# A Posteriori

## Attempts to grapple with and elucidate empirical knowledge

### Mixes made equalNovember 9, 2009

Filed under: Uncategorized — Rāhul @ 16:38
Tags: ,

Today, I shall explore a simple puzzle which illustrates the advantages of extreme conditions in getting pointers towards the answer. Consider this scenario found at the Times’ Tierneylab blog a few weeks ago. Essentially, the question is-

If we have two vessels, X and Y, X containing a volume “a” of liquid A and Y containing  a volume “b” of liquid B which mixes with liquid A uniformly, and we pour a volume x of liquid A from X to Y first followed by pouring a volume x of the mixture of liquids in Y back to X, will there now be more of liquid A in Y than of liquid B in X? Assume that b is greater than a.

Before solving this algebraically, let’s think about the problem. Since the volume b is more than a, there is more of liquid B than liquid A. When x of b is poured into X, X will now have a+x amount of a mixture of A and B whereas Y will have b-x amount of just liquid B. When x amount of the mixture in X is poured back to Y, X will go back to having just a amount, but of the mixture of A and B, and Y will go back to having b amount, again of the mixture of A and B. There doesn’t seem to be any immediately compelling reason why the amount of B in X and A in Y should be the same or for the percentages to be the same either. So, let’s consider an extreme case for some help with the arguments.

Let us consider the case where x=0. Then, we can see that no liquid is exchanged between X and Y and so at the end the amount of B in X and A in Y are the same, which is zero.  Now consider the case where x=a. After the first exchange, we’ll have 2a amount of liquid in X, which is made of a amount of A and a amount of b, i.e. 50% of each. When x=a amount is poured back from X, this would involve a/2 amount each of A and B. Adding this to the b-a amount of B that was left in Y, we’ll have b-(a/2) amount of B and a/2 amount of A there. At the same time, X will have a/2 amount each of A and B. So, the amount of B in X and A in Y are the same- a/2.

In both the special cases we considered, the amount of B in X and amount of A in Y are the same. It is interesting to see that the concentration of B in X and of A in Y are not the same because the volumes a and b are not the same. Given these two special cases, which give the same result for very different values of x, it seems likely that the result might hold for all x. These examples do not prove so, but it does motivate us as to the answer to look for. So, while we solve it algebraically, not only will we have a higher confidence in the result if we prove it, it is likely that we are more likely to get to it since we know what we are looking for!

Getting to the algebra now, in the general case, after the first exchange of liquids, X will now have a-x amount of A and Y will have a total amount of b+x, of which there is b of B and x of A. When x amount of the liquid from Y is now poured back to X, since the liquids in Y mixed well, the amount poured back will involve [b/(b+x)]*x of B and [x/(b+x)]*x of A. In the proportion of their volumes, this now adds up to x amount poured back from Y to X.  So, the amount of A left in Y is x-[(x*x)/(b+x)] which is bx/(b+x). The amount of B now in X would just be the amount of B poured from Y to X, which again is bx/(b+x). Voila, we have the same number!

This means that the concentration of A in Y is much less than the concentration of B in X. So, if I gulp in a little sea water and spit out the same amount, there will be the same amount of sea water left in my mouth as the amount of my spit in the sea, but my mouth will taste much saltier than the sea will of my spit!

### Martin Gardner and the Jones sistersOctober 26, 2009

Filed under: Mathematics — Rāhul @ 20:04
Tags: ,

Today’s puzzle at New York Times’ Tierneylab blog was interesting enough to whet my appetite and yet not hard enough to elude me more than a tiny bit.  I used elementary probabiltity theory and algebra, and with a little help from ever-dependable OpenOffice Calc, I got results most amusing. So, let’s dive straight into it.

Quoting the first problem,

If you happen to meet two of the Jones sisters (this assumes that the two are random selections from the set of all Jones sisters), it is an exactly even-money bet that both girls will be blue-eyed. What is your best guess as to the total number of blue-eyed Jones sisters?

To attack this, I first realise that there isn’t one unique answer as the existence of one almost surely means the existence of others greater than it.  Assuming the total number of Jones sisters to be n and the subset of n that is blue-eyed to be m, we have- Probability of the two Jones sisters being blue eyed= (Probability of the first Jones sister being blue eyed) X (Probability of the second Jones sister being blue eyed). As the probability is 0.5, we have,

$(n/m) ((n-1)/(m-1))=0.5$

This is because first part follows because the first Jones sister could be blue eyed if any n of the m is chosen. But since 1 blue eyed sister has already been chosen as the first, the second could be blue eyed only if any one of the n-1 blue eyed sisters among the m-1 sisters remaining is chosen as the second. We are able to do this only because the choices of sisters are random. Attempting to solve this as a quadratic equation in n, we see that n has a different value for each value of m. Since the Jones sisters can only exist in whole numbers, the solutions should be the ones where both n and m are whole numbers. Do such solutions exist at all?

Starting from n=1 and counting all natural numbers, we see that $n=3$ and $m=4$ is one combination that agrees. In fact, that is the only combination of n and m under 10. So, the answer must be that there are 4 Jones sisters of whom 3 are blue eyed. Then, the problem goes on to say-

A simplified version of eye color inheritance, long held to be true, went as follows. Whether we are blue or brown-eyed is decided by a single pair of eye-color genes, one of which we get from our mother and the other from our father. For a person to have blue eyes, both genes must be for blue eyes; if either gene is for brown eyes, the person’s eye color is brown. The brown eye-color gene is dominant over blue. Each parent randomly passes on either one of his or her two eye-color genes to any given child. Thus if Mr. Jones has one blue and one brown eye color gene, he will be brown-eyed, and any given child will get either the blue or brown eye-color gene from him. We now know that human eye-color inheritance is much more complex than this, but for the purposes of the next question, assume that this simple model is true.

Based on the solution of problem 3, what would you guess are the colors of Mr. and Mrs. Jones’s eyes? What is your best guess about the eye-color genes that Mr. and Mrs. Jones have? How confident can you be about this, assuming that the only information you have is the colors of their daughters’ eyes?

Since the Jones’ have 3 blue-eyed and one brown-eyed daughter, the 4 eye-genes they have are not all blue or all brown. If one of them has both genes brown, then all the daughters will be brown eyed. So, that is ruled out too. So, at least one of the parents needs to have one blue and one brown gene. The other parent may have either both blue genes or one brown and one blue genes. Let us consider the former case first. The probability of each daughter being blue eyed is 0.5, that is when she gets blue-eye genes from both parents. Using binomial probability distribution (which I won’t elaborate here), the probability then of having 3 blue eyed and one brown eyed daughter is 0.25. In the latter case, the probability of each daughter being blue eyed is 0.25, when she gets blue eyed genes from both parents which in this case happens only in one case out of four. Using binomial probability distribution again, the probability now of having 3 blue eyed and 1 brown eyed daughter is 0.046875. Comparing the two probabilities, 0.25 and 0.046875, we can be reasonably confident that one of the parents is blue-eyed and the other is brown-eyed with one brown-eye gene and one blue-eye gene. Our confidence, in my opinion will be the ratio of 0.25 to the combined probability, which is close to 84.21%. Not, the widely accepted 95%, but very high.

Then, the puzzle goes on-

In the same neighborhood there are a whole bunch of animal couples who want to keep up with the Joneses and produce offspring that have some prized trait that we will metaphorically call being “blue-eyed.” The animals (and the range of eggs they produce) are: snakes (10-40), tortoises (50-150), frogs (500-800), beetles (1000-5000), lobsters (5000-50,000), bees (100,000-500,000), termites (500,000-1 million), codfish (1 million-8 million) and oysters (10 million-80 million).

It turns out that all these animal couples produced a bunch of eggs within their expected range, such that (as with the Jones sisters) if you picked any two eggs from the bunch randomly, there is an exactly even chance that they would both be “blue-eyed.” How many “blue-eyed eggs” and total eggs did each of these animal couples produce? Can you further extend this series of numbers using some analogy other than animal reproduction (which we may have exhausted, notwithstanding nature’s amazing profligacy)?

Here, we have to find values of n and m beyond 3 and 4. For this, I sought the help of MS excel and using the relationship connecting n and m and iterating n through increasing natural numbers while searching for whole number m I got the following list.

 n m (n+1)/n (m+1)/m) m/n 3 4 1.3333333333 15 21 5.0000000000 5.2500000000 1.4000000000 85 120 5.6666666667 5.7142857143 1.4117647059 493 697 5.8000000000 5.8083333333 1.4137931034 2871 4060 5.8235294118 5.8249641320 1.4141414141 16731 23661 5.8275862069 5.8278325123 1.4142011834 97513 137904 5.8282828283 5.8283250919 1.4142114385

Although, I didn’t go into the millions as the question demands, we can see patterns emerginng already. The ratio of any value of n or m to its preceeding value increases quickly at first and then plateaus at some point a little above 5.8. Also, the ratio of m to n also increases first and then plateaus near 1.414. The puzzle goes on to explore these numbers.

Not surprisingly, the ratio between “blue-eyed” eggs to the total number of eggs approaches a finite limit as we go towards infinity. But, unexpectedly, so does the ratio between consecutive numbers of eggs in the series, both blue-eyed and total. Why should this be? Can you figure out what these limits are and come up with a general formula that will yield all possible numbers of even-probability blue-eyed babies?

The last question asks for the general formula for all possible values of n, at which my limited math skills seem stumped. But, to find the finite limit for the ratio of m to n, let’s go back to the first equation. As n and m tend to infinity, 1 can be neglected in comparison to n or m. So, n-1 and m-1 tend to n and m respectively. Thus, the equation reduces to-

$(n/m) (n/m)=0.5$

which can be solved to get

$(m/n) = sqrt{2} = 1.41421...$

which is indeed what we saw it plateauing at. After some thought, I am still at a loss why the ratio of consecutive values of n and m also converge to a number as n and m tend to infinity. I suppose I’ll have to wait till other Tierney lab readers solve that! But, since the ratio of n and m converges to a values, if the ratio of the consecutive values of n converges to a number, we can see that the ratio of consecutive values of m also converges to the same number.