Math Is Fun Forum

ondine

Member

Registered: 2010-02-21

Posts: 3

Oscar Prediction Calculator

Hi there!

I hope it's okay to post this scenario to your forum. I will also email it directly just in case this needs to be taken down. (not sure how you feel about non-ranked pools---as opposed to picking NCAA March Madness winners for example)

I'm interested in Math, but it's not my strongest subject. Looking at your forum on Combinations and Permutations: "Order does/does not matter" and "Repeats are/are not allowed" I was wondering if you could help formulate how you would break down the following:

Every year we do an educated guess on picking the 24 Oscar Winners in a local email group forum.
You can enter as many times as you like (for a theoretical $10) and the winner with the most correct picks wins.

There are only 3 winners selected which share the theoretical amount $500 as follows:
60%
30%
10%

My question is this:
If I have ballot #1 with my "Best Guess" of 24 answers, but want to 'Hedge my Bet" mathmatically what is the best way to go about it?

For example: For every entry you are putting in $10. ONLY ONE BALLOT may be correct. OR several Ballots may be near correct--and you could take 1st, 2nd and/or third.

It is a RANKED CHOICE system, thus If BALLOT 1 and 4 are both 18/24
Per the pool rules, the tie is broken by weighting the importance of each entry according to the number at which it appears on our ballot, in categories 1 through 24.  That is, to resolve a tie I check which of tied ballots is the first to have an incorrect pick; if both ballots have the same first incorrect pick, I go to the second; if that's the same, I go to the third, and so on.

So Mathmatically, what is the best # of Ballots to submit?
In a Second Ballot: How many variables would be the optimal number (IE: Change 2 answers or more?)
In a Third Ballot: How many variables and would you create these variables in the same categories?
At what point does offering MORE THAN ONE ballot create a disincentive---due to odds that only one is correct and the per ballot fee is more than what you would win?

Ballot #1 = 24 answers
Ballot #2 = change 2 answers (for questions 4 and 16)
Ballot #3 = change 2 or more answers? (for questions 8 and 17)
Ballot #4 = change 4 answers?
Ballot #5 = change 1 answer only?

Thanks for entertaining this notion and helping figure out the logic/math!

Best,

o

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Oscar Prediction Calculator

Hi ondine;

Due people in this pool typically buy more than one ticket? If so what is the average number of tickets purchased per person? What was the total number of tickets put in play in last years pool?
Can you provide any of that?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

ondine

Member

Registered: 2010-02-21

Posts: 3

Re: Oscar Prediction Calculator

People usually only buy 1 to 3 ballots (tickets). Most folks only fill out one, a few do two, and rarely does a person enter 3 ballots.

ondine

Member

Registered: 2010-02-21

Posts: 3

Re: Oscar Prediction Calculator

Also, there were only 51 people participating last year.