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Topic: Biased Coin (Read 2051 times) |
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marsh8472
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Biased Coin
« on: Sep 14th, 2012, 2:17am » |
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A biased coin is flipped k times. Heads comes up x times. Tails comes up k-x times. What is the probability that heads is more probable than tails on this coin? (in terms of k and x)
I just made this up and am new here. If this falls within the hard category feel free to move it there. 
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Grimbal
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Re: Biased Coin
« Reply #1 on: Sep 14th, 2012, 5:32am » |
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You need to know the prior distribution of the bias.
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towr
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Re: Biased Coin
« Reply #2 on: Sep 14th, 2012, 6:17am » |
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I'd say, when in doubt, take a uniform prior distribution. That's the only one that doesn't contain information.
Or you could add the prior as an extra free parameter in the equation and see where that leads you.
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marsh8472
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Re: Biased Coin
« Reply #3 on: Sep 14th, 2012, 9:21am » |
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on Sep 14th, 2012, 6:17am, towr wrote:I'd say, when in doubt, take a uniform prior distribution. That's the only one that doesn't contain information.
Or you could add the prior as an extra free parameter in the equation and see where that leads you. |
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Yeah I would assume a uniform in this.
I found a similar problem when I looked up "Maximum_likelihood_estimate" for wikipedia
Maybe this would help
http://izbicki.me/blog/how-to-create-an-unfair-coin-and-prove-it-with-ma th
alpha = number of heads
beta = number of tails+1
Then I think the answer is the integral from 0 to 0.5 of that pdf?
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| « Last Edit: Sep 14th, 2012, 9:42am by marsh8472 » |
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towr
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Re: Biased Coin
« Reply #4 on: Sep 14th, 2012, 11:48am » |
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Something doesn't seem right with that equation; unless it represents something other than what I'd expect.
I'd expect that the exponents of x and (x-1) should be the number of heads and tails respectively; and that the value you get is the probability of getting h heads and t tails when the probability of heads is x.
I'm not much of a fan of the Gamma function, I prefer just using factorials and related functions (because I'm more familiar with them, and they're clearer in this case, since the parameters match the values you're working with, which avoids off-by-one errors).
So, I'd say f(x, h, t) = (h+t)!/h!/t! * xh * (1-x)t
And then what we're actually after is:
 11/2 f(x,h,t) dx / 10 f(x,h,t) dx
The sum of all outcomes (h,t) where the coin is biased in heads favor (x > 1/2), divided by the sum of all outcomes (h,t)
Note that we can just eliminate the factor (h+t)!/h!/t! above and below the dividing line, since it is a constant (and thus it doesn't matter in the end whether we use the Gamma function or factorials). However, I haven't been able to simplify the equation further.
You could use the incomplete Beta function, but that's just begging the question imo.
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| « Last Edit: Sep 14th, 2012, 12:02pm by towr » |
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marsh8472
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Re: Biased Coin
« Reply #5 on: Sep 15th, 2012, 5:58am » |
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on Sep 14th, 2012, 11:48am, towr wrote:Something doesn't seem right with that equation; unless it represents something other than what I'd expect.
I'd expect that the exponents of x and (x-1) should be the number of heads and tails respectively; and that the value you get is the probability of getting h heads and t tails when the probability of heads is x.
I'm not much of a fan of the Gamma function, I prefer just using factorials and related functions (because I'm more familiar with them, and they're clearer in this case, since the parameters match the values you're working with, which avoids off-by-one errors).
So, I'd say f(x, h, t) = (h+t)!/h!/t! * xh * (1-x)t
And then what we're actually after is:
11/2 f(x,h,t) dx / 10 f(x,h,t) dx
The sum of all outcomes (h,t) where the coin is biased in heads favor (x > 1/2), divided by the sum of all outcomes (h,t)
Note that we can just eliminate the factor (h+t)!/h!/t! above and below the dividing line, since it is a constant (and thus it doesn't matter in the end whether we use the Gamma function or factorials). However, I haven't been able to simplify the equation further.
You could use the incomplete Beta function, but that's just begging the question imo. |
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Yes that's it  That was easier than I thought it would be
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