Losing Your Marbles

Logic Level 3

You have a large jar filled with colored marbles. It contains 1001 $\color{#3D99F6}{\text{blue}}$ marbles, 1000 $\color{#D61F06}{\text{red}}$ marbles, and 1000 $\color{#20A900}{\text{green}}$ marbles.

You randomly draw two marbles out of the jar, then put one or two marbles in (decreasing the total number of marbles in the jar by one or keeping the number constant) using the following rule:

If the two marbles are $\color{#3D99F6}{\text{blue}}$ and $\color{#20A900}{\text{green}}$ , put a $\color{#D61F06}{\text{red}}$ marble into the jar.
If the two marbles are $\color{#20A900}{\text{green}}$ and $\color{#D61F06}{\text{red}}$ , put a $\color{#D61F06}{\text{red}}$ marble into the jar.
If the two marbles are both $\color{#D61F06}{\text{red}}$ , put two $\color{#3D99F6}{\text{blue}}$ marbles into the jar.
If the two marbles are anything else, put a $\color{#20A900}{\text{green}}$ marble into the jar.

Repeat this process until one marble remains. What color is it?

\color{#20A900}{\text{Green}}

\color{#D61F06}{\text{Red}}

\color{#3D99F6}{\text{Blue}}

3 solutions

A Former Brilliant Member
Aug 10, 2015

Although it says that you draw randomly two marbles, but by considering that the solution will always the be same, one can solve this by controlling which marbles are drawn each time to simplify the problem, instead of the need to solve it generally. Furthermore, if the problem is truly random then the following is a possible outcome.

Consider the following:

You draw 2 red everytime, this means that you will add 1000 blue marbles.
Then you draw only blue marbles, you will be left with 1 blue and 2000 green marbles.
Finally only draw green marbles, you will be left with 1 blue and 1 green, which means I get a red marble as the solution.

Well, it works out the same, but you made a simple mistake. Two blues only make one green. So, you would be left with 1500 green marbles and 1 blue in your second consideration.

Aaron Stockton - 4 years, 10 months ago

I did it the same way!

Rohan Gupta - 5 years, 10 months ago

Shavindra Jayasekera
Aug 6, 2015

Let the last ball be X.

If the last ball is X, in the previous turn he must have taken two balls out and replaced it with either 1 red or 2 blue or he could have taken no balls out and put in a green. In the penultimate turn if he took two balls and put in 2 blue balls, the number of balls in the jar would be greater that 1, which is not possible because X is the only ball left. Therefore X is not blue.

At the start there was an even number of red and an odd number of blue. On the first turn:

If you take out two reds and put in two blues, there is still an odd number of blues and an even number of reds.
If you take out a blue and green out and put in a red, there is now an even number of blues and an odd number of reds.
If you take out a blue and red and put in a green, there is now an even number of blues and an odd number of reds.
If you take out 2 green and put in a green, the number of reds and blues aren't affected.
If you takes out two blue and put in a green, the number of blue is still odd.

Therefore, after each turn, the number of red and blue are never both even or odd.

Therefore, on the last turn is X is not blue, there are 0 blues. Hence, the number of blues is even. As a result, the number of red is odd. Therefore, there is at least 1 red. Therefore, X has to be red.

Ah, sorry - you can't take out no balls and put in a green. The otherwise means, "If the two marbles are not (blue, green), (green, red), or (red,red) then put in a green"

Maggie Miller - 5 years, 10 months ago

A green marble being put into the jar has a probability of occurring per round of around one half. (Blue, Blue) (Green, Green) (Blue, Red) vs (Blue, Green) (Green, Red) (Red, Red). Red's ability to be replaced by one marble every 1/3 of the rounds and Blue's ability to be replaced by two marbles every 1/6 of the rounds seems to show green having an advantage to persist.
However, Blue + Red is held constant in intervals of 2. The 2001 marbles which are represented by blue and red can only be decreased each round by two or zero. If a blue marble is taken out in (Blue,Green), a red is inserted, reducing blue+red by zero. If a red is taken out in (Green, Red), a red is reinserted, again reducing blue+red by zero. If two reds are removed, two blues are inserted. If two blues are removed a green is inserted, reducing Blue+red by 2. If one of each is removed in (Blue, Red) again a green is inserted, reducing Blue+red by 2. Therefore the answer cannot be green as you will always be stuck with one blue or red which cannot be taken from the jar without another blue or red. Red is the only one between it and blue which is added back as one marble. That alone doesn't seem to eliminate the possibility of blue, however, the statistical advantage green has to persist should give red an advantage also as green containing rounds give a red marble.

Joe Hillman - 5 years, 10 months ago

You can't be left with one blue marble, since if you just added a blue marble, you must have added two. You're on the right track with green - to nail it down, note that the parity of the total number of marbles and the parity of the number of green marbles are always opposite. So when there's one marble left, there must be an even number (i.e. zero) of green marbles!

Maggie Miller - 5 years, 10 months ago

Yes, that's right. I'll edit the solution

Shavindra Jayasekera - 5 years, 10 months ago

Joel Toms
Aug 7, 2015

Given the rules and initial conditions, the following holds true:

If there are an even number of blue marbles, there will be an odd number of red marbles, and vice-versa (the parity of each of these two numbers is always different).

This is easily verified for the three specified rules and initial conditions; the other possibilities for the two marbles removed are as follows:

B,B — parities of R and B remain unchanged
B,R — parities of R and B both reverse
G,G — numbers of R and B are unaffected

In any of these cases, the parities of red and blue remain opposite. Therefore, we cannot be left with 0 red and 0 blue marbles: i.e. the last marble cannot be green.

The last marble is left when you add it at the end of the last step. Since none of the rules allow for one blue marble to remain, but it is possible to be left with one red marble, we conclude that the last marble must be red.

Losing Your Marbles

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