# Statistical fallacy from a Japanese light novel

The following is a snippet of dialogue from the first volume of the Japanese light novel series Combatants Will Be Dispatched:

"Listen here, number 6. This teleportation machine hasn't failed even once, so the success rate is currently 100%. But what if I kept running trials, and an accident occurs? That's right, the 100% success rate will be no more. In other words, if I keep running more trials, the success rate will drop. That means the odds of you safely teleporting will drop as well."

…I, Combatant number 6, thought about Lilith's words.

"There’s no way that's the case! There's something off about the way you calculate the odds! You're definitely not a genius scientist! They say that there's a razor-thin line between an idiot and a genius, and you're definitely an idiot!!"

So, there is a statistical fallacy in the scientist's reasoning, but I don't think we covered it yet in our high school statistics classes. I know it's a fallacy, but I would certainly appreciate getting the proper terminology for the fallacy, and an explanation for why it is a fallacy. (References will also be appreciated.)

• Oct 8, 2023 at 10:26
• I think Ben's and @Silverfish 's and others' rich answers about statistical reasoning shows that 夏目漱石's question has a there there. Am voting to reopen. Oct 9, 2023 at 15:15
• @Alexis With luck, perhaps it would be the start of a series of questions about statistics in literature. I recall "The Woman in the Dunes" has a whole paragraph on the Normal distribution. Oct 9, 2023 at 15:20
• @Silverfish In my opinion “…it's not straightforward to map such statements onto a taxonomy of "fallacy X" or "bias Y…" is exactly part of a good non-opinion answer to a good question which belongs on the site. Oct 9, 2023 at 16:59
• +1 I imagine something like this can be used in teaching to point students to the dangers of mixing up the estimator with the estimand. Oct 10, 2023 at 14:57

Accepting the premise that no failures have occurred in the sample so far, I think there are three separate steps in the reasoning here.

(A) The best estimate of the probability of failure is currently zero.

(B) The estimated probability of failure cannot decrease. But a failure will occur one day, so the estimated probability of failure is certain to increase eventually.

(C) Since the probability of failure cannot decrease and may increase, then (assuming you must use the device at some point) it is safer to do so now than to wait.

Does anyone actually read this stuff? I have put some funny pictures at the bottom instead. Scroll down if you like, you may prefer them.

For (A), two related posts show how to estimate the probability of failure, when no failures have yet occurred: see here and here. The true probability of failure remains unknown. These posts show it is not correct to deem the current probability of failure to be zero, merely because no failures have occurred in the sample observed to date. Since the device could have failed, even though it didn't, claiming the probability of failure was zero in those attempts is a confusion of ex-ante and ex-post. Focusing on the empirical (observed) probability of success of this particular device, over its recent uses, rather than considering the general rate of success among all such devices over a longer time period, is a classic example of the base rate fallacy, and to some extent the availability heuristic and recency bias. (Even if the device is "unique", we can still consider a wider set of somewhat similar devices, or analyse the likely performance of the device based on the performance of its components and the way they interact. This answer lists some references for how Probabilistic Risk Assessment is used for novel spacecraft or nuclear power plants where engineers face similar issues of lack of long-term data about new or unique designs.)

For (B), the very fact that a future failure is even being countenanced is sufficient to show that the speaker does not believe the current probability of failure is zero, despite what they previously stated! So there is a glaring inherent logical contradiction in what the speaker is saying, regardless of the probability model we might use to analyse the situation. But this is also revealing a flaw in the statistical analysis: why would the estimated probability of failure get worse if a failure occurs, but only stay the same if there is another success? If a Bayesian has the prior belief that the probability of failure is truly zero, then any subsequent successes will not cause them to update their belief. But this would be a very poor choice of prior. See Cromwell's rule: "I beseech you... think it possible that you may be mistaken." Moreover, the fact the speaker countenances the possibility of failure shows this case does not apply here. Nor would the more sensible estimators linked to in the paragraph above behave this way. The probability of failure at the next attempt might already be estimated to be very low if lots of successes have been observed so far, but it will get even lower (without hitting zero exactly) if there is another success. This would even be the case for a simple estimator like empirical probability, if you had already included some failures in the "track record". For example, if there had been 2 failures in 9 attempts, then one more success would bring the empirical probability of failure down to 2 in 10 (a reduction from about 22% to 20%).

Provided you accept its premises, (C) seems more reasonable. Considering what I said about (A) and (B), could those premises ever be fulfilled? I previously argued against the premise that "the probability of failure cannot decrease and may increase" if we are measuring it properly, but I was making some implicit assumptions when I did so: that the true probability of failure is an unknown constant, that events are independent, and so on. There are other models where this premise is true, and (assuming we also accept the premise that we must use the device at some point) it is indeed safer to do so now rather than wait.

Here is one such model. An interesting feature of the claim made here is "once a failure has occurred, failures become more likely in future". The reasoning behind it was incorrect (just because your estimated probability rose, it doesn't mean the true probability rose — you may simply have been underestimating the risk and received a reality check!) but it isn't absurd to claim events are not independent. For mechanical devices, "it's more likely to fail once it's failed before" seems a decent heuristic — many of us have had a car that was never the same after its first breakdown. Let's suppose the device's state of maintenance is either adequate (state $$A$$, fails 20% of the time) or shoddy (state $$S$$, fails 60% of the time) and your faith in the engineering team is such that you initially believe it's 50:50 whether the device started in the adequate or shoddy state. The device's state stays the same except, after any failure, its state permanently becomes "shoddy" — so if it was previously adequate then all subsequent use becomes more risky, but if it was shoddy already then, well, plus ça change...

If the device has a 100% track record of two successful uses out of two attempts, how would you (in Bayesian parlance) "update your prior belief" that the device is 50:50 to be in the adequate or shoddy states? We apply Bayes' theorem! Given that the device was shoddy, the probability of observing two successes uses (which I'll denote $$UU$$) out of two is $$\Pr(UU|S) = 0.4 \times 0.4 = 0.16$$. If the device was adequate, that rises to $$\Pr(UU|A) = 0.8 \times 0.8 = 0.64$$. So what was the probability of observing two successful uses? We need to weight our previous answers by your prior probabilities for the state of the device:

\begin{align} \Pr(UU) &= \Pr(UU|A) \Pr(A) + \Pr(UU|S) \Pr(S) \\ \Pr(UU) &= 0.64 \times 0.5 + 0.16 \times 0.5 = 0.4 \end{align}

How likely do we now think it is that the device is in an adequate state, given its two successful uses?

\begin{align} \Pr(A|UU) &= \frac{\Pr(A\text{ and }UU)}{\Pr(UU)} \\ \Pr(A|UU) &= \frac{\Pr(UU|A)\Pr(A)}{\Pr(UU)} \\ \Pr(A|UU) &= \frac{0.64 \times 0.5}{0.4} = 0.8 \end{align}

Given the new evidence, our new (posterior) belief is that there's an 80% chance the device state is "adequate". How do we feel about using it on the third attempt? There's an 80% chance it's adequate and fails 20% of the time and a 20% chance it's shoddy and fails 60% of the time, so the probability of a failure $$F$$ on the third attempt given the available evidence is

$$\Pr(F|UU) = \Pr(F|A)\Pr(A|UU) + \Pr(F|S)\Pr(S|UU) = 0.2 \times 0.8 + 0.6 \times 0.8 = 0.28$$

Contrary to (A), despite the 100% observed success rate, we don't think there's zero chance of failure. If we don't fancy a 28% chance of failure, what if we decide to stand back and observe another use of the device first? This will give us more information to update our beliefs. I'll reset things so our new priors for adequate and shoddy are 0.8 and 0.2. If we observe a failed attempt, then even if the device wasn't shoddy before, we know for sure it's shoddy now! But if we observe a successful attempt,

$$\Pr(A|U) = \frac{\Pr(U|A)\Pr(A)}{\Pr(U|A)\Pr(A) + \Pr(U|S)\Pr(S)} = \frac{0.8 \times 0.8}{0.8 \times 0.8 + 0.4 \times 0.2} = \frac{8}{9}$$

So the evidence suggests we should update to an even higher (about 89%) probability the device is adequate, and our estimate for the probability the next use will fail drops to $$\frac{8}{9}\times 0.2 + \frac{1}{9}\times 0.6 \approx 0.244$$.

So contrary to (B), the estimated probability of failure can fall as we observe extra successes. (To see the relevance of Cromwell's criterion, rerun the above calculations with a prior where you're 100% certain the device is adequate. You'll find no number of successes will cause you to change that belief, so your estimated probability of failure gets stuck at 20%.)

Does the fall in your estimated probability that the next use of the device will fail, mean that you made a wise choice to delay using it yourself? Well, if you'd observed a failure instead, I suspect you'd also say the decision to delay was very wise, since it means you avoided an early demise! In truth it sounds like you'd rather delay than use, because you just don't want to use it at all — and who could blame you? But if you have to go, you're still better to go early, as (C) suggests. With each use, there's a 20% chance an adequate device becomes shoddy. If the probability the device was initially adequate is $$p_A$$, then the probability it is still adequate after $$n-1$$ uses is $$0.8^{n-1} p_A$$, so clearly decreasing (provided $$p_A > 0$$ i.e. there was at least some chance it was initially adequate). The probability of failure must be rising: the chance of failing on the $$n$$-th use is $$0.2(0.8^{n-1} p_A) + 0.6(1 - 0.8^{n-1} p_A)$$. Using the device at the third attempt is better than using it at the fourth, regardless of the true probability that the device was initially in an adequate state. Collecting more data lets you better judge which "state of the world" you are in, but by itself doesn't improve the state of the world. Since (C)'s premise of an increasing probability of failure is, in this model, actually fulfilled, then its exhortation was correct. Things are only going to get worse so you might as well get on with it!

Now, I might be being overly generous in the way I wrote (C). I interpreted references to increasing probability of failure as being to the true probability, but given the context, perhaps they mean it's better to go now merely because the estimated probability of failure is increasing. If that is the case, it's another logical error, because it's the true probabilities that really matter. Let's alter our toy model so that even if the estimated probability of failure might improve if you wait an extra use, the actual probability always gets worse. At the moment a successful use of an adequate device does no harm, but we could introduce a small probability that the device transitions to the shoddy state despite, to all appearances, performing its job successfully. Or a continuous model of wear and tear, where the 20% and 60% failure rates of adequate and shoddy devices gradually deteriorate with use, towards e.g. 30% and 70% respectively. Something like $$\Pr(F|A)=0.3 - (0.3 - 0.2)\lambda^{n-1}$$ would work, where $$0 < \lambda < 1$$ is a damping parameter (near one means slower deterioration). If we observe a few more successful uses, we will become more confident that the device is an adequate state. If this overwhelms the effect of deterioration, our estimated probability of failure will decrease.

But you'd still be irrational to wait for that to happen before using the device! An omniscient observer with access to the true probabilities would know that by waiting you are taking a worse chance now than if you'd not waited. You just feel more comfortable with the risk you were taking due to your lower estimate for it. And if you yourself reassessed what the probability of failure was on the turn that you skipped, given your new estimated probabilities for the state of the device, even you would see it would have been better to go then! In the end it's the true probability of failure which gets you. The fallacy of replacing a concept in your reasoning by your estimate or measure of it, is called surrogation.

These estimated probabilities become much more decision-relevant, though, if I remove the premise in (C) that you must use the device eventually. Instead you can weigh up the payoffs of chickening out, using the device and it catastrophically failing, or successfully using the device and engaging in your mission (which in a decision tree may branch out into outcomes of various degrees of success and failure, each to be weighted by its own payoffs and probability of occurrence). We are in the realm of decision making under uncertainty and the value of (imperfect) information. Another option is to see how well the device works a couple more times before deciding to use it, or not, yourself. This lets you make a better estimate of the probability the device will fail. If there's no way for this to change your decision — either you have no choice or it can't influence the expected payoffs enough — then this information is of zero value. If it can influence your choices, then it has a positive value, which means you'd be prepared to pay a price for it. In business analytics we'd express that in monetary terms. In this context, you would accept a certain reduction in expected payoff: perhaps the delay allows a key target to escape (reducing the payoff of a successful mission) and the enemy to strengthen their defences (reducing the probability of a successful mission). You'd even accept some added personal risk from additional wear and tear to the device.

If better knowledge of what state of the world you're in lets you make better decisions, the value of that information can justify you entering a somewhat worse state in order to obtain it. So when you have the option of postponing a go/no-go decision until more evidence is available, we are no longer so persuaded by the argument we should "do it now, because the situation will only get worse while we collect more data" — even if every word after the "because" is true. To make my formulation of (C) reasonable, I emphasised the lack of choice. But if the characters believe it's possible (albeit cowardly) to decide not to go ahead, their version of (C) is erroneous. Choice matters: choices give information a value, which we can trade off against the costs (including additional risk) of obtaining the information. Those costs are no longer the not be-all and end-all, but must be set against the benefits of being able to change your mind based on the information: if the extra data suggests the state of the device is likely unacceptable, you can choose not to go even though you'd previously been inclined to try; or if previously unwilling to go, but would if further evidence suggests the device is good enough.

My aim in this answer was to explore some of the rational ways to approach decision-making under uncertainty, in the hope it illuminates some of the irrationality in your quote. Showing how those irrationalities map onto a taxonomy of Fallacy X or Bias Y is trickier. It would be nice to see an answer that looks in more detail at whether probability estimates are well calibrated as that's clearly an issue in this case. Another quick thought experiment: we could imagine an early prototype device that generally improves over time, as each use gives valuable real-world data to your engineer. By sheer luck it hasn't failed yet, but we might model the probability of failure falling as further improvements are made. This example really puts the lie to the claim "since the current failure rate is zero, the probability of failure is doomed to increase", even though it's true the empirical probability cannot improve. This illustrates why we shouldn't rely on the empirical probability to make our decisions. While this means we'd benefit from delaying our use of the device, the extent to which this goes against the advice in (C) to use the device ASAP is due to the premise of an increasing risk of failure being completely reversed, rather than a flaw in (C)'s logic. But nor would the analogous exhortation "it's always getting better, so you should always delay" be very helpful: you'd just postpone indefinitely. This is where you need a decision-theoretic framework that lets you trade off the benefits and costs of delaying.

# Funny pictures

Best viewed full-size in a new tab (browsers usually let you do this if you right-click on a computer or press and hold on a mobile device). Read a decision tree left-to-right, but payoffs flow through right-to-left. The expected value of a "chance" node is found by weighting its payoffs by their probabilities; the value of a "decision" node comes from selecting the optimum choice. State transition diagrams (see: Markov chain) show how the state of the device changes with each use. Numbers on arrows coming out of a state represent probabilities: arrows coming out of a state must sum to 100%, and arrows can loop back round if the state doesn't change. These diagrams are not "solutions" to the problem, but rather illustrations to show how (A) and (B) can fail and to explore why the exhortation to "go as soon as possible, before the success rate gets worse" is surprisingly thorny.

## Figure 0: If the device improves with use, you're better off delaying

I numbered this differently as it's the only scenario I explored where the device tends to gets better with use: every time the device fails, it gets repaired. This returns an 'adequate' device to an adequate state, and one third of repairs on 'shoddy' devices fix an underlying bug, so now they work adequately. The longer a device has been in service, the more likely all its bugs have been ironed out (though even 'adequate' devices fail 20% of the time, it's still better than 60% for shoddy ones).

Contrary to (A), the 2 out of 2 success rate doesn't mean our estimated probability of failure is zero. Contrary to (B), the estimated probability of failure can improve if we observe another success. Another test run risks breaking out 100% record of success, but that doesn't stop us preferring to go later than go now — the true probability of success is what matters, and that never gets worse and sometimes improves.

## Figure 1: If the device deteriorates with use, you're best to go ASAP

In this (and subsequent) figures, shoddy devices stay shoddy, while adequate devices become shoddy if they fail. The more uses it's had, the more likely a device is to have degraded to the shoddy state.

Unsurprisingly, now the true probability of failure really does get worse with time, we prefer to take (C)'s advice and go early. But just when everything seems cut and dried...

## Figure 2: Choice changes everything, by giving extra information a value

In previous figures, the results from an extra trial run weren't useful to us, because we couldn't respond to them. But what if we got the choice to go or not, depending on whether we thought the device is in a favourable state? The value of extra data, yielding a clearer picture of what state we're in, means we may now prefer to delay!

The above tree is "abbreviated" to focus on the decision-making. Here's the full tree.

Of course we wouldn't always choose to delay: it depends on the numbers chosen for the payoffs and probabilities. If devices deteriorate quickly that would put us off delaying. The extra data can be of no value if it wouldn't change our mind. E.g. in the bottom branch of the tree, if we'd choose to go on the mission regardless of whether the trial run was a success or failure, then the trial run was pointless and not worth risking a deterioration of the device for.

## Figure 3: extra risk can be part of the price you're willing to pay for more information

Even when the probability of failure really is getting worse over time, having choice means we no longer believe "the increasing risk means you're better to go now, than to delay". You might quibble that in the above example, we only use the device if we see a third success, in which case the risk of the device deteriorating hasn't materialised (regardless of whether the device was truly adequate or shoddy all along). So I now add an extra element of risk if you choose to delay teleporting: even if you arrive in one piece, the guards are more likely to get you. As an extra disincentive to delay, I also reduced the mission pay-offs. Again it all comes down to the exact numbers used, but in this case delaying has a price you're (only just) prepared to pay. So yes, even when delay definitely makes things riskier, it can still be rational to delay and collect more data... provided you have the ability to respond to what you learned by doing so.

• I am only in high school, so this might be a bit hard to digest for me, but I am still thankful for you bringing up "Cromwell's rule", which seems very apropos to Lilith's fallacious claim. Oct 9, 2023 at 9:09
• @夏目漱石 I have rewritten my answer to incorporate a high-school level example, Let me know if this helps. Oct 9, 2023 at 13:28
• "it's more likely to fail once it's failed before" seems a decent heuristic This is because failures. Failure in one component might cause additional stress in another, causing cascade of failures. Or fixing one part may damage another (this is a common problem in software). Oct 10, 2023 at 15:42
• I came down here looking for funny pictures but all I could find were some incredibly informative flowcharts. Oct 11, 2023 at 13:06
• @user2296603 Luckily "puffery" is a valid defence in English law against false advertising charges: Carlill v Carbolic Smoke Ball Co [1892] EWCA Civ 1. :-) I teach decision trees to MBA students ("do we run another geological survey before our go/no-go decision to drill for oil?") or medics ("if we delay the diagnostic test, patients who really have the debilitating disease progress to a more advanced stage before we can treat it"). Nice change to write "go now or the👽leader escapes🛸"! Oct 11, 2023 at 17:32

#### Conflating an estimator with its estimand is the reification fallacy—it leads to the conclusion that "ignorance is bliss"

This is quite a humorous statistical approach. The main flaw in the reasoning here is that it treats an estimator of an underlying reality as if it were more important than the underlying reality that is subject to estimation! As Luca Citi points out in the excellent answer below, this is the reification fallacy (sometimes known by the saying that "the map is not the territory").

The proper way to perform statistical inference is to concern yourself with some unknown aspect of reality of interest (the estimand) and then value the relevant inference/estimator only as an imperfect means of knowing that estimand. When we do this, we want as much data as we can, because more data tells us more about the true estimand. However, the speaker here treats the estimator as if it is the determinative thing of importance, rather than the estimand itself. This leads to the absurd idea that a person is more safe if the statistical point estimator for success is more favourable, rather than if the underlying success rate being estimated is more favourable.

Since the estimator (rather than the estimand) is here being treated as the determinative thing of interest, and since the present value of that point estimator is as favourable as it can be, the speaker effectively adopts an attitude that "ignorance is bliss"—i.e. we are better off stopping when we have an estimator using less data, since the estimated success rate is presently at its highest possible rate. The speaker actually recognises the reality that the underlying true success rate (the estimand) is likely to be lower than the estimator, and it is precisely this fact that leads to the desire not to get more data, lest it expose this reality.

• "This is quite a humorous statistical approach." - the novel is intended to be a farcical comedy, after all. :) (This event is also why number 6 enacts revenge on Lilith a few volumes later, but to say more would be spoiler territory.) Oct 9, 2023 at 8:55
• I agree a key problem here is treating the empirical probabilities as if they are the thing that really matters - "ignorance is bliss" is a great way to phrase it in this case, but I wonder if there's a name for the phenomenon in general. The idea that the estimated probability of success can't improve ("it's already perfect, don't wait for it to get worse!") is a separate statistical error, that flows from a poor choice of estimator! Oct 9, 2023 at 13:36
• Estimated probabilities can be valuable in their own right, if they can help you make better decisions (eg analyses like expected value of imperfect information). There's a real irony in the way their importance has been elevated here, though: if you think probability estimates provide valuable information about the decisions you face, you generally would prefer more data, not less! The claim here is that more data might make the picture look less rosy, so it's better not to look... not such a rare attitude! Oct 9, 2023 at 13:46
• I had a hard time choosing between your answer and @Silverfish's, so in the end, I tossed a dime into the air... Oct 9, 2023 at 18:05
• I really like this answer because, in my opinion, it hits the nail in the head. I have expanded upon it in an answer below: stats.stackexchange.com/a/628377/146890 Oct 10, 2023 at 10:58

The post that J-J-J gave is indeed related, but that's more about finding the actual probability.

The flaw in the reasoning is that of treating a sample as a population. They have a sample of however many (it doesn't give a number). But Lilith then reasons as if it were a population -- it hasn't failed, therefore it can't fail.

But she contradicts herself. She acknowledges that, at some point, there will be a failure. That means that the failure could be the next event and poor number 6 will .... whatever happens when the device fails.

If the likelihood of failure on any trial is independent, then this is a variation on the gambler's fallacy.

• Also, since they only had one teleportation device at that point in the novel, I am guessing the likelihood of failure cannot be independent, since each use would certainly contribute to wear and tear. Oct 8, 2023 at 11:07
• @夏目漱石 "I exercise my right to remain silent on the number of trials". One might wonder if the number of trials is really small, like 1. In addition, what is Lilith's definition of "success"? Nothing says that the previous trial(s) were conducted on living organisms. I'm sorry to say that Lilith's pep talk might deviate a bit from usual informed consent guidelines. Oct 8, 2023 at 11:32
• @J-J-J I am pretty sure that what you suppose was exactly what Akatsuki (the novel's author) was implying with Lilith's dialog. Oct 8, 2023 at 12:07
• This almost seems an "inverted gambler's fallacy", since in the classic form the gambler believes we are "owed" an event - the fact it hasn't happened as often as expected makes it more likely in future. "The device keeps succeeding - hurry up and use it, all the successes mean it's getting ever more likely to fail!" would fall into that form. In contrast, the claim here seems to be "once it fails, it's more likely to fail again!" so the reverse of what a classic fallacious gambler would say... Oct 9, 2023 at 0:12
• ... To be fair, "once it fails, it's more likely to fail again!" seems a decent heuristic about mechanical devices in general (many of us have had a car that was never quite the same after a breakdown), as your answer hints at by stressing the importance of the independence assumption! But the reasoning supporting it is erroneous: treating empirical probabilities as the true probabilities (I wonder if this is a fallacy with a name in its own right?), while - as you say, self-contradictorily - treating eventual failure as a certainty when the empirical probability so far is zero. Oct 9, 2023 at 0:23

This is quite a humorous statistical approach. The main flaw in the reasoning here is that it treats an estimator of an underlying reality as if it were more important than the underlying reality that is subject to estimation!

It made me think of a more general type of comic trope that is rather common, for example in cartoons. It's when a character tries to affect a process by acting on a measurement/indicator of it (the case of the statistical estimator is an instance of this). For example, I immediately though of Wile Coyote trying to stop a car with broken breaks by turning the speedometer needle to zero (I don't think I have actually seen this sketch but I imagined it based on my internal model of the type of humour in that cartoon). While I have seen this type of humour a lot, I could not name it so I have asked ChatGPT about it. Here is its answer:

The situation you described, where someone tries to affect a process by manipulating an indicator or measurement of that process, is a form of a logical fallacy known as the "fallacy of misplaced concreteness" or the "reification fallacy." This fallacy occurs when an abstraction (in this case, the speed measurement) is treated as if it were a concrete, tangible reality.
In the context of comedy or absurd situations, it can also be seen as an example of "magical thinking" or a "comic exaggeration" where the character believes that changing the measurement will directly impact the reality of the situation. The humor often arises from the absurdity or incongruity of the action.
In the case of Wile E. Coyote turning the speedometer needle to zero to stop a car with broken brakes, it's a classic example of cartoon physics and comedic exaggeration, where the character's actions defy the laws of physics in a humorous and exaggerated way.

Note: As suggested by Silverfish, this type of error is often called surrogation, especially in management/business analytics context. See their comment below.

Additional note: I stumbled upon a website about tropes in cartoons and they have a page dedicated to elevators. One such trope falls into our category:

Elevators sometimes have clockface-style indicators above the doors with an arrow that points at the number which correspond to the floor the elevator is currently on. In comedies, it is possible to control elevators by manipulating these indicators — pulling the arrow to move the elevator car, or, alternately, trapping the elevator on or between floors by blocking the arrow.

• Yes I agree with this (though I think it is one of several flaws in a very short space, impressive for such a short piece of absurdist humour!) ... not only is the measure of a "100% success record" being treated as unduly important, at times (though glaringly not at others) it's being treated as if it really is the true probability. While editing my answer I realised that this error is called surrogation, though I've mostly seen that used in management/business analytics context. Oct 10, 2023 at 22:22
• I hope this proves I cannot be replaced by ChatGPT yet :-) Classic example is a manager taking steps that massage the results of the firm's customer satisfaction survey, instead of focusing on things that will improve actual customer satisfaction! But there's a similar issue with potentially dodgy use of surrogate outcomes in medical stats - something Ben Goldacre has written a lot about (see eg this paper) Oct 10, 2023 at 22:42

There is no fallacy.

The act of testing the machine must stress its workings in some way. Once too much testing stress has accumulated, it's possible that a part of the machine would break, resulting in a failed test and increased risk to future users.

This picture is perfectly consistent with the scientist's words.

There is only a fallacy when one additionally assumes that there can be no causal link between testing and machine health. In that case, we have a logical fallacy, not a statistical one, since the scientist is directly asserting such a causal link (but doesn't mention the mechanism).

• +1 for pointing out that using the machine can and will increase the probability of a failure. That is a very good point. However I disagree that there is no fallacy, as the reasoning in the other answers still applies. So apparently we are facing a trade-off: testing more has the advantage of getting better estimates of the risk of failure but the downside of increasing this risk. What, under these opposing pressures, is the optimal amount of testing is a very interesting and very hard question and it might even be different for the two characters as they value number 6's life differently. Oct 9, 2023 at 9:56
• There is no reason this must be the case, though - some machines have what's called a "burn-in" period where the chance of failure goes down as the number of trials increases (a lightbulb, for example, is most likely to fail either the first time you turn it on, or after thousands of hours). Seeing the observed failure rate increase above 0% as more trials are conducted may not be statistical evidence that the actual failure rate is changing. Oct 9, 2023 at 13:46
• @NuclearHoagie The scientist's claim is of the form $A$ causes $B$. It may not be true, as you point out. On the other hand, it may be true. So it is not automatically a fallacy.
– Mark
Oct 9, 2023 at 15:50
• @Mark Aren't we Affirming the Consequent? We know that "machines with increasing failure probability" is one class of machines, and that the scientist's apparatus is a machine, but from that alone it is fallacious to conclude that the scientist's apparatus has increasing failure probability. The conclusion that the machine becomes more dangerous the more you use it is not logically valid based on what we know. That it may be true doesn't make it a valid conclusion. Oct 9, 2023 at 17:20
• @NuclearHoagie I wonder if we are talking past each other a little. I claim that "My car is yellow." You reply, "Yes, but there are many classes of cars, and many of them are not yellow." You are correct that I have not proven my claim. But my claim itself not in the category of logical fallacies.
– Mark
Oct 9, 2023 at 19:57

Besides the reasoning flaws explained in the other answers, other possible flaws in Lilith's presentation of the situation are:

• Talking about "this machine". If there are other machines not really different from this one but that have experienced failure, this is possibly a case of cherry-picking.
• Not defining what a success is. Is a success teleporting someone and keeping them alive at the end of the process? Or is it just teleporting the body, no matter if the person is dead or alive at the end of the process? Has the machine even been tested on living organisms before? This might be a case of persuasive definition.
• In a comment under another answer, you say that she mentions her "right to remain silent on the number of trials". She uses a concept from a judicial context (right to silence) that is not relevant here, to dismiss the idea that she might have a duty to give complete and accurate information. This rhetorical device is sometimes called "Diminished Responsibility". Not giving complete and accurate information to someone participating in an experiment or a study may be in violation of several laws and ethical guidelines.
• The novel seemed to imply that the teleportation machine is a prototype, so number 6 and his partner were sweet-talked by his superiors into the machine only after however few trials. (Also, I did not bring this context up in the question, but the quoted conversation happened while number 6 was already in the machine and the machine was already running...) Oct 9, 2023 at 9:15
• @夏目漱石 Yes, I guess you did not mention everything, but my point is that with the information you provided, the whole thing seems quite shady :) Anyway, this is quite fun to find how much information is missing for someone to be really informed about the risks they're taking by using this machine. Oct 9, 2023 at 9:49

Three are 5 ways to communicate a mathematical idea, and while some of them have been shared above, one has not. It's a kludge, so its absence is unsurprising.

The 5 modes are: words (word-problems), tables of numbers, graphs/plots, flowcharts/programs, and algebraic/symbolic structures. Here, I am looking at the programmatic style of engagement with the problem.

Also, seeing how you are a high-school student, an accessibly approach can serve you well over your lifetime.

Think about flipping a coin. Do you get heads or do you get tails? Some folks choose who starts sports games using a coin flip.

For a non-cheat coin the ratio of rate of occurrence of either face coming up to all outcomes is very close to 50%. If you flip it randomly 1000 times you would expect about 500 heads and about 500 tails. NOT exactly the same, and NOT exactly 500 every time.

So lets use the R-language and make a numeric coin flipper that truly random flips truly fair coins to see how many come up one side out of 1000 trials. I will assume you have worked through "getting started with R".

Here is some rough code:

coin_weight <- 0.5
num_trials<- 1000

y <- sample(x = coin_faces,
size = num_trials,
replace=TRUE,
prob = c(coin_weight, 1-coin_weight) )

summary(y)


The output it gives, for me, for this one time, is this:

> summary(y)
504   496


If I rerun the code the results are different.

So lets wrap it in a loop, and run this 30k times, and get a sense of how the number of "heads" varies. We will add a variable for the number of repeats, and initialize a store for the counts. Then we will also need a way to visualize the results, and I will use a ggplot for it.

Here is the updated code:

coin_weight <- 0.5
num_trials <- 1000
num_repeats <- 30000

store <- data.frame(prob=coin_weight %>% as.character(),

for(i in 1:num_repeats){
y <- sample(x = coin_faces,
size = num_trials,
replace=TRUE,
prob = c(coin_weight, 1-coin_weight) )

store$count_heads[i] <- summary(y)["heads"] } ggplot(store, aes(count_heads)) + geom_histogram(bins=35 ) + xlab("number of heads found") + ylab("number of times drawn")  Here is the graph: We aren't exactly to your situation, but lets adjust some numbers. What if there have been 10 trials of the machine with no failures? The "prevalence" of failure is 0%, but the binomial confidence interval includes numbers that are not zero. The "Jeffreys prior" says the 95% CI for the proportion includes 21.72%. This means that something with a ~20% failure rate can rarely (but not never) pretend it has no flaws with only 10 trials. I like to use this webpage. https://epitools.ausvet.com.au/ciproportion So lets update the prob to be 10%, the mid-interval, and lets update the number of draws to be 10, and see what fraction of the time the result is zero heads. Here is the code: coin_weight <- 0.1 num_trials <- 10 num_repeats <- 30000 coin_faces <- c("heads","tails") %>% as.factor() store <- data.frame(prob=coin_weight %>% as.character(), count_heads=(1:num_repeats)) for(i in 1:num_repeats){ y <- sample(x = coin_faces, size = num_trials, replace=TRUE, prob = c(coin_weight, 1-coin_weight) ) store$count_heads[i] <- summary(y)["heads"]
}

geom_histogram(bins=35 ) +
summary(store$count_heads %>% as.factor())  Here are the graphic results: Here is the result of the "summary": > summary(store$count_heads %>% as.factor())