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# Magic beans math problem (stock portforlio)

Discussion in 'Trading' started by virtualmoney, Sep 25, 2011.

1. ### virtualmoney

You have 2 bags of beans containing 10 beans each. In each bag, there 3 magic beans. How do you maximize the number of magic beans selected if you are allowed to pick 10 beans out of the 20 altogether? You can mix them if you want.
I.e. If you do not mix the bags and just pick any one bag, you know you will get a minimum of 3 magic beans out of 10 but can you achieve more?

2. ### zwib

idk what this has to do with trading, but this seems like a simple statistics problem.

My guess would be to keep the two bags separate. Draw from one bag to start out, then reevaluate the probably of picking another magic bean from each bag and choose the one that has a higher probability. Repeat this process until you are out of 10 tries.

Example:

1st draw: from bag 1 you get a normal bean.

2nd draw: bag 1 now has a higher ratio of magic beans, so pick from it again. This time you get a magic bean.

3rd draw: bag 1 has 2/8 ratio and bag 2 has a 3/10 ratio, so you draw from bag 2. You get a magic bean.

4th draw: bag 1 has 2/8 ratio and bag 2 has a 2/9 ratio, so you draw from bag 1.

repeat until you are out of draws

ahchoo!

4. ### HATEtheRisk

Hmm?? quick thought, i think the probabilities are the best, if i pick 5 beans of each bag, than i could get more than 3 magic ones...
could....not must....

If i pick 3 beans and two are magic. than i would pic the other 7 from the other bag, odds would be higher than to get more on the other bag.

So it would be depending of how much magic beans i pick of the first bag 5 beans.....

But for sure minimum is to split it two 2x 5 beans of each bag and decide after your success - bean by bean.

So the odds would be, with picking 5 beans of each one.
= 100% to pick all 3 magic ones.
+ 50% to pick 4 to 6 magic ones.

What you think, am i wrong ?

5. ### virtualmoney

Actually my question is about picking best 3 performing stocks from 2 different sectors. The problem is you won't know which 6 will perform best until end of the year but you are allowed to pick 10 from 20 and you want it to include as close to all 6 best as possible.
I.e. You can't draw one at a time to see if it is normal or not because you will only know whether it is magical after it grows into a beanstalk at the end of the year.

6. ### HATEtheRisk

Actually, it would be
= 100% to pick 3 magic beans.
= 50% to pick 4 magic beans.
= 25% to pick 5 magic beans.
= 16,67% to pick 6 magic beans.

Am i wrong ? Dont know....

7. ### Hook N. Sinker

I'd plant all the beans, water them every day, and pick the ones that grow.

8. ### oldtime

plant your garden in such a way that you can break even with 3 magic beans (minus costs). So every year you will at least break even (minus costs) but some years you will make a profit.

And then one year you will hit all 6 beans. Make a big deal out of it.

Then when you walk down the street people will say, "There goes oldtime. I don't know how he does it. He has some secret system. This year he had 6 magic beans in his garden. Best bean picker I ever saw."

9. ### N54_Fan

If the whole point is to maximaize return by picking the best 10 stocks in 2 sectors then you have to do some technical and fundamental analysis to pick the best performing ones based on that data/info. You reevaluate every week or so to see which ones are performing best. As it becomes clearer as time goes by you rotate out of the lowest performing stock and into a better performing stock. Keep doing this until you have the 10 best performing stocks overall.

You can't do any better because your sample space has doubled.

Bag 1 P(x) = 3/10 Bag 2 P(x) = 3/10

You have 10/10 (100 %) chances to pick 3 beans if you just focus on 1 bag...

If you focus on 2 bags combined then you have P(x) = 6/20 = 3/10.

However, your prob. of picking the red beans drops from 100 % to 50% or 10/20 chances of actually getting the beans or 1/2...

Then your prob. of picking the red beans combined with your chances at picking the red beans drop significantly...

P(A*B) = 1/2*3/10 = 3/20

If you try to pick out of both bags you only have a 15 % chance of picking three beans while if you just pick from 1 bag you have a 100 % chance of picking three beans.

#10     Sep 25, 2011
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