I've entered the local pub's footy tipping competition which normally I'd stand no chance at winning. However, I'm going to let a coin flip decide my tips each week and I'm hoping that strategy will return a good result!.
The entry fee was $25 and so far 44 people have entered meaning the prize pool is $1100.
1st place = 70% of entry fees ($770)
2nd place = 20% of entry fees ($220)
3rd place = 10% of entry fees ($110) + $100 bar tab!
There are 23 rounds of (normally) 9 games. Games are between two different competing teams.
If you tipped every game correctly, picking all the winners, the maximum score you could achieve is 9 x 23 = 207.
Last year, a computer program run by Swinburne Uni scored 133 for the regular AFL season (excluding finals). That's an average of 5.7 correct tips per-week, over 50%.
I wonder if my approach of coin flipping will produce a score of 207/2=103.5?
Wait, did I just waste $25? My applied statistics knowledge might be to shit for this.
If I flip a coin 100 times, I should expect a result of roughly 50 heads and tails because each outcome is equally likely. But in this case, I don't think each outcome (team A or team B wins), is equally likely. This is probably covered in the first class of statistics.
Anyway, let's see how we go!
Correct tips & position (updated weekly):
Round 1: 8, first place
Round 2: 7, first place
Total: 15, first place