JJG
Reputation
239
Top tag
Next privilege 250 Rep.
 Sep24 awarded Autobiographer Jul13 accepted Quaternion Group as Permutation Group Jul13 asked Quaternion Group as Permutation Group Jun3 awarded Yearling Mar11 awarded Popular Question Jan30 comment Nonlinear Second Order Differential Equation Wow. So simple. I feel stupid. Jan30 accepted Nonlinear Second Order Differential Equation Jan30 asked Nonlinear Second Order Differential Equation Jan16 awarded Nice Question Jan9 comment Taking balls from a bag with replacement Oh, thanks. Never heard of this problem, so it was difficult to google it. Jan9 asked Taking balls from a bag with replacement Aug17 awarded Supporter Aug17 accepted Modern formula for calculating Riemann Zeta Function Aug17 asked Modern formula for calculating Riemann Zeta Function Dec30 revised Probability and the Collatz Problem added 561 characters in body Dec30 comment Probability and the Collatz Problem I would like to offer a thought experiment: If you pick one thousand positive even integers randomly and do one iteration, $C(n)$ of each, then on average, half the results will be even. If you do this again $C^{2}(n)$ then 5/8 will be even, and so on. As you keep going, I think you will find that the average settles at 2/3. I don't think Chebyshev's inequality matters, as that is asking a different question. Dec30 revised Probability and the Collatz Problem added 1741 characters in body Dec30 revised Probability and the Collatz Problem edited body Dec30 comment Probability and the Collatz Problem @Alex Becker I don't think the probability $C^{k}(n)$ is even depends on anything other than the fact that $n$ is even (and positive). The probability tree takes into account all possible paths the Collatz iterations have on $n$, so the probability is just the sum of all paths that end in a positive number, which can be found with a geometric sum. Dec30 comment Probability and the Collatz Problem @Roupam Ghosh. We only need assume $n$ is even. All possible paths (e.g. even, odd, even,even,even,odd...) are accounted for when caclulating the probability that $C^{k}(n)$ is even.