why do you have to wait till the end of the year to determine if the bean is magical? Plant 10 beans and cull your crop replanting the culls.
Not true though. Pick from bag 1 until you get 3 magic beans, then pick from bag 2. You guarantee 3 magic beans (if you're unlucky enough to pick a magic bean as the last one out of 10) and have a chance of more than 3. This is clearly better then ignoring the second bag or picking randomly. A valid observation (which applies to both the stocks and the beans) is what are we trying to maximise? Are we trying to guarantee a certain number of magic beans or maximise the expected number of magic beans while carrying the risk we might end up with none (or fewer than we could have guaranteed)? This depends on our risk tolerance.
It would be best if we can actually increase the expected no. which seems to be mathematically fixed as 3 but if we can change the variance(risk) to occasionally have 5 or all 6 magic beans, it would be good too. In stock portfolio management, we want low variance, the beta, for any given expected no.,the alpha. However, for lottery fully hedged system, we want Maximum variance. Consider the case of a 44no.s (6+1balls) lottery. If we split it into 2 sets of 22no.s, covering all no.s, and employ 4of4, 5of5 or 5of6 wheeling systems and create variance large enough to occasionally trap 5+1 no.s or more to either side of the sets, we stand a chance to win top 3 prizes. All the possible outcome scenerios of 2 fully hedged wheels covering all no.s. (3,4) , (4,3) , (5,2) , (2,5) , (6,1) , (1,6) , (7,1) , (1,7) If variance swings are large enough, there is no escape, we will win top prizes. So the key question is how to create large swings to either side?
Um... not quite.... Pick from Bag 1 until you hit the first bean. In the worse case, all three beans are at the very bottom; Bad luck - you got three. Say, you hit the 1st bean in Bag 1 on the second try. The probability of hitting a second bean in Bag 1 next is 2/7 = 28.514%. The probability of hitting a bean in the 2nd Bag is 3/10 = 30%. So you switch to the second bag and pick from there. The solution is simple: at every turn, evaluate the probability of hitting a bean in either bag; and Select from which ever is the greatest. You *can* do worse than 3 beans; But you are expected to do better.
Yes, as I said it depends on your aim, I was just showing that it is possible to do better than random selection which someone was claiming it wasn't possible to do. Your solution is higher EV but with a lower guarantee, whether that is "better" depends on your utility function.
Consider one bag: Pick 10: 3 magic beans Pick 9: 2 or 3 beans Pick 8: 1, 2 or 3 beans Pick 7: 0,1,2, or 3 beans Pick 6: 0,1,2, or 3 beans Pick 5: 0,1,2, or 3 beans Pick 4: 0,1,2 or 3 beans Pick 3:0,1,2, or 3 beans Pick 2: 0,1,2 beans Pick 1:0,1 beans Pick 0: 0 beans The expectation is maximum for the following combinations: 7 and 3 6 and 4 5 and 5 in this case the expectation is: 0,1,2,3,4,5,6 These are the combinations that can offer you the maximum 6 magic beans at the risk of getting 0 beans. If you want to play safe, you pick 10 from one bag but you have only 3 beans. So this is an incomplete problem. You have to specify the risk aversion profile. If risk aversion is not to have less than 2 magic beans then you have the following combinations: 10,0 9,1 If 1 is minimum then: 10,0 9,1 8,2 So the answer should depend on the constraint of minimum number of magic beans. This is a boundary value constraint problem. Otherwise, it is incomplete.
I think thats the best safest way to have the 3 for 100% and a good chance to get more. And its easy to understand, also stupid magic beans like me can understand that.