What are the odds of drawing a particular rune spread help? In a book about runes, there are 25 runes, and they can be put in a sequence of 5 (1 rune, then another, another, another, another). There is a section that reads, "If you select five runes, and placed them one below the other in front of you, the odds against drawing this particular spread are 607,614 to 1. If, however, you decided to mark down the tube you select an then return it to the bag, you will be making each selection from a full set of runes, and the odds against drawing this particular spread soar to 312,500,000 to 1."
I'm interested in how the author came to this conclusion. The author was trying to make a point about the difference in probability with there is replacement and there isn't, but where did the numbers come from? 
Should a formula like n!/n-r! and n^r be used? 
 A: It makes absolutely no sense.  If you draw $5$ runes with replacement from a set of $25$, with probability more than $0.65$ your sample will have no duplicates, so the notion that the odds "soar" by a factor of nearly $500$ is ludicrous.
A: The first number should be 6375599, and the second number should be 9765624.  I have no idea what faulty reasoning led them to their numbers, which make little sense.
A: I think the consensus will be that Ralph H. Blum's The Book of Runes: A Handbook for the Use of an Ancient Oracle is not a trustworthy source of mathematical information. It sounds like the 25 runes can appear either "upwards" or "downwards", except that 9 of them are the same when rotated by 180 degrees. This means that there are 9 + 2 * 16 = 41 different possibilities for an individual "rune", or 415 = 115,856,201 different combinations for drawing five of them when returning runes to the bag.
However, this is not a good measure of odds (the odds of a typical configuration are not 1 in 115,856,201) because not all of these combinations are equally probable! One way of getting approximate "odds" of this form is to calculate an "entropy" and exponentiate it. In this case that process yields $$\exp \left (-9 \cdot \frac{2}{50} \cdot  \ln \frac{2}{50} - 32 \cdot \frac{1}{50} \cdot \ln \frac{1}{50} \right ) = 50^{32/50} \cdot 25^{18/50} = 38.958...$$So this says that if you draw a single rune from the bag, there are still 41 different outcomes, but because 9 of them are twice as common as the other 32, the "odds" of choosing that particular rune are not quite 1 in 41 in the typical case, but rather about 1 in 39, because very often you'll have one of the "more common" possibilities. Similarly, taking this decimal to the fifth power, it's "like" there's one chance in 89,742,059 of getting your particular 5-rune configuration, because "usually" you'll have one of the 9 reversible runes and then your configuration is a little more probable than other configurations. Bottom line: "averaging" these sorts of odds gets a bit messy.
Now let's handle the more complicated conditional case, where you subtract rune-stones from the sample as you're going. It's not as simple as $$41 \cdot 40 \cdot 39 \cdot 38 \cdot 37$$ because sometimes when you draw out a runestone for the first element you eliminate two possible runes for the next draw: that rune, but also its reverse! So we need to keep track of the reversible and nonreversible stones separately, and we'll want a computer to do the counting. Here's the Haskell program:
Prelude> let count level rev nonrev = if level == 0 then 1 else rev * count (level - 1) (rev - 1) nonrev + 2 * nonrev * count (level - 1) rev (nonrev - 1)
Prelude> count 1 9 16
41
Prelude> count 5 9 16
72913680

So there are actually 72,913,680 distinct possibilities in this case! But of course, that's not the odds of a typical configuration. Let's instead try to figure out the entropy again; this will manifest in sending a new parameter down to the lowest level (the probability of the current thing happening) and then summing up the entropies on our way back up. We hand a probability of 1.0 in, we calculate the probabilities for all of the lowest cases, then we add them back together on the way up.
Prelude> let entropy l r nr p = if l == 0 then (-p) * log p else r * entropy (l - 1) (r - 1) nr (p / (r + nr)) + 2 * nr * entropy (l - 1) r (nr - 1) (p / (2 * (r + nr)))
Prelude> exp (entropy 1 9 16 1)
38.95822898302497
Prelude> exp (entropy 5 9 16 1)
5.8589129856952876e7

In other words, the typical real "odds" are something like one in 58,589,130.
