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montadhar said:Ahh yea, that is right. If outcomes were picked at random, and I predicted no draws for any match, and for a big enough sample, is the maximum I can get about 66% or 50% of the predictions ?
My apologies for making this into a maths thread
Anyway, let's answer your question.
A game match is denoted by a variable ##X##. So ##X## can have three outcomes: A wins, B wins and draw (denoted by A,B,D). Based on the data of the group matches I propose the following probability model. Team A wins with probability 2/5, Team B wins with probability 2/5 and there is a draw with probability 1/5. So ##X## takes on ##A## with probability ##2/5##, it takes on ##B## with probability ##2/5## and it takes on ##D##with probability ##1/5##.
Your prediction are denoted by a variable ##Y## which has the same distributioin as ##X##. So it is again random.
The question is to find the probability ##\mathbb{P}\{X = Y\}##. Since ##X## and ##Y## will be assumed independent, we can calculate it as
[tex]\mathbb{P}\{X = Y\} = \sum_{k\in \{A,B,D\}} \mathbb{P}\{X = k\}\mathbb{P}\{Y = k\} = \frac{4}{25}+ \frac{4}{25} + \frac{1}{25} = \frac{9}{25}[/tex]
So if you would pick the teams totally randomly, you would expect to get a percentage of 36%.
Another strategy is never to choose draws, but always to choose randomly a winning and losing team (or equivalently: always choose that ##A## wins). Then we get
[tex]\mathbb{P}\{X = Y\} = \frac{2}{5}\frac{1}{2} + \frac{2}{5}\frac{1}{2} + \frac{1}{5} 0 = \frac{2}{5}[/tex]
In this case you would get a percentage of 40%. So we should expect Borek to get a percentage of 40% eventually
The worst strategy is probably to consistently predict a draw. We get
[tex]\mathbb{P}\{X = Y\} = \frac{2}{5} 0 + \frac{2}{5} 0 + \frac{1}{5} 1 = \frac{1}{5}[/tex]
This will get you 20% correct outcomes.