Friday Puzzle -- marbles and probability

A start, maybe

Larry Barrett

Suggestion: Try a simpler problem to see how this works. Since the original problem has 729B and 243W, try starting with 3B and 1W (same ratio). Even this is complicated to analyze (and even more complicated to describe in words).

Here are some of the paths:

1. If first marble drawn is W (Prob=1/4) then discard W, leaving 3B0W;

Draw B (Prob=1), return, draw B (Prob=1), discard, draw B (Prob=1), discard;

End with 1B0W; Prob for this path =1/4.

2. If first marble drawn is B (Prob=3/4) then discard B leaving 2B1W;

If draw B (Prob=2/3) then discard B leaving 1B1W;

If draw B (Prob=1/2) discard B;

End with 0B1W; Prob for this path is (3/4)(2/3)(1/2) = 1/4.

3. If first marble drawn is B (Prob=3/4) then discard B leaving 2B1W;

If draw W (Prob=1/3), return W leaving 2B1W;

If draw W (Prob=1/3) discard W leaving 2B0W;

Draw B (Prob=1) and return;

Draw B (Prob=1), discard,

End with 1B0W; Prob of this path is (3/4)(1/3)(1/3) = 1/12

There are recursive paths which may simplify the analysis. I have not finished yet, and can't even draw a tentative conclusion.

On the right track!

Question.

Gary Smyth

Step four in example. After the white marble is placed in the bag, a black marble is drawn. shouldn't it also be put into the bag as it is different than the white marble also placed in the bag. Won't this affect subsequent rounds of where the marbles are placed?

Re: Question.

Alex Y

That would be a reasonable, but different, puzzle. As described, when you pick a different color, that ends the previous string of throw-ways, but does not start a new one. It is returned to the bag, and the process begins all over, in the example with a bag that now contains 727 black marbles and 243 white ones.

It's not

Gary Smyth

the number but the ratio. So if the first round is 3 to 1, 75% to 25%, even though the interim numbers of marbles will vary, subsequent rounds ought to be irrelevant provided the procedure is continued in the same fashion to the end. No matter how many tries every time a marble is discarded the overall ratio of black to white over time never changes until the end. So too are the discards, using the same ratio so that at the end there are two marbles. If I draw a black it is discarded or returned to the bag. If white, it is discarded or returned. No matter which is selected next the probability of it being black is 50%

Re: It's not

Larry Barrett

I don't follow Gary's reasoning, but I think I agree with his conclusion - that the probability of there being one Black marble at the end is 1/2, and the same for one White marble.

I base this on having worked out the paths for my simple example (start with 3B, 1W).

There are 8 possible outcomes, 4 ending with a B marble at the end, 4 with a W marble. And the probabilities for each of the outcomes for B are the same for W; each add up to 1/2. The distribution of B and W at each step do not remain the same, however, which is why I can't follow Gary's reasoning.

It is not evident that the results for this simple example extend to the original problem (729B, 243W), so have to continue thinking about this.

My answer may not be correct

Gary Smyth

but my reasoning is that the number of marbles thrown into the fire will diminish at the same rate as the original ratio 3 -1. At the end over N number of repetitions the probability is of two choices in the bag. Two black or one white and one black. Drawing a black is 50%

Hard to argue with success!

Alex Y

Very good!

Like Larry, I didn't follow your reasoning (nor the subsequent explanation--if BB or BW is more likely than WW as the last two balls, why is the probability of the last one left being black 50%?) But your intuition has served you well -- the answer in the source for this puzzle was 50%.

Their "demonstration" was along the lines of Larry's. They showed that a similar game, starting with 3B and 1W would end up with a Black ball 50% of the time. I'm not sold on the generalization to the larger game, and feel like I should be able to prove it by induction, but I get bogged down in the effort.

Re: Hard to argue with success!

Larry Barrett

I also do not see how to generalize the results from the 3B 1W case to the original problem (or even to intermediate cases). There is a surprising result from the 3B 1W case that might help with a generalization. The analysis of the 3B 1W case showed that there are eight possible outcomes, four leading to a B at the end, and four leading to a W, which was surprising in and of itself. But even more surprising was the symmetry of the probabilities - they are exactly the same for the four B outcomes and the four W outcomes. Without attempting to describe each possible outcome, the probabilities are: 1/4, 1/8, 1/12, and 1/24 (with sum = 1/2).

Solution

Alex Y

Congrats to Gary and Larry for getting this one. I was finally able to convince myself with an informal proof that the answer was correct, and in the process found a surprising result.

We are going to consider games with non-zero counts of black and white marbles, and prove that no matter what the size of game or relative size of the counts of white or black marbles, the probability of ending with a black marble is .5. Note that in this game, every time you draw a different color from the previous string, you put it back in the bag and start a new game with a reduced count.

The smallest such game consists of two marbles, one black and one white. Obviously for such a game, the probability of ending with a black marble is .5.

Now consider a larger game with n = b+w marbles, n>2. There are three ways for this game to end:

• restart at a smaller size game with m marbles, 2 LE m LT n, or


• with a string of b black marbles in a row, or


• with a string of w white marbles in a row,

The probability of choosing b black marbles in a row is b/n * (b-1)/(n-1) * ... * 1/(n-b+1) = b! / (n!/b!) = 1/n!

Similarly the probability of choosing w white marbles in a row is 1/n!.

Thus, this game ends in a smaller game or leads directly to a"victory" for white or black, with equal probability for either of the last two possibilities.

By strong induction, we get that the probability of any game with non-zero number of black and white balls has a 50% probability of ending on black, which seems very counter-intuitive (at least to me).