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Shuffling Cards:
Foster's first feat of mathematical statistics produces the figure of:

1 in 8.066 x 10⁶⁷
This is the chance of getting a specific arrangement of cards by spontaneous ordering. Now, a deck of cards is not a good model for natural selection because it is a strict set of 52 items that can never vary or increase, whereas in nature the number of available materials and the way they can be arranged is all but unlimited, so you could get, say, a deck of 52 aces of spades in nature, but you cannot simulate this in a deck of cards where only 4 different aces exist, and no more.
Consequently, all we can do is show the effect of some simple selection rule on successive shufflings of a deck of cards. On page 39 Foster looks at the odds of a deck shuffling into a complete sequence from high card to low, i.e. from the Ace of Spades down to the Two of Clubs. To simulate 'natural selection,' you need to account in some way for reproduction, mutation, and selection. For example, we could use a shuffling rule such that if the top two cards ever turn up in the right sequence they get to 'live' by reproducing themselves, and these are thus removed from the deck to begin adding up toward the final result. This simulates the link 'reproducing' itself and all other links being 'selected' out of the gene pool and reshuffled (i.e. killed by the hostile environment, to which only certain organizations are suitable). Then, whenever the shuffled deck turns up the next two cards in the right order, granting a greater survival advantage, they attach to the previous two and the sequence grows, and the rest are killed and the deck is shuffled again. The former will represent beneficial mutation, the latter harmful mutation. With only these simple rules, how many shuffles will it take to produce the outcome Foster wants?
The odds of getting exactly two cards in the right order depends upon the Law of Permutations:

1 / (_nP_r) = 1 / (n! / (n-r)!)
And since:

₅₂P₂ = 52!/(52-2)!
It follows that the odds are 1 in 2652, or 00.038%.
The odds of getting another shuffle with the right top two cards would then be:

1 / ₅₀P₂ = 1 / (50! / (50-2)!) = 1 / 2450
And so forth. The odds get better as the deck thins out. But let's cheat for Foster, and pretend the odds remain 1 in 2652 with every draw, as if the shuffled part of the deck were to refill itself with an endless supply of useless jokers. Even with these odds stacked against us (this produces a probability even worse than Foster's of reaching Foster's sequence in 26 straight draws: i.e. about 1 in ten to the ninetieth power), time is our friend. Because we are relying not on random assembly, but slow and methodical assembly over time, the more times we reshuffle, the less time it takes to reach our goal.
In fact, there is more than a 97% chance that we will reach Foster's sequence in only 100,000 shuffles, which means:

10⁵ shuffles
In contrast to Foster's prediction of:

10⁶⁸ shuffles
This is calculated using sophisticated math perhaps beyond Foster's ken, although the details can be found in Mario Triola's Elementary Statistics (5th ed., 1993), pp. 250-6. I will remind you, I am referring to an introductory statistics textbook! This is not something only experts know. This is a method taught in the very first semester of statistical mathematics. The odds can be precisely determined using the binomial equation, solved for all necessary values, but the laws of normal probability distribution allow us to arrive at a reliable estimate with much less work. If you want to see why we would prefer the easy estimate, I will show you what the binomial equation looks like:

P(x) = n! * p^x * q^(n-x) / (n-x)! * x!
Where x is the number of successes needed (in our case at least 26), n is the number of tries (I have arbitrarily chosen 100,000 tries), p is the chance of any one try being a success (as we already have figured, this is 0.00038), and q is the chance of any one try being a failure (i.e. based on the Law of Complementarity, this is 1 minus 0.00038, or 0.99962). To really press this home, the '!' symbol means 'factorial,' or the value of the number given times every whole number between that and zero (i.e. 6! equals 6*5*4*3*2*1, or 720). To calculate the odds, we would have to solve for P(x) for every value of x between 26 and 100,000. Since no one wants to work through such a monstrous equation nearly a hundred thousand times over, much less calculate the factorials of numbers in the tens of thousands, we will go the easy route.
If we charted all the possible results of this equation on a graph, including the results for all values below 26 as well as above, they would form a bell curve, with a mean value in the center equal to n*p, or 100,000*0.00038, which equals 38. This means that the most probable number of successes in 100,000 tries will be 38. We only need 26, but we are dealing with whole numbers, and so 26 is really the whole range from 25.5 to 26.5 (we will not bother with this distinction in the future, since it becomes insignificant when n exceeds 1000). The difference between 25.5 and 38 then gives us what is called a z-score, and that z-score tells us (via any standard z-score table) the probability of getting any value between the two (in actual fact, this equals the area under the graph, where the z-score equals the value along the x-axis). Since 38 is in the middle of the graph, the odds of getting any number of successes from 38 to 100,000 is a flat 50%. We will add that to the odds of getting any number of successes from 25.5 to 38 to arrive at a good estimate of the odds of getting any number of successes of 26 or more. The z-score equals (x-m)/s, where x is the minimum needed result (25.5), m is the average result (38), and s is the 'standard deviation,' which equals the square root of the product of n (100,000), p (0.00038), and q (0.99962). Doing the math, we get a z-score of -2.02. On any z-score table, this reveals a final result of 47.83%. Added to 50%, we get a 97.83% chance of getting at least 26 successes (which is all we need to complete Foster's sequence) after 100,000 shuffles. Indeed, the chances only begin to drop below 50% when you have fewer shuffles than 68,000.
How might this translate into biological terms? We can assume that each shuffle represents a mutation, and each successful shuffle a beneficial mutation. To show what this means in terms of a population of organisms, if we assume that only 1 in 5 billion births suffers a mutation, and that each generation consists of only 1 billion births, then there will be only one mutation every 5 generations. But if there is one generation every hour (as there would be in any colony of bacteria), then it will only take 57 years to produce Foster's sequence! That is because there is about a 98% chance that it will be produced after 100,000 mutations, and if there is a mutation every five hours, that works out to 57 years. Now, we have chosen what is perhaps an unfairly high chance of beneficial mutation in this example, because we are limited by the awkward fiction of a deck of cards. But it should already be clear that the sort of math needed to actually get anywhere on this problem is far more complicated than what Foster uses, and the results are quite different. We have also assumed that only one arrangement of cards has survival benefits----in reality, and in the information space of DNA genomes, the number of viable and advantageous sequences of genes is incalculable and could very well be nearly infinite, and the range will differ for every different environment.

Typing Monkeys:
Foster claims, beginning on page 52, that a million monkeys typing 318 random words per minute for a million years could never produce Wordsworth's poem Daffodils, which Foster describes as a sequence of 159 letters (he ignores spaces and punctuation). His calculations are correct as far as determining the odds of this occurring purely by chance. But, as I've already pointed out, evolution does not occur purely by chance----it occurs as the result of natural selection. So let's factor in what Foster has left out: natural selection. As already noted, natural selection is the combined effect of three forces: replication, mutation, and selection. How might this be simulated with the monkeys and their typewriters?
Let us imagine that the monkeys are typing at computers which are simulating a process of natural selection. Each keystroke represents a mutation, and any incorrect keystroke represents an unviable mutation----a failure, a genome that is easily and quickly killed by the environment----whereas any three consecutive correct keystrokes represents a robust survivor (in this arbitrary analogy, the only kind of order that can survive----an unrealistic limitation, but applicable to our abstract case). The computer will automatically erase ('kill off') any incorrect generations, but let live any correct one. How long will it take for the monkeys, aided by this natural selection, to produce Daffodils?
I have chosen three-letter sets, instead of single letters or letter pairs, so as to simulate the reality that most mutations are fatal, and only a scant few beneficial----nevertheless, we see that this does not matter, because only the viable ones reproduce and multiply anyway. That is the beauty of natural selection. This rule produces a rate of viable mutation of only 1/17,576, or 0.006%, much lower than in the card deck example above. Foster does a more accurate job by accounting for the variable probability of the letters in the poem. I am assuming the same odds for every letter in the poem, which is not as exact, but it is a more than reasonable approximation, and this is necessary for what we have to do. Foster uses the figure of one million monkeys typing for one million years, for a grand sum of:

1.67 x 10²⁰ keystrokes
We'll make it even harder on ourselves. With that amount of work, if the conditions are assumed to be correct (i.e. if the computer is in fact selecting for Daffodils----i.e. if that and all its simpler ancestors are the only 'genomes' that can survive in our imaginary 'environment'), then success is actually guaranteed. So we'll stack the odds even more against us. What are the odds of one lone monkey, aided by natural selection, producing Daffodils after only 3 million keystrokes, or just one week of random typing according to Foster's generous assumptions? It will be easier to work in three-keystroke units, and with that in mind, using the same normal probability distribution as in the card deck example, we get the following values:

n = 1,000,000 triple-keystrokes
p = 0.00006 (the chance of three consecutive correct keystrokes)
q = 1 - p = 0.99994
m = n*p = 60
s = (n*p*q)^0.5 = 7.75

Now, if x = 53 (the number of correct triple-strokes needed to complete Daffodils), then z = (x-m)/s = -0.90, producing a percentage of 0.5 + 0.3159, or 81.59%. So, while Foster wants us to think that it will take a million monkeys an untold trillions of years to produce such a result, in fact, if we actually account for natural selection, it could take as little as one week for a single monkey! And this was assuming an even lower rate of beneficial mutation than in the card example above.
We might go further and solve another monkey problem dealt with by Foster, inherited from (supposedly) Huxley [1]: the claim, paraphrased on pages 54 and 55, that six monkeys randomly typing for 'millions and millions of years' would produce all the books in the British Museum. Now, this claim is, as stated, false, and Foster rightly demolishes the assumption. What Huxley apparently forgot to consider was that such a result is not likely to happen by chance, but it is likely given the operation of natural selection. How likely? Foster's assumptions are these:

"In 1860 there were 700,000 books in the British Museum"
We can estimate "9.8 x 10^9 lines of type in all the books"
We can assume about "50 letters per line"

From this we can guess that the total number of correct keystrokes needed is about:

4.9 x 10¹¹, or 1.63 x 10¹¹ triple-strokes
We will use exactly the same 'natural selection' simulation as above, only now with 6 monkeys typing for only nine million years, in all producing about:

8.55 x 10¹⁵ keystrokes, or 2.85 x 10¹⁵ triple-keystrokes
So:

n = 2.85 x 10¹⁵ triple-keystrokes
p = 0.00006 (the chance of a correct triple-stroke)
q = 1 - p = 0.99994
m = n*p = 1.71 x 10¹¹
s = (n*p*q)^0.5 = 413,509

Now, if x = 1.63 x 10¹¹ (the number of correct triple-strokes needed), then z = (x-m)/s = -19,346.62. Now, a z-score of just -3.5 produces .4999, for a total probability of 99.99%. A z-score in the negative hundreds or thousands indicates a probability so near 100% that the odds are astronomical for the event not to happen! Now, if we give the monkeys a little over just eight million years, this score starts to reach and exceed a zero z-score, i.e. the odds begin to drop below 50%; and if we allow only seven million years, the odds become effectively zero. What does this mean? While randomly typing monkeys would be unlikely, as Foster says, to type even a single line from all the books in the British Museum, even given the life of the universe to type away, nevertheless those same six randomly-typing monkeys, when aided by natural selection, would be guaranteed to do it in only nine million years! In fact, we can be even more accurate than that: they would be virtually guaranteed to succeed some time between seven and nine million years, and probably neither sooner nor later.