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Odds of picking games correctly

Posted on 2016-11-03
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Last Modified: 2016-11-03
Let's say there are 12 games being played.  One entry in the contest consists of picking the winners of any 7 of the 12 available games.  I need all 7 to be correct to win.  For the purposes of simplicity every team is evenly matched.

What would the odds be for selecting 7 winners from the 12 games being played with a single entry? Two entries? Three entries?  How would I calculate this?


Now let's say I am submitting multiple entries.  I am submitting every combination of picks for 4 games (2^4 = 16 combinations) to ensure at least one of the entries has all 4 picks correct. This would leave only 3 picks to chance. I would submit the same 3 picks with all 16 combinations.  Therefore as long as my 3 guesses were correct, one of my 16 entries would have all 7 picks correct.  How would I calculate these odds?

I hope this makes sense.


Thanks.
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Question by:Tom
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by:Kimputer
ID: 41871902
Sounds like a homework assignment (which EE experts can't help you solve). Therefore, to get this question moving, you have to have at least _some_ idea how to solve it, and explain all your steps. EE experts will then correct you on some of your reasoning or the steps. EE experts will NEVER finish your whole homework assignment.
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by:Tom
ID: 41871910
I'm 33 yrs old and haven't been in school for many years.  This is not a homework assignment but rather a question of curiosity to see what my odds are betting on NFL games for this weekend.

Thanks for your answer though.
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by:Kimputer
ID: 41872019
In that case:
12 games has 4096 unique outcomes (2^12) when only considering the winner. It means you have 1/4096 chance of guessing the correct 12 games.
At any time, any combination of 5 wrongs (2^5)  will still make you a winner, thereby upping your changes by a 32 factor
32/4096 still is a sad 0.008 chance to win though. If you make 1000 bets, 8 times you will win. Multiply the entries you're making (obviously, the entries has to be different), so  10 unique entries, is still 0.08, meaning about 8% chance you will win.
Obviously, you can calculate the winning odds against cost for 10 entries, and you'll probably know it's a lost case to bet.

The second part of calculating will probably take more time to calculate. However, it doesn't matter how smart you think you are, how many entries you make, the odds are always against you. That's why these games exist.
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BigRat earned 500 total points
ID: 41872025
Since all the games have an equal chance of ending one way or another and there is nothing else to distinguish them, the picking 7 from 12 is the same as picking 7 of 7, For example how many ways are there of picking seven identical red balls from 12 identical red balls? The answer is one, since all the balls are identical there is no way to distinguish one ball from another.

Now you need all seven to win. If x is the probability of winning (x is less than 1) then to get all seven correct the probabilities just multiply so it is x to the power seven. If the probability of winning one is 1/2 (and here we are again dealing with IDENTICAL objects) then the chances of winning become 1 in 2 to the power 7 which is 128. So if make another entry that entry would have the same probability, so you'd need to make 128 entries to get the probability near to one.

I don't think that helps much because the teams are not equal and the choice is not about picking identical objects. It reminds me of a chap I knew in Britain many, many years ago. He put one pound on the first horse, which lost, then he put two pounds on the next. After loosing his weekly wages he tried to borrow from me using the argument that eventually he must win. Of course he was correct, but one would need a lot of money to continue this scheme and he was forgetting the betting tax and the real probabilities. A one shilling flutter on a horse which took me most of Saturday morning to choose was much more fun - even if it lost. Good luck!
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