Maximize Probability on a Weighted Coin
Daniel has a weighted coin that flips heads 5 2 of the time and tails 5 3 of the time. If he flips it 9 times, the probability that it will show heads exactly n times is greater than or equal to the probability that it will show heads exactly k times, for all k = 0 , 1 , … , 9 , k = n .
If the probability that the coin will show heads exactly n times in 9 flips is q p for positive coprime integers p and q , then find the last three digits of p .
The answer is 888.
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1 solution
Changed. Thanks, I didn't realize that both yielded the same probability.
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Great. Yeah, I was surprised that the two yielded precisely the same probability. I'll delete the comment in my solution now that the change has been made.
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Here's a challenge: If I flip it x times, what is the appropriate value of n , in terms of x ?
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@Daniel Liu – I was thinking of the same generalization, including making the probability of heads p , (and thus of tails 1 − p ). As we've seen for x = 9 , we can't just take the closest integer to p x as being the "obvious" solution. This is a challenge; I'll have to give it some thought. :)
@Daniel Liu – O.k., what we're after here is the mode of the distribution, which is outlined in Proposition 1, item (5) of this paper .
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@Brian Charlesworth – Dang, looks like someone else already wrote a paper on it. The general form of this problem is part of a paper I wrote (not yet published)
It addresses the coefficient of x that is the maximum, in the expression ( m x + b ) n
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@Daniel Liu – Sorry about that. I'm impressed, though, that you're already writing papers for submission.
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@Brian Charlesworth – Not formally though, it's more of a hobby. Although, I've been trying to get AwesomeMath to publish my papers, but it's hard getting something accepted there.
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@Daniel Liu – I hadn't heard of AwesomeMath before. I've just had a look; they have some great content, both articles and problems/solutions. Hope that you manage to get something accepted there someday. :)
The probability that it will show heads exactly n times is
( n 9 ) ∗ ( 5 2 ) n ∗ ( 5 3 ) 9 − n .
Since 9 ∗ ( 5 2 ) = 5 1 8 we can be certain that the value for n we are looking for is either 3 or 4 . Now it just so happens that both of these values yield a probability of
q p = 1 9 5 3 1 2 5 4 8 9 8 8 8 ,
so the desired solution is 8 8 8 .