__Summary__

We present below copies of the expert mathematical testimonies from Professors Field and T'Mont, submitted to the court martial of Captain Jonathan Archer. Captain Archer stood accused of falsification of logs, conduct unbecoming an officer, and the ill-treatment of subordinates. All three charges related to a single incident, the drawing of lots for shore leave on the planet Risa on Earth-date 3.5.2152.

The prosecution argues that Captain Archer had wilfully manipulated what was intended to be, and had been described as, a random selection process, in order to ensure that he and his entire human command staff would be among the approximately 50% of the crew chosen to engage in shore leave. If is further argued that, should this behaviour be proven to occur, it would constitute both conduct unbecoming an officer and, by decreasing each subordinate's chance of being selected, ill-treatment of those subordinates. Finally, it is argued that Captain Archer, by stating in his captain's log that the process had been random, had falsified said log.

With no direct evidence to support these contentions, the prosecution's case rests heavily on the expert testimony of Professor Field, Tycho City University, who agrees with prosecution's counsel that the results of the lottery represented evidence in themselves - that the probability Archer and all four of his bridge crew were selected was so low in a fair lottery, that the process could be stated without reasonable doubt to not have been fair. The prosecution further notes that, if the lottery had indeed been tampered with, only Captain Archer himself has access to the command codes necessary.

In response, Professor T'Mont, University of Selaya Range, argues, in effect, that the admittedly very unlikely event under consideration had to be viewed in the context of how often the opportunity for such events presented themselves. By definition, we expect "one in a hundred" events to happen once in every one hundred opportunities for that event to manifest. How many lotteries have been run on Starfleet vessels in the decades since the institution's founding?

Sumont additionally argues that the true comparison to be made in terms of probabilities is not the chances of the result seen versus a less unusual result. It lies in comparing the likelihood such a result would be seen in an untampered lottery with the likelihood any Starfleet Captain - and particularly one with Captain Archer's service record - would choose to engage in such tampering in order to secure an opportunity his own first officer testified he had argued against taking advantage of.

__Expert testimony, Professor Regina Field, Tycho City University__

The court has asked me to provide the best answer possible to a single question: can we credit that, in a lottery selecting forty persons from amongst seventy-nine (the number of crew to be awarded leave and the total human crew complement of USS Enterprise, respectively), not only would the three most senior human officers be selected, but additionally the two bridge officers most commonly cited by Captain Archer for their performance, one of whom was directly approached by Archer himself about joining his crew? Put another way, could such an event plausibly occur, were the Captain - for whatever reason - not secretly playing favourites.

My answer is no.

I am aware, from my conversations with prosecution counsel, that the answer to this question is unusually important in context - that the (at least apparently) sheer implausibility of the result of what I shall refer to henceforth as the "Risan lottery" constitutes the entirety of the case against Captain Jonathan Archer. I am neither qualified nor inclined to comment on any broader implications of this approach, but it certainly requires my testimony be as complete as possible. A man's career and even freedom hang in the balance, and the recognition that I hold the only available weights and measures, if I might be permitted a moment of levity, hangs heavy.

With this in mind, then, I shall endeavour to explain my methodology to the fullest extent. The following calculations belong to a category of mathematics known as combinatorics, which we might call - perhaps to some bafflement - the mathematics of counting. More specifically, and perhaps less confusingly, combinatorics contains the study, as the name suggests, of combinations. It is possible to calculate, and with actually very little effort, both the total number of ways in which the Risan lottery could have been resolved, and how many of those resolutions would have included five specific people - in this case, Captain Archer, Commander Tucker, Lieutenant Reed, and Ensigns Mayweather and Sato. By comparing these values, we can determine what percentage of results would include those five people and hence how likely the result we actually observed was to happen.

We will begin with the total number of possible resolutions for the lottery. Let us assume, initially, that there are *n* crew aboard a ship, of which *r *will be selected by a lottery. Naturally, we assume *r*<*n*, as *r*>*n* is impossible, and *r*=*n* would select all crewmembers, rendering the exercise superfluous, except perhaps for those who insist the simple act of existing should somehow earn one a trophy for participation.

Under the assumption that *r*<*n*, then, we can proceed. Our next step is to make an additional assumption, one which we know without question is not in fact the case, but which allows us key insights which will remain valuable once the assumption is dismissed. Let us pretend that each available space as a passenger on a shuttlepod to Risa is individually numbered, and that the numbering was in some way important ; the crew might disembark in numerical order, say, making lower numbered tickets (fractionally) more valuable. There are *n* possible crewmembers to be assigned to that first, most coveted seat. There are then n-1 possible crewmembers to be assigned to the second seat, because one crewmember has already been assigned. In total, then, the number of ways in which those first two seats can be assigned is *n* x (*n*-1), because every possible assignation of the first seat can then be followed by *n*-1 different assignations for the second.

We can follow this logical path to its terminus: with r numbered spaces to assign, the total number of different ways to assign those seats is equal to *n* x (*n*-1) x (*n*-2) x ... x (*n*-*r*+2) x (*n*-*r*+1), with the final number in the product justified by the fact that *n*-*r* crewmembers will remain on the ship, meaning *n*-*r*+1 are contesting the final seat.

It is worth noting that, in this fictitious example where the ordering of seats matter, it actually would be reasonable to consider the case where *n*=*r*. In that scenario, everyone would get to travel to Risa eventually, but a strict order would exist regarding to how soon they can play in the surf. In such a situation, then, in which a ship finds itself emptied and useless for the sake of a little temporary self-indulgence, the number of possible seat assignations becomes *n* x (*n*-1) x (*n*-2) x ... x 2 x 1. We give a specific name to this product, which is *n* factorial, denoted n!. This is the product of the first *n* positive whole numbers multiplied together.

Using our new toy, we can see that the expression *n* x (*n*-1) x (*n*-2) x ... x (*n*-*r*+2) x (*n*-*r*+1) is similar in construction to* n*!, but contains fewer elements in its product. In fact, in order to get from this result to *n*!, we would need to multiply it (*n*-*r*) x (*n*-*r*-1) x ... x 2 x 1. But that product is itself is equal to (*n*-*r*)!.

We now use the fact that division is the inverse of multiplication. By this, I mean simply that multiplying a value by some number, and then dividing by that same number, will always return us to that same original value (provided we did nothing so foolish as to multiply and then divide by zero). In other words, choosing not to multiply by (*n*-*r*) x (*n*-*r*-1) x ... x 2 x 1 is exactly equivalent to performing that multiplication, and then dividing by that exact same amount. Therefore:

*n* x (*n*-1) x (*n*-2) x ... x (*n*-*r*+2) x (*n*-*r*+1) = n!/(n-r)!

We can apply this to the specific case of the *USS Enterprise* by setting *n*=79 and *r*=40, learning in the process that there are over one thousand trillion *quintillion* such combinations.

Fortunately, we need not consider nearly so many possibilities. If we remove the assumption that each shuttlepod seat be individually numbered, then large groups of previously distinct assignations suddenly become equivalent. To see this, consider that for any specific assignment we previously considered, swapping around the first and second crewmembers to receive a seat would result in a different assignation. Once we stop caring who gets which seat, being interested instead only in who receives a seat and who does not, such a swap would make no difference to us.

The question then becomes how many assignments in our previous approach (where order matters) collapse down into a single one in our new approach (where it doesn't)? Put another way, if we imagine a specific set of *r *crewmembers receiving places, how many ordered combinations of those *r* crewmembers are there?

Once again, the solution lies in the use of factorials. For a specific group of *r *crewmembers, there are *r *choices for the first seat in the situation where ordering matters. There are then *r*-1 choices for the second seat, and so on. This means that for every unordered group of *r* crewmembers, there are *r*! ways to order them. Equivalently, in order to find how many *actually *different groups of crewmembers the lottery could have chosen, we need to divide the total number of *ordered* groups by *r*!. In mathematical notation, we denote this as *n*C*r* (shorthand for "take *n* objects and choose *r* of them").

nCr = (*n* x (*n*-1) x (*n*-2) x ... x (*n*-*r*+2) x (*n*-*r*+1))/(r x (r-1) x ... x 2 x 1) = n!/(r!(n-r)!)

Applying this to the case of Archer's vessel results in a far more reasonable total of distinct groups. Choosing 40 crewmembers from a total of 79, without troubling ourselves with ordering? That's not even sixty thousand quintillion combinations. We are making progress.

The next question to answer is simple: how many of those 5 400 000 000 000 000 000 000 or so distinct was to pass out shore leave permission include all five officers the prosecution have asked me to take an interest in? The good news is that we have already done all the hard work necessary to find the answer. We proceed now by dividing the lottery into two - a lottery for five seats which has been fixed, and a second lottery for 35 seats which was allowed to proceed fairly. With the order of seat assignments held to not be relevant, there are 5C5 = 1 ways to assign the seats in the first lottery. This makes intuitive sense - with five seats to offer to five people, there is only way to choose which people get a seat.

Meanwhile, for the 35 remaining spaces, and the 74 remaining crewmembers, there are 74C35 combinations of crewmembers who could be given the spaces; that's a number just above one thousand quintillion. In general, having performed this split, we would calculate the total number of possible combinations by multiplying the totals from each lottery together (see my comments above about how to calculate the number of possibilities for filling the first two seats). In this case, though, that isn't necessary; there is only one way for the fixed lottery to play out. Hence, there exists no difference between the overall number of possible assignations, and the number of possible assignations from the fair lottery.

Under the assumption of a totally fair lottery, then, where Captain Archer had not interfered in any way, and the computer programme used to select crewmembers gave each person an equal chance of receiving a seat, we can easily calculate the probability of all five named officers receiving a seat. We simply take the number of combinations that include all five officers, and divide by the total number of combinations. This gives us

74C35/79C40 = 0.0292

(Technically the actual expression should be (74C35 x 5C5)/79C40, but I have explained above how this makes no difference, since 5C5 = 1).

To convert to the percentages that laypeople often find more comprehensible in discussing probability, the chance of all five named officers receiving a seat on a shuttlepod in a fair lottery is just under 3%. By way of comparison, the probability of four of the five named officers being selected is almost 16% ((74C36 x 5C4)/79C40), and that of three being selected is almost 33% ((74C37 x 5C3)/79C40).

It is not for me to advise the court on whether so low a percentage can be taken as evidence of guilt - indeed I am legally prevented from doing so. That decision must be for the court to make. All I can do is reaffirm my initial comments; the event witnessed does not in my opinion reach the definition of plausible, given the very low probability of it occurring.

__Expert testimony, Professor T'Mont, University of Selaya Range__

Members of the court, my thanks for asking me to present evidence to you.

To a Vulcan, there can be no such thing as an unpleasant task. The provision or otherwise of pleasure in the completion of activities is something we neither consider, nor even experience. Nevertheless, some aspects of academia prove more easy to engage with than others. One of the most difficult tasks a scholar may find themselves required to perform is to demonstrate precisely where a colleague's logic has failed them.

Nevertheless, it is precisely this activity to which I must devote myself here. Doing so is necessary work. To insist one's logic is beyond criticism is in itself not logical, and a scientific discipline which refuses to scientifically analyse itself will not long remain robust or even functional. Further, I trust both the court and the eminent Professor Field herself to recognise I do not, and indeed *can* not, take pleasure in explaining where a fellow academic has misapplied the knowledge our shared discipline has granted us. I offer these criticisms not as a competitor, but as a fellow traveller, with an eye not on my reputation, nor anyone else's, but only upon the future,

Highlighting where Professor Field has made her errors is made all the more important here by the fact that she is *almost* correct. That is, the method by which she calculates the probability of all five named crewmen being picked by a fair lottery under the circumstances described contains no errors. The mistake comes in the extent to which Professor Field considers this probability relevant to the question of the of Captain Archer's guilt.

What Professor Field is dealing in here is what we refer to as conditional probabilities. In plain language, we often intuitively shift our beliefs about the likelihood of various future events occurring, when we acquire new information about what has recently taken place. To take a simple example, imagine a two-player game in which a fair coin is tossed three times. If there are more heads than tails, Player A wins. If there are more tails than heads, Player B wins. Before the game begins, each player has a the same chance of victory. There are eight possible combinations of coin tosses, listed below.

HHH HHT

HTH

THH

HTT

THT

TTH

TTT

We can see four of these eight combinations have more heads than tails, and the other four have more tails than heads. This justifies the claim that each player has the same chance of winning, though really this follows immediately from the construction of the game - if the coin is fair, and both players have exactly equivalent winning conditions (that is, at least two of a certain event, with those events being equally likely), neither player can begin with an advantage over the other.

After the first coin toss, however, the situation changes. If that first coin came down heads, Player A will win if *either *of the second two tosses come down heads - three of the four possible combinations of the next two coin tosses will therefore grant them victory. Player B can only win if *both *tosses come down tails; only one possible combination of the next two coin tosses will see them win. Player A sees their probability of winning increase from one half to three-quarters. Player B sees their own probability of winning shrink to one quarter.

The idea that one can and should update one's beliefs about the future as we learn more about what that future could look like is, then, entirely sensible. Sometimes we can even use the present to update what we believe about the past - revelatory new evidence regarding a historical event might prompt us to radically change our beliefs about what most plausibly might have happened. Such, indeed, is the approach Professor field takes, by suggesting observing the outcome of a lottery on a Starfleet ship should cause us to re-examine our assumptions about how likely a Starfleet captain - highly venerated individuals within your own culture - would be to fix such a lottery.

What these examples demonstrate is that the laws of conditional probability allow us to reconsider the probability of a specific event based on other events, which might happen even before or after the event we are interested in. They are not directly bound by the rules of causality, in that sense - all we require is that the event we take interest in does not have a result that is already known to us. To return to our example, just as we can calculate the probability of Player A winning *given *the first coin came down heads, we can calculate the probability of the first coin coming down heads *given *Player A won the game (assuming we neither witnessed the game, or were given details about the results afterward). As discussed previously, there are eight combinations of results for three consecutive coin tosses, of which four will contain more heads than tails, and result in victory for Player A. Of those four combinations, only one does not begin with a result of heads. As a result, the probability of the first coin having come down heads, given a victory for Player A, is three in four.

This is an example in which swapping the event of interest and the event held to be known does not result in a change of probability. Professor Field's testimony relies, however tacitly, on the assumption that this is always true - that the probability of innocence given the observed evidence is equal to the probability of the observed evidence given innocence. This is incorrect in general. Indeed, it is *obviously *incorrect. Many societies run variants on recreational lotteries which involve probabilities of winning that are vastly lower than 3%. Are we therefore to hold that every winner must have cheated? Professor Field is narrowly correct in describing the event of (effectively) winning a lottery as "implausible", *for any given player*, but the idea that winners *exist* cannot be taken as evidence of anything, except the fact that extremely unlikely events will eventually occur, if given enough opportunities to do so.

One response to this realisation might be to begin a search through Starfleet records for other lotteries conducted aboard ship, whether for pleasurable purposes as here, or in far less positive circumstances, with regards to remaining rations, life-pod access, and so forth. By Professor Field's calculations - which, as I note, are in themselves perfectly accurate - results so favourable to senior staff and bridge crews (or in more dire straits, the most senior officers remaining) will be witnessed more than once in every forty such lotteries. Are we to hold that every one of such lotteries were tampered with? If so, I would suggest the court may find itself very busy in the near future, as it processes the need for many more such courts martial, of which no small amount may require delivering posthumous sanction.

Even this, however, would not be the correct approach. To truly consider what we believe the chances of Captain Archer's guilt is, given the lottery result, we must compare how often lotteries are honestly won through extremely good fortune (a concept I take exception to, but which has useful rhetorical properties in this context) with how often a Starfleet captain of Captain Archer's calibre will knowingly commit a serious offence in order to secure two days of shore leave for him and his most senior/most favoured officers. If the former explanation seems the more plausible, it would be ridiculous to say an event which has only those two explanations must point to the latter.

We arrive at the necessity to introduce algebraic notation. For two events, which we label E and G, we express the probability of each happening as P(E) and P(G), respectively. We further express the probability of event E happening *given* we know event G has happened as P(E|G). The reverse of this, the probability of event G happening *given *we know event E has happened, is denoted P(G|E). It has been known on Earth since that world's 18th century that:

P(G|E)=(P(E|G) x P(G))/((P(E|G) x P(G)) + (P(E|G') x P(G'))

where G' is the circumstance where event G has *not *occurred. This result holds in general, but in this case we shall use G to mean the event that an accused person is in fact guilty, and use E to mean the event that is being used as evidence in a bid to determine whether or not guilt is present. In this case, E is the fact of the five named officers all receiving permission to land on Risa.

This equation shows that, in general, P(E|G) is not equal to P(G|E) - you cannot swap round the events under consideration and expect the probabilities to remain the same. Doing otherwise is referred to by my people as "The Shame Of Tyrok", though the human term of "Prosecutor's Fallacy" has perhaps more weight in the current context.

From Professor Field's testimony, we have that P(E|G') = 0.0292 (more precisely, the figure is 0.02919612). The value of P(E|G) = 1, since were Captain Archer guilty, we most definitely have seen the related evidence (a lottery which seemed to be fixed). Feeding these values into the equation above gives us

P(G|E)=P(G)/(P(G) + (0.02919612x P(G'))

This clearly demonstrates that we cannot judge the probability of Captain Archer's guilt after seeing the lottery results, without having some idea of how likely we would find Captain Archer to be guilty *without *knowing that result. The rules of legal process within your culture do not allow me to speculate on how likely a Starfleet court should consider such a breach regulations, but I am permitted to present you with alternatives. In the table below I offer a series of values for P(G), and the corresponding values for P(G|E), so you may see for yourselves how changes in the former are reflected in changes in the latter.

P(G) | P(G|E)

----------------------------------

0.01 | 0.2570419

0.05 | 0.6432000

0.1 | 0.7919129

0.5 | 0.9716321

Note then that, for instance, if the court believes that one in twenty (that is, 5% of) Starfleet captains would fix a ship-wide lottery to their own advantage (or perhaps that a Starfleet captain would succumb to such a temptation one time in twenty), the additional evidence of seeing a lottery that could only have been so favourable to the captain 2.919612% of the time had it been fair increases the probability of that captain's guilt to 0.6432000. Guilt would therefore be more likely than innocence, but not to the extent required or conviction, according to your culture's commonly held definition of "beyond reasonable doubt". The point at which we reach an actual probability of innocence around 3% - the value implicitly given by Professor Fields - is under the assumption that Starfleet captains are precisely as likely as not to cheat their crew. This is to be expected, as Professor Field's mistake in failing to consider the *a priori *probability of Captain Archer choosing to rig the lottery is essentially an unconscious assumption that him doing so was precisely as likely as not.

Again, it is for the court to decide for themselves what kind of person Captain Archer is, taking into account any and all aspects of his temperament, his service record, and the extreme stress of the mission undertaken by the *USS Enterprise*, into consideration as it pleases. All I can provide is the theoretical justifications for that holistic approach to be necessary. The results of what my colleague calls "The Risan Lottery" are not irrelevant to the question of whether Captain Archer is sufficiently plausibly guilty. That said, it does a great disservice both to our profession and Captain Archer himself to suggest, even by implication, that it is the only evidence which need be taken into account.

**Addendum**: Ultimately, Archer was cleared of all three charges, thereby avoiding both the loss of his command, and up to two years in a Starfleet correctional facility.

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