Balerion wrote:I am unsure on that second assumption; given the number of sides we have seen and the number of casters per side, the fact we have yet to see a duplicate on a side starts to seem unlikely. Admittedly, it is within the realm of believably still (i think), but i think we could have expected to see one by now
Holy probability of repeating digits Mathman!
To answer that properly, someone (and it ain't gonna be me) needs to tally exactly all we know of which side
popped which caster (as opposed to acquired one by other means). If that field survey were done, the method to interpret the data would be -
There's two problems here. First one is, suppose a side popped n casters. What's the probability that at least one caster type appears at least twice (or just as well, that there is no caster type repetition)?
+23 caster types: 22 of them have the same chance of being popped provided the event that a caster popped has occurred, while thinkamancers are x times as likely as another caster type to pop provided the event that a caster popped has occurred
+the event that a caster pops may grow more unlikely as it is repeated, but the probability P(caster type | caster has popped) (aka, probability to pop a certain caster type conditional on the event that a caster has popped) remains unchanged regardless of any caster pop history
The fact that the Thinkamancers are of a different probability|caster pop is throwing an apparent spanner in the works of an otherwise simple assignment ...
Were all casters equally likely|caster pop, then the reasoning would go, it doesn't matter which type of caster was first popped, the probability that the second caster popped is different from the first is 22/23; the probability that the third is different from the first two is 21/23 etc. Since we assume caster probabilities|caster pop are independent of history, for 0<n<23, the probability that no repetition of caster type occurs at all is 23!/((23-n)!*(23^n)); for n=0 or n=1, the formula correctly produces a probability of no repetition of one; for n>=23, the probability of no repetition is, trivially, zero. Then, the probability that repetition occurs is one minus the probability of no repetition, as the two are mutually exclusive events.
However, not all casters are equally likely|caster pop ...
Assume the first caster is a thinkamancer. Then the probability that the second caster is different from the first is 22/(x+22); the third is different from the first two with a chance of 21/(x+22); the fourth is different with a chance of 20/(x+22) etc. In general, the formula for the probability of "if the first caster was a thinkamancer, no repetition of caster types occurs" is 23!/(23*(23-n)!*(x+22)^(n-1)). Combining into one event, "the probability that first caster is thinkamancer, and after that no caster repetition occurs" is x*23!/(23*(23-n)!*(x+22)^n).
Now assume the k-th caster popped is a Thinkamancer, and that no Thinkamancer popped before that. This particular event has probability x*(22^(k-1))/((x+22)^k). What I'm essentially doing is counting possible sequences, in this case, from the set of sequences with the first thinkamancer popped at the k-th caster pop. How many of those sequences have the property I seek (no repetitions) is not affected by me rearranging all of them in exactly the same fashion, say, so as to consider the Thinkamancer as having popped first. Therefore the probability of "if the k-th caster was a Thinkamancer and there was no Thinkamancer before that, no repetition of caster types occurs" is again 23!/(23*(23-n)!*(x+22)^(n-1)), while the combined event "the probability that the k-th caster was a Thinkamancer, and that there was no thinkamancer before that, and that there are no caster type repetitions" is x*(22^(k-1))*23!/(23*(23-n)!*(x+22)^(n-1+k)).
Finally, assume that no thinkamancer popped, ever, in all the n trials, which is the familiar (22/(x+22))^n. Combining again into one event, as above, "the probability that there was no thinkamancer, and no caster type repetition", is (22^n)*23!/(23*(23-n)!*(x+22)^(n-1+n)).
Now, either a thinkamancer pops on the first caster pop, or the second, or the third etc, or not at all. These are mutually exclusive events and likewise all events of the form considered above. So we can add the probabilities of the combined events above to get the probability of no repetition
sum{k from 1 to n} {x*(22^(k-1))*23!/(23*(23-n)!*(x+22)^(n-1+k))} + (22^n)*23!/(23*(23-n)!*(x+22)^(n-1+n))
... which leads to the somewhat compact probability that, in n caster pops, there is not even one repetition of caster types to be
23!/(23*(23-n)!*(x+22)^(n-1)) for 1<n<23
For n >= 23, no repetition probability is trivially zero. For n=0, the probability of no repetition is 1. For n=1, the formula correctly produces a no repetition probability of 1.
(remember, x is the amount of times a thinkamancer|caster pop is more likely than any other one particular type of caster).
Notice that in the end, the formula looks just like the one obtained for "if thinka first, no repetition"; it doesn't really matter where, or whether, a thinkamancer appears in a sequence when counting the probability of no repetition, but this isn't obvious, not to me anyway.
The second problem is, given m sides, how to combine the above in the event "there is no repetition in the caster roster of each side" (we allow the same caster type to appear to various sides). We assume caster pops are independent of history, including that of other sides, so we can treat "side X has no caster repetition" as a completely independent event to "side Y has no caster repetition", for any sides X and Y (and Z etc). So, by multiplying the probabilies of no caster repetition for each side, we get the probability that no side has caster repetition.
Now all that's left is to tally this with actual Erfworld data. For some flavour of the results though -
if Thinkamancers are twice as likely than any other one caster type, and the side in consideration has popped 5 casters, it has about 0.53 chance of no caster repetition. Increase that to six casters popped, and the chance drops to about 0.4. Supposing we know of two sides, one having popped 5 and the other six casters, the probability of no repetition in either of them is then about 0.21.
The whole point of this is lost if you keep it a secret.
