Red Marbles and Blue Marbles puzzle...goggle puzzle

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Problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar. Solution: Chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we’ll pick a jar, and each jar will have a 1/2 chance of being picked. then we’ll pick a marble, and depending how we stack the marbles, we’ll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one. for example, say we put all the red marbles into JAR A and all the blue ones into JAR B. then our chances for picking a red one are: 1/2 chance we pick JAR A * 50/50 chance we pick a red marble 1/2 chance we pick JAR B * 0/50 chance we pick a red marble do the math and you get 1/2 chance for a red marble from JAR A and a 0/2 chance for a red marble from JAR B. add ‘em up and you get the result = 1/2 chance for picking a red marble. think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from JAR A, but we didn’t have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could’ve just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in JAR B to help the odds out there. let’s look at those chances: 1/2 we pick JAR A * 1/1 we pick a red marble 1/2 we pick JAR B * 49/99 we pick a red marble do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4. we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in JAR A and maximizing the odds of picking a red marble in JAR B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we’ve maximized the odds in JAR A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we’ve also maximized the number of red marbles in JAR B. if we added any more red marbles to JAR Bwe would have to take them out of JAR A which reduce the odds there to 0 (very bad). if we took any more blue ones out of JAR B we would have to put them in JAR A which reduce the odds there by 50% (very bad). it wasn’t really a good proof, but QED anyway :-P POST YOUR SOLUTION IF YOU HAVE BETTER SOLUTION

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Problem:

you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you’ll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.

Solution:

Chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we’ll pick a jar, and each jar will have a 1/2 chance of being picked. then we’ll pick a marble, and depending how we stack the marbles, we’ll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.

for example, say we put all the red marbles into JAR A and all the blue ones into JAR B. then our chances for picking a red one are:

1/2 chance we pick JAR A * 50/50 chance we pick a red marble
1/2 chance we pick JAR B * 0/50 chance we pick a red marble

do the math and you get 1/2 chance for a red marble from JAR A and a 0/2 chance for a red marble from JAR B. add ‘em up and you get the result = 1/2 chance for picking a red marble.

think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from JAR A, but we didn’t have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could’ve just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in JAR B to help the odds out there.

let’s look at those chances:

1/2 we pick JAR A * 1/1 we pick a red marble
1/2 we pick JAR B * 49/99 we pick a red marble

do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.

we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in JAR A and maximizing the odds of picking a red marble in JAR B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we’ve maximized the odds in JAR A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we’ve also maximized the number of red marbles in JAR B. if we added any more red marbles to JAR Bwe would have to take them out of JAR A which reduce the odds there to 0 (very bad). if we took any more blue ones out of JAR B we would have to put them in JAR A which reduce the odds there by 50% (very bad).

it wasn’t really a good proof, but QED anyway :-P

POST YOUR SOLUTION IF YOU HAVE BETTER SOLUTION