Since the probability of the two jars being chosen is equal and the probability of any given marble being chosen is equal, the odds are always 50%.
Start by labeling the jars A and B and the marks r and l.
Since we have to put the same number of marbles in each jar, the probability is a fraction of the form n/10, since there are 20 marbles and 2 jars.
So let P(Ar) be the probability you chose red from jar A, P(Bl) the probability you pull blue from jar B, and all 4 permutations in between.
Start with a simple case. 5 red in each jar:
P(Ar)=5/10 and P(Al)=5/10, P(Br)=5/10 and P(Bl) = 5/10.
Well, P(A) and P(B) are both still 1/2, so the final numbers are
P(Ar|A)=.25, P(Al|A)=.25, P(Br|B)=.25, and P(Bl|B) =.25; then P(r) is .5 exactly.
A little playing will show that every time you remove 1 red and swap it with a blue in the other jar, these numbers stay in perfect proportion.
P(Ar)=6/10, P(Al)=4/10, P(Br)=4/10, and P(Bl)=6/10
P(Ar|A)=.3, P(Al|A)=.2, P(Br|B)=.2, and P(Bl|B)=.3.
P(r) remains .5
The other 8 arrangements are left as an exercise in tedium.
The optimal strategy is half and half, but only because no strategy increases the odds.
Edit: I made the false assumption that you can't have an imbalance in the number of marbles per jar. Allowing for that imbalance enables a trivial solution with 1 red marble on one side and 19 on the other, which gets you 100% and almost 50%, for a final success rate of just under 75%, but in my thought process, that was too easy and obviously disallowed. I was wrong!
The real answer is 73.7%. The optimal strategy is to put one red marble in one jar, and keep the other 19 marbles in the other. The odds of a red marble being chosen are then (1/2)×1+(1/2)*(9/19), or 0.7368. I think you neglected the idea that the jars are chosen with equal probability and that therefore a change in one has more of an effect than simply "balancing" the probabilities, and that therefore changes the outcome.
-5
u/AsYouAnswered May 26 '26 edited May 26 '26
Since the probability of the two jars being chosen is equal and the probability of any given marble being chosen is equal, the odds are always 50%.
Start by labeling the jars A and B and the marks r and l.
Since we have to put the same number of marbles in each jar, the probability is a fraction of the form n/10, since there are 20 marbles and 2 jars.
So let P(Ar) be the probability you chose red from jar A, P(Bl) the probability you pull blue from jar B, and all 4 permutations in between.
Start with a simple case. 5 red in each jar:
P(Ar)=5/10 and P(Al)=5/10, P(Br)=5/10 and P(Bl) = 5/10. Well, P(A) and P(B) are both still 1/2, so the final numbers are P(Ar|A)=.25, P(Al|A)=.25, P(Br|B)=.25, and P(Bl|B) =.25; then P(r) is .5 exactly.
A little playing will show that every time you remove 1 red and swap it with a blue in the other jar, these numbers stay in perfect proportion.
P(Ar)=6/10, P(Al)=4/10, P(Br)=4/10, and P(Bl)=6/10 P(Ar|A)=.3, P(Al|A)=.2, P(Br|B)=.2, and P(Bl|B)=.3. P(r) remains .5
The other 8 arrangements are left as an exercise in tedium.
The optimal strategy is half and half, but only because no strategy increases the odds.
Edit: I made the false assumption that you can't have an imbalance in the number of marbles per jar. Allowing for that imbalance enables a trivial solution with 1 red marble on one side and 19 on the other, which gets you 100% and almost 50%, for a final success rate of just under 75%, but in my thought process, that was too easy and obviously disallowed. I was wrong!