Monday, May 18, 2009

NDP Balloting and My Strategy this Year

Ticket application is from 16 May 2009 to 25 May 2009 and is NOT based on a first-come-first-served basis.
P(ballot x tix on y) = P(ballot x tix on z), where y and z are between 16 May 2009 to 25 May 2009
Each applicant must choose one out of the two choices given, either NDP or Preview Ticket. (NDP is on 9 August 2009; the Preview is on 1 August 2009).
I'll ballot for the NDP ticket as I prefer it over the preview.
Each applicant must choose 2, 4 or 6 tickets. The chance of obtaining 2 tickets will be higher than that of obtaining 4 or 6 tickets. Similarly, the chance of obtaining 4 tickets will be higher than that of 6 tickets.
P(2 tix) > P(4 tix) > P(6 tix)

As usual, to maximize the chance of getting a ticket, I'll ask my parents and sister if I can use their NRICs to ballot for the tickets. Last year I balloted for 2, 4 and 6 NDP tickets and didn't get it. It was a flawed strategy, based on what I hope to get. I do not want to get as many NDP tickets as possible. I just want to get 2 (at least 1, but they come in 2s anyway) NDP tickets. Let's analyse why my strategy last year was wrong and what balloting strategy I will be using this year:

First of all, we have to make a few assumptions:
  1. They are using a pseudo-random generator to select the winning ballots. There's no need to rush to ballot since any time within the given time will have equal chance for same ballot. This means that no one can game the system. Incidentally, the company that is running the ballot system is 10tacle Studios, a game development company. Gaming the system, game company.. both got game.. mm.. OK nvm, I'm just being lame.
  2. P(2 tix) = 2*P(4 tix) = 3*P(6 tix) = d, where d is a small positive number closer to 0 than 1. I guess that this is a reasonable assumption. It satisfies "P(2 tix) > P(4 tix) > P(6 tix)", and it's somewhat intuitively fair since the mathematical expectation are the same: E(2 tix) = E(4 tix) = E(6 tix). I mean if I were to set the parameters, I would set it as such.
So last year I balloted for 2, 4 and 6 tickets as I felt since E(2 tix) = E(4 tix) = E(6 tix), it doesn't really matter which I ballot for, and 2, 4, 6 sounds nice and fun. WRONG! I was a weakling who went along with my feelings without thinking logically. I hope I'm better now. The flaw was as result of misusing expectation. Mathematical blasphemy!

Expectation is mostly useful only when dealing with many many many trials. In this case, I only had three ballots. Also, I was seeking to have at least 2 ticket instead of having as many tickets as possible. In fact, with a fixed number of ballots at one's disposal, one cannot improve the expected values of tickets possible based on assumption 2. So what's my strategy for this year's ballot? It's to shoot low in order to maximize P(having at least 2 tix with 3 ballots): P(2 tix) + P(2 tix) + P(2 tix) >= P(a tix) + P(b tix) + P(c tix). There you have it, my NDP ballot strategy for this year~!

Of course, a yet better strategy is to ask around for the tickets or for people to ballot with their NRIC for you. But I don't know many people and don't like to ask for favours.

Have you balloted?
Promoting the ndp website actually reduces the value of d, so I should not do it if I were really desperate to get a ticket. Then again, the value of d is already so small, and my blog probably has a readership of 1 (me) at the moment.

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