Recent content by Zafa Pi

Is this better than coin flipping?

I agree with what you say about this space and function. However, the same is true if 0 is eliminated, i.e. you don't need the fixed point. Again, if we are on the circle group and this time θ1 = 1, θn+1 = θn + 1, we get density, f is continuous, no fixed point, and no separation. There are...

Let M = {p, x1, x2, x3, ...} be a metric space with no isolated points. f: M → M is continuous with f(xn) = xn+1, and f(p) = p. We say f separates if ∃ δ > 0, ∋ for any y and z there is some n with |fn(y) - fn(z)| > δ, where fn+1(y) = f(fn(y)). QUESTION: Does f separate?

I think that the way the problem was formulated is problematic. I attempted to point that out to @StoneTemplePython in post #72, but received no response. For example, If an urn contains 3 whites and one black, then what is the probability of drawing (with replacement) three balls such that they...

Let N be the number of balls in each urn, W the number of white balls in urn 1, b and w the number of black and white balls in urn 2, n > 2 the number drawn from each urn. With my interpretation in #73 of the problem we get: (W/b+w)n = (b/b+w)n + (w/b+w)n. Multiplying by (b+w)n we have Wn = bn...

The only way I can make a reasonable problem from the way it was posed is as follows: The probability of getting all white or all black from urn 2 is prob all white + prob all black. In that case it is impossible due to Fermat's Theorem and n ≥ 3. That's what Fermat told me.

I find #6 ambiguous. If an urn has 3 whites and 1 black what do you think is the probability of drawing 3 that are either all white or all black?

If I understand you correctly it seems as though you will eventually end up with all of the same paths as I did, and your approach isn't any faster.

OK,OK, you are correct. Nice observation. My error was in round 4. I omitted a 3p4 = .0768 which, with exact calculations brings the total probability to 1. The expected number of rounds then becomes 3.0831744. So what is easier than enumeration? Run a zillion trials on a computer and take an...

Wait, I had round 4 correct (except the first bold + shouldn't have been bold). So where is the error?

Change ⅕ to p/2. 1•p/2 = .2 2•2pp/2 = .32 3•(2p3 + p3 = 3p3) = .576 4•((p3 + 3p3 +3p3(corrected))p/2 + 6p4 = 9.5p4) = .9728 5•(5p4p/2 + 4p5 = 6.5p5) = .3328 6•(6p5p/2 + 11p6 = 14p6) = .344 7•(p6p/2 + p7 = 1.5p7) = .0172 8•(p7p/2 + 2p8 = 2.5p8) = .013 Adding the bolds = 2.7758 ≈ 2.8. You're...

I have a picture of Fermat and he talks to me, so I can't take full credit. Solution to optional: The expected number of rounds for the game to terminate = ∑n⋅prob game ends on nth round, n ∈ [1,8] (see post #60). Prob game ends on nth round = prob "other" is rolled at the nth turn + prob game...

Solution to #3 The average win for heads or tails is 2 with probability 4/5. A risk neutral player will play as long as her expectation is positive. She will quit if her expectation is ≤ 0. If she has accrued x, then her expectation for the next play 2⋅4/5 - x⋅1/5. That is positive for x < 8 and...

"Let's assume that the entangled photons move in opposite directions and have the same frequency and same polarization." The book is in error. With any of the usual entanglement states, neither photon has its own polarization state. Furthermore, whatever axis Joe chooses he will see ±1 with...

That won't do at all. 1. The quart is too voluminous for a photon. 2. The quart has been supplanted by the liter (litre) in physics. If quarks can have beauty or charm why not call photons cuticles? Joking aside, after reading the above interesting posts and watching Feynman, where he uses...