which are multiples of 2)
    3000-(2000/2)=2000
    remove multiples of 5 ( removing multiples of 2 and 3, there are 4 between 5-60, therefore 400 up to 6000)
    2000-400=1600
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    (Original post by Dmon1Unlimited)
    Sorry, I didn't understand the bold bit, how did the 6 change to an 8?

    Also could you explain the underlined if you can
    Okay sorry I didn't make the calculations I did clear:

    By "So the probability of an unbiased coin doing what has been done is 1/258" I meant the probability of an unbiased coin having got heads 8 times is (128/129)*(1/256) = 1/258


    As for the underlined bit:

    We have a situation in which we have gotten heads 8 times. The probability of this happening is 1/258 for the unbiased coin (as shown above) and 2/258 for the biased coin. Therefore, it is twice as likely that the coin being talked about in the question is a biased coin. So then we can say the probability that the coin the question is referring to is a biased coin is 2/3, and 1/3 for unbiased.

    Sorry I'm not great at explaining - I'm not very good at probability
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    (Original post by Dmon1Unlimited)
    is question 4, 1600 by any chance?

    6000 numbers
    remove multiples of 2 (half of them)
    6000-3000=3000
    remove multiples of 3 (half of which are multiples of 2)
    3000-(2000/2)=2000
    remove multiples of 5 ( removing multiples of 2 and 3, there are 4 between 5-60, therefore 400 up to 6000)
    2000-400=1600
    Sounds right, I got this question in interview

    I did it in a pretty round-about way (from now on I'll write number of multiples of x as M(x)):

    M(2) = 3000
    M(3) = 2000
    M(5) = 1200

    If we take away all of these from 6000 we are also taking 2 groups of M(6), M(10) and M(15) - so we need to add these back in - however this will add in an extra M(30) so we must take that away.

    M(6) = 1000
    M(10) = 600
    M(15) = 400
    M(30) = 200

    So the answer is 6000 - M(2) - M(3) - M(5) + M(6) + M(10) + M(15) - M(30)= 6000 - 3000 - 2000 - 1200 + 1000 + 600 + 400 - 200 = 1600

    Interview stress made me overcomplicate things I think haha
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    (Original post by hassi94)
    Sounds right, I got this question in interview

    I did it in a pretty round-about way (from now on I'll write number of multiples of x as M(x)):

    M(2) = 3000
    M(3) = 2000
    M(5) = 1200

    If we take away all of these from 6000 we are also taking 2 groups of M(6), M(10) and M(15) - so we need to add these back in - however this will add in an extra M(30) so we must take that away.

    M(6) = 1000
    M(10) = 600
    M(15) = 400
    M(30) = 200

    So the answer is 6000 - M(2) - M(3) - M(5) + M(6) + M(10) + M(15) - M(30)= 6000 - 3000 - 2000 - 1200 + 1000 + 600 + 400 - 200 = 1600

    Interview stress made me overcomplicate things I think haha
    This is basically inclusion exclusion. Anyway a slightly different method is to consider the numbers up to thirty and multiply by 200
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    (Original post by TheMagicMan)
    This is basically inclusion exclusion. Anyway a slightly different method is to consider the numbers up to thirty and multiply by 200
    Never heard of inclusion exclusion Yeah I think that is the best way to do it (the way you said).
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    (Original post by hassi94)
    Never heard of inclusion exclusion Yeah I think that is the best way to do it (the way you said).
    Inclusion exclusion is exactly what you did there...it's just the fancy name for that method of alternating subtracting and adding when calculating the size of sets

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