Talk:Bertrand's ballot theorem

Too technical?
I am interested in this topic, but the phrasing is strange enough that I (complete layman) can't even tell if it is using semantic conventions I am unfamiliar with, or is just badly written. I have been reading the first example for ten minutes, and it looks like a fallacious argument, but I am too turned around by the phrasing to be certain. Also, starting the first example of the page with "clearly" is rather insulting, given that I reached this page in an attempt to understand a reference on another. Making this, and the other probability pages, more accessible to a layman who has a firm grasp of intermediate math (geometry, algebra) would be greatly appreciated. Nightsmaiden (talk) 19:46, 3 June 2009 (UTC)
 * This article seems like the sort of stuff secondary school pupils work on, so labeling it "too technical" for lay persons seems like a stretch. As for "clearlyl", I've looked at the version of this article that existed on 3 June 2009, the date Nightsmaiden's comment, and what it says is that "clearly" if one of the candidates gets no vote at all, then the other is ahead throughout the whole count.  It's hard to imagine someone reading that sentence all the way through without that being clear. Michael Hardy (talk) 18:42, 13 January 2010 (UTC)
 * Perhaps what isn't clear to you is the statement of the problem (it took me a bit to understand what the theorem is saying). Perhaps an example might help. Suppose there are two candidates A and B, and three voters x, y and z. With x and y voting for A and B, and z voting for B. Then in this case, p = 2 and q = 1. Then the question is, as each vote is counted (in random order) what is the probability that, as the partial results are computed, the eventual winning candidate A is always strictly ahead. To compute this probability by hand we would need to list all of the ways the votes might be counted and count the number of ways in which A is always ahead. For example if the votes are counted in the order x, y, z, the three results are A = 1, B = 0 (after counting x 's vote, A=2, B=0 (after counting y 's vote) and finally A=2, B=1 (after counting z 's vote). So counting the votes in this order, A is always ahead. But if we count the votes in the order z, x, y, then B starts out in the lead with one vote to none for A, so A is not always ahead. Here is a complete tabulation:
 * A always ahead: (x, y, z) and (y, x, z)
 * A not always ahead (i.e. sometimes behind or tied): (z, x, y), (z, y, x), (x, z, y) and (y, z, x).
 * So there are six total cases, with 2 cases in which A is always ahead giving a probability of 1/3. Which agrees with the theorem (p-q)/(p+q)=(2-1)/(2+1)=1/3. In the case P > 0, and q = 0, That is in the case where A gets all of the votes, and B gets none then (clearly no?) A is always ahead, no matter what order the votes are counted in? Does this make sense? Paul August &#9742; 16:50, 13 January 2010 (UTC)
 * I tweaked this example a bit and added it the article so hopefully at least the statement of the problem is clearer.--RDBury (talk) 18:43, 13 January 2010 (UTC)
 * Thank you, this page is very clear and concise now, and much easier to follow. Nightsmaiden (talk) 19:20, 16 June 2010 (UTC)

Which Bertrand?
The article fails to identify the person the theorem is named after. My guess is that it is Joseph Louis François Bertrand. Can anyone confirm? Marc van Leeuwen (talk) 09:13, 19 January 2010 (UTC)


 * I guess the cited Monthly article confirms this (at least the J.) so I'll put this in. Marc van Leeuwen (talk) 14:28, 19 January 2010 (UTC)

Redundant section
The section "Bertrand's and André's proofs" seems at first glance like it should be dismembered; the stuff that isn't totally redundant with the rest of the article should be pitched, and the other stuff should be worked in more naturally. Thoughts? --Joel B. Lewis (talk) 15:41, 17 June 2012 (UTC)