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#1 Re: Help Me ! » Oscar Prediction Calculator » 2010-02-22 08:53:04

Also, there were only 51 people participating last year.

#2 Re: Help Me ! » Oscar Prediction Calculator » 2010-02-22 08:51:43

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.

#3 Help Me ! » Oscar Prediction Calculator » 2010-02-21 09:28:28

ondine
Replies: 3

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

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