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For an unfair coin toss that produces heads with probability P, what is the value of P that will result in 0.5% (i.e. 0.005) chance of getting exactly x tails out of y tosses?

i.e. is there a general solution to ${y\choose x}P^{y-x}(1-P)^x=0.005$ for $P?$

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    $\begingroup$ When you say .5% do you mean the probability multiplied by 100 to get a percent value, or is that just the probability with a percent put at the end? $\endgroup$ – Dunka Dec 13 '14 at 18:29
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    $\begingroup$ When you say getting $x$ tails, do you mean exactly $x$ tails or at least $x$ tails? $\endgroup$ – peterwhy Dec 13 '14 at 18:30
  • $\begingroup$ Dunka, I meant what I wrote 0.5% = 0.005. peterwhy, I mean exactly x tails. $\endgroup$ – user1745038 Dec 13 '14 at 21:23
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Since the probability of getting $x$ tails out of $y$ tosses is ${y\choose x}P^{y-x}(1-P)^x$, we just need to solve ${y\choose x}P^{y-x}(1-P)^x=0.005$ for $P.$

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  • $\begingroup$ Thanks user105013, but this is now a y degree polynomial with y solutions. I was thinking that some of them would be imaginary or negative, but for x=1,y=2 there are two real solutions between 0 and 1: 0.0025062814466900174 and 0.9974937185533099. Is there a general solution to this system i.e. some smart application of the binomial theorem that yields all of the solutions for a given y? $\endgroup$ – user1745038 Dec 13 '14 at 21:27
  • $\begingroup$ For the $x=1, y=2$ case, your two values are complementary, which makes sense as you can switch the roles of "heads" and "tails" in this case due to symmetry. $\endgroup$ – mathmandan Dec 16 '14 at 18:25

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