Timeline for A balls-and-colours problem
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2023 at 19:13 | comment | added | Hugo van der Sanden | @TengLong see the second paragraph for explanation. | |
| Jul 20, 2023 at 12:11 | comment | added | Teng Long | Could you explain why the probability from k to (k+1) is (k+1)(n-k)/n(n-1)? Wouldn't it be k(n-k)/n(n-1) for both k+1 and k-1. | |
| May 10, 2020 at 23:17 | comment | added | Hugo van der Sanden | You asked "By weighted, do you mean conditioned on the chance cc will be the final color?" I guess so; more precisely I mean the wording from my reply: "we need to take account of the fact that not all selections are equally probable: each selection must be multiplied by the probability that it results in cc being the eventual colour." | |
| May 9, 2020 at 19:51 | comment | added | anonuser01 | My $E[N_j]|F_1]$ is equivalent to your $E_j$. But yeah, it took me a lot more work to get to where you got. | |
| May 9, 2020 at 19:49 | comment | added | anonuser01 | Then I just worked with E[N_n|F_1]. So we can write $$ E[N_i|F_1] = 1 + \sum_{j = \{i+1, i, i-1\}} p_{ij}|F_1 E[N_j|F_1] $$ $p_{ij}$ is the transition probability from state $i$ to $j$ and we are conditioning this on $F_1$ | |
| May 9, 2020 at 19:47 | comment | added | anonuser01 | By weighted, do you mean conditioned on the chance $c$ will be the final color? I arrived at the same equation, but it look me like a lot of work :(. I basically did $$ E[N_n] = \sum_{i=1}^{n} E[N_n|F_i]P(F_i) $$ where $F_i$ is the event that all balls will end up the $i-th$ color and $N_n$ is the number of steps needed to make all balls the same color. Thus we have $$ E[N_n] = E[N_n|F_1] = \cdots = E[N_n|F_n] $$ | |
| May 9, 2020 at 18:23 | comment | added | Hugo van der Sanden | The essence is: we have k balls of our target colour, what will the effect of the next pick be? We're making a pick, hence the +1, and then calculating the chances that we'll then have respectively k+1, k-1 or k balls coloured c after the pick, in each case weighted by the chance that c will be the final colour in this case. | |
| May 8, 2020 at 15:32 | comment | added | anonuser01 | It took me a while to arrive at the same expression as you for $E_k$. Could you expand on how you determined this relation so quickly? | |
| Oct 13, 2010 at 15:31 | history | answered | Hugo van der Sanden | CC BY-SA 2.5 |