Not sure, does anyone know if alternating bean selection from the 2 bags in varying groups sizes helps to improve overall chances of the final 10 having most magic beans.
It doesn't... The fact that they're in two different bags makes no difference. You're looking at the sample size. You can spill both bags out unto the floor and pick beans blindfolded. The result will be the same.
If you start with one bag and keep picking until you have 3 magic beans then move to the other bag you are guaranteed at least 3 and maybe more. If you dump them all on the ground and start picking aren't you running the risk of possibly picking no magic beans?
That's a good idea... If you get lucky and pick three magic beans then by all means use your remaining choices to pick from the second bag. This does in fact increase your chances but it is a special case as it is dependant on first picking your 3 magic beans and on how many picks you still have left. It changes the math and you can't predict it from the beginning. Yes... The exact same thing happens when you pick from the two bags instead of just picking from one bag... You've now added sample spaces... You've gone from 3/10 to 6/20... That is 3 magic beans out of ten to 6 magic beans out of 20.
You can always start with one bag and keep picking as you described but in this problem you can't identify if any bean is magical until end of the year.
Another insightful answer .i.e. pick only stocks or sector that you personally make most $ historically.
If you can't know which beans are magical until well after all the beans are selected, it says to me you don't actually have a way of knowing that there are in fact three magical beans in each bag. Which turns the entire selection process into a roulette wheel. So pick however you want - it's going to be a function of luck anyway.
You can simply call them the annual top 3 stocks/candidates/hedge funds if magical beans sounds abstract. Its not about knowing now but the ranking will happen to become a fact. This is an optimization problem.