Problem of the Week 17: Prediction markets
By Nigel (Chief News Junkie)
The United States Military Academy’s Department of Mathematical Science runs a weekly Problem of the Week. Last week was a really interesting problem involving making predictions on Hubdub! Unfortunately the deadline is now closed for entries however kudos to the first commentator who posts the correct answer below.
Hat tip to jenniandboys.
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Tags: hubdub, math, prediction markets, problem
April 12th, 2008 at 4:14 am
I’d say H$180 ($34 on Giuliani in Q2 and $146 on NO in Q1).
“Why?”, you ask. Well, I’m assuming that the best strategy is to buy the cheapest on the McCain-Giuliani ticket, which is obviously the 12% in Q2, and the cheapest on the hedge which is the 52% on NO in Q1 (versus covering the equivalent combined 88% on the two options in Q2). Also, it helps that the question lacks sufficient data to figure out how much buying 2 or 3 options in Q2 would work - ie the odds would change after each bet, and we don’t know exactly what they would become. But I’m “hand-waving” here, not really proving anything.
Anyhow, that assumed, let G be the bet in Q2 on Giuliani and let N be the bet on NO in Q1. We require:
1) G/.12 >= G + N + 100 and
2) N/.52 >= G + N + 100
Rearranging things you get the two inequalities (for N)
1) N =13/12 G + 1300/12
The solution for a profit of at least $100 can be seen by graphing N as a function of G to be the infinite area to the right of N=22/3 G -100 and above N=13/12 G + 1300/12 as well as the boundary lines. In particular, the sum G+N is in this case minimized where G and N are minimized, namely at the intersection of the 2 lines.
So 1) 12N=88G-1200 and 2) 12N=13G+1300. Subtracting you get 75G=2500 or G=100/3. Since you bet in dollars this means G=$34. Substitute this in 2) gives you N>=13/12(G+100)=13/12(134)
=871/6 which rounds up to $146. Note that this obeys 1), namely 146 < 22/3(34)-100=448/3=149 1/3
Checking
34/.12 = 283.33 ~ $283
146/.52=280.76 ~$280
so you’ve met the conditions.
Thoughts?
April 12th, 2008 at 4:23 am
This ate part of my answer (bad dog!) when I pasted it in.
Between “Rearranging things….” and “The solution for a profit….”, it should read:
1) N= 13/12 G + 1300/12
Sorry about that (it’s pretty important to understanding the rest).
April 12th, 2008 at 4:27 am
One last try. (It’s thinking it’s HTML code.)
1) N is less than or equal to 22/3 G -100 and
2) N is greater than or equal to 13/12 G + 1300/12
There! That’s what I meant to say.
April 12th, 2008 at 6:13 am
The solution at usma turns out to be wrong, from a true HD perspective. They give the G bet at 33.34 and the N bet at 144.45. As we know, there’s no pennies in HD wagers. So my $34 is best for G. Now the N bet I gave as $146, but it’s arguably $145, because 145/.52 - 100 - 145 - 34 = -0.15385, which in HD would round to 0. In other words, $34 for G and $145 for N total to $179 and in both cases pay $279 or more.
To be clear, I didn’t see the usma solution until I’d posted my solution here. (It’s quite different in its approach.)
The question:
http://www.dean.usma.edu/departments/math/activities/potw/problems08/POTW17.pdf
Their solution:
http://www.dean.usma.edu/departments/math/activities/potw/problems08/POTW-0802-17-Solution.pdf
April 12th, 2008 at 11:45 pm
Still in my to do box unfortunately. A couple of goes on Midas Oracle: http://www.midasoracle.org/2008/04/05/hubdub-math-problem/
and Reddit: http://reddit.com/info/6eqd5/comments/