# Qua locus

Keeping both eyes on the long game.

## Avoiding the death marble…

This blog has been split in three and all content moved to the new blogs – Qua Locus LifeQua Locus Tech and Qua Locus Puzzle – comments are disabled here.

A slightly easier puzzle this week…

I have 50 white marble and 50 black marbles. You can divide these marbles between two jars in any way you like. Then I will blindfold you, toss a coin to pick one of the jars and give it to you to pick a marble. If you pick a white marble you go free. If you pick a black marble you are sentenced to death.

How do you divide the marbles between the two jars to maximise your chances of survival?

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Written by CGF

June 10, 02012 at 08:04

Posted in Puzzles

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### 8 Responses

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1. 1 white marble in one jar other 99 marbles in the other.

Jonathon

June 10, 02012 at 08:11

• Oh yes – v. good.

Matt Wall

June 11, 02012 at 10:57

• How does that help, given that you do not get to choose the jar?

qwandor

June 12, 02012 at 22:44

2. Firstly: what happens if you do not pick a marble at all? Or if one of the jars has no marbles at all?

That aside, surely you can just put all the black marbles in the jars first, with the while marbles on top? Either all in one jar (if that is allowed), or else split evenly between both jars.

qwandor

June 12, 02012 at 22:46

• You’re right, my wording does not cover this situation. Since picking a marble is necessary for the puzzle to make sense, it seems reasonable to me that if one of the jars is empty I would not toss a coin but would just give you the jar with all the marbles in it.

Christo Fogelberg

June 13, 02012 at 10:18

3. I agree with Jonathon’s answer. I’m struggling to give a decent proof why it is the optimum though – feel free anyone to help complete the argument below!

Observation/framework:
$p(\textrm{live}) = \textrm{average}(p(\textrm{live}|A) + p(\textrm{live}|B)) = \frac{p(\textrm{live}|A) + p(\textrm{live}|B)}{2}$

Hypothesis:
The arrangement which maximises p(live) is 1 white marble in jar A, 49 white and 50 black marbles in jar B, i.e. A = {1W}, B = {49W, 50B}. In this situation p(live|A) = 1, p(live|B) = 49/99 and p(live) = 0.747.

An argument showing that this is a local optimum:
Moving a white marble from B to A does not increase p(live|A) but decreases p(live|B). Therefore it decreases p(live). Moving a black marble from B to A decreases p(live|A) more than it increases p(live|B). Therefore it decreases p(live). Therefore A = {1W}, B = {49W, 50B} is a local optimum.

An incomplete argument showing that this is a global maximum:
This part is tough because it’s not immediately clear to me how I can extend the local optimum proof to show which other divisions of marbles it holds for. Because in some cases moving a black marble increases p(live), and in some cases moving a white marble increases p(live).

As a starting point, equation (1) shows p(live) for an arrangement with k white marbles and m black marbles in A, excluding a constant factor of z=2:

(1) $p(\textrm{live}) = \frac{k}{k+m} + \frac{50-k}{100-k-m}$

If a white marble is moved from B to A then:

(2) $p(\textrm{live}) = \frac{k+1}{k+m+1} + \frac{49-k-1}{99-k-m}$

And if a black marble is moved from B to A then:

(3) $p(\textrm{live}) = \frac{k}{k+m+1} + \frac{50-k}{99-k-m}$

Doing some comparison and cancellation of these equations shows that p(live) is increased when a white marble is moved from A to B iff 2k + m < 74, and that p(live) is increased when a black marble is moved from A to B iff k < 25.

Noting also that if |A| = |B| then p(live) = 0.5 (because any surplus of white marbles in one jar is balanced by an equal shortage in the other and likewise for black marbles), and this all feels very close to a final proof. However, to be honest I’m tired and don’t think I’m going to crack it tonight… if genius strikes you please feel free to take this and run with it!

Christo Fogelberg

June 17, 02012 at 21:50

• .. in laymans terms (which is how I tend to think) – for any jar containing only white marbles, there is a probability of 1 that should that jar be chosen, you will live. For any jar containing the remaining white marbles, and black marbles, the probability will tend towards 0.5 as the total number of marbles at play increases. Given that there is a .5 probability of each jar being chosen, the best outcome you can hope for is a probability trending towards 0.75 based on two variables – a) the number of marbles in the game and b) the fewer white marbles within the first jar. So there are two rules if you are playing the game – ask for more marbles, and only put one white marble in the first jar.

Ash Fogelberg

June 17, 02012 at 22:39

• Makes sense to me, I’d ask for only white marbles but agree that if they have to be balanced (white/black) then the more you have the closer you’ll get to p(live) = 75%

Christo Fogelberg

June 18, 02012 at 07:00

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