A friend in need
Is it safe to help a competitor before an exam or interview?
The relationships we form during studying and working are important during our medical careers. Many medical students and clinical trainees will instinctively help each other when under pressure.
Yet tougher economic times and narrower career pyramids might see doctors’ better instincts subverted by the fear that helping a competitor could harm their own success. Sadly, many students see “competition, rather than cooperation” as the defining characteristic of medicine. Although the practice of studying in groups is widespread among medical students, a strong element of competition still exists, and some students may attempt to undermine one another.
So how should you respond when a “competitor” asks for help in preparation for an exam? “Can you help me make sense of nitrogen metabolism? How can I remember the lifecycles of different parasites?” they might ask. Or, when competing for a job, “What weaknesses can I safely admit to that are actually strengths? What questions can I ask at the end to sound intelligent?”
The question of when it is safe to help a competitor has not been dealt with quantitatively in the literature. Statistics can help to quash the idea that helping your fellow students is detrimental to your own success. For brevity, we will present a simplified calculation and then consider how real life might differ.
The dilemma: a friend in need in a competitive setting
You are a random candidate facing a competitive selection process, such as a job application or interview, or a prize exam. You will succeed only if ranked among the top T candidates among a total of N. Another random competitor asks you for help. You are worried that helping her might displace you from success. What do you do?
What conditions will cause me to fail if I help?
For the act of helping your competitor to cause your own failure, three conditions must be satisfied.
The first condition is that without giving help you must be ranked exactly Tth out of N (if you were ranked higher, she could not displace you from success; if you were ranked lower, you were going to fail anyway). If you are a candidate chosen at random, the probability of this is 1/N.
The second condition is that she must be ranked somewhere from T+1 to N (otherwise she was going to beat you anyway). If she is a random candidate, the conditional probability of this, given that you are at position T, is 1-(T/(N-1)).
The third condition is that your help must be sufficiently powerful to cause her to overtake you in the rankings. The conditional probability of this occurring depends on two variables: how good your help is and how close she is to your rank of T. This conditional probability is difficult to establish and subject to considerable uncertainty because you have no way of knowing what either variable is.
A straw poll of colleagues indicates that the help they have received before exams or interviews had effects ranging from 0 (most common) to an estimated rise in ranking by 20 percentile points. (If you raise people’s rankings by more than 20 percentile points by giving brief assistance, this article may be not only incorrect but irrelevant—you could give up medicine and nurture your world class talent in coaching for interviews or exams.)
For simplicity, let us suppose that the distribution of the effect of your help runs uniformly from 0 (no effect) to raising your competitor’s ranking by H[max] percentile points. By how many percentile points does she need to rise to beat you? If she started at the bottom, the gap she must close is 100(1−(T/(N-1))), which can be approximated for simplicity as 100(1−(T/N)). If, at the other extreme, she started just one rank below you, the gap is nominally just 100/N (box 1). But we must recognise that, in real life, the competitive differences between individuals are not sized equally: some are almost ties, while others are widely spaced. To incorporate this, the calculation should preferably consider the gap between your competitive positions to be a continuous variable ranging from almost 0 (almost neck and neck) to 100(1−(T/N)).
Box 1: An example of the percentile gap
Suppose there are 75 candidates (N), and the top scoring 25 candidates (T) will succeed. If your competitor starts at the bottom, position 75, the percentile gap she must close is:
Unless N is very small, this can be approximated for simplicity as:
If, instead, your competitor starts just one rank below you, at position 26, we can estimate the percentile gap she must close as:
If we assume the distributions of each variable are uniform and independent, the conditional probability that the help will be sufficient depends on the relative size of the maximum helpfulness H[max] and the amount of help needed 100(1−T/N). For reasons shown in box 2, if H[max]<100(1−T/N), the conditional probability is equal to (1/2)H[max]/100/(1-(T/N)); otherwise it is 1-(1/2)(1-(T/N))/(H[max]/100).
Box 2: Working out the conditional probability that helping another candidate will cause you to fail
Let us start with a simple thought experiment. Imagine you are given a random amount of money, uniformly distributed continuously between £0 and £A. You want to buy an object, whose cost is an independent random amount uniformly distributed between £0 and £B. What is the probability that you can afford it?
The probability space of the money you have and the cost of the object can be imagined as an A×B rectangle, as shown in figures 1 and 2. 1 2 The probability of “success,”—that is, having enough money to buy the object—is the area of the dark triangle plus, if A>B, the area of the dark rectangle. Mathematically, this is one of two formulas, depending on whether A is larger (fig 1) or B is larger (fig 2).
Our friend in need question is a parallel to this.
The size of the help is somewhere between 0 and H[max], equivalent to the amount of money being between 0 and A. The amount of help needed to tip the balance is somewhere between 0 and 100(1−T/N), equivalent to the price of the object being between 0 and B. If we replace A and B in the probability formulas in figures 1 and 2 with our H[max] and 100(1−T/N), we obtain the formulas in our article.
Graphical analogy such as this can be useful in rapidly obtaining formulas. These can be confirmed by integral calculus, albeit a little more painfully.
What is the probability that all three conditions will occur together?
All three conditions must occur together, so multiplying the first probability and the two following conditional probabilities gives the probability that a random candidate helping another random candidate will cause failure of the candidate giving the help. The table shows the probabilities for a range of values of N, T, and H[max].
|With limited help||With infinitely powerful help|
|No of candidates competing (N)||How many at top will be successful (T)||Maximum likely effect of help (Hmax) (ranking uplift, %)||Approximate probability that help you give will harm you||Worst case probability that help you give will harm you|
Calculation of the probability of the third condition involves many assumptions and is vulnerable to error. If you are pessimistic you might wish to consider a truly worst case scenario of that conditional probability being simply 1—that is, your help is infinitely powerful. Even so, you can cause yourself to be displaced only if the first two conditions are met. The final column in the table shows this worst case probability.
As shown in the last two columns in the table, you are unlikely to harm your own success by helping others. It may be intuitive for nationwide competition, but this analysis shows that in random pairings harm is rare even when the size of the competition is only 100, such as between fellow students in a medical school.
Even when applying for core training, or higher specialist training, with just dozens of competitors, you can now see that helping a friend brings little risk of harming yourself. More surprisingly, even when the competition is among a handful, such as five, and the helping hand is worth somewhere between 0 and 20 percentile points, the resulting probability of your being harmed is only 0.02. So, even when competing against friends for consultant jobs, do not hesitate to assist them. Word of your ability to work as a team player may reach the ears of employers as you enter into a career long relationship with your colleagues or employer.
What are the limits of this analysis?
We are the first to admit that this analysis is simplistic and focuses on selfishness; therefore, it may not apply to all readers.
Firstly, it assumes that generosity is not reciprocated. In practice, mutual help carries even less risk of net loss: this analysis of one sided help should be ignored by such cooperating pairs. Helping a fellow candidate may cost time (time that might otherwise be spent on one’s own studying), but in real life you may receive reciprocal help. Even without reciprocation, the act of giving help can reinforce and even enhance one’s own knowledge and skills.
Secondly, it assumes that one’s own success is the sole motivation, with no place for joy in the success of friends, delight in collaborative enterprise, or even the simple pleasure in keeping good company.
Thirdly, this analysis applies only to competitive scenarios in which a fixed number of candidates will succeed—for example, for prizes at medical school or jobs beyond it. It does not apply to scenarios in which the aim is to exceed a predefined threshold to pass.
We recognise that most of us rate our ability highly and rate our friends also fairly highly (but perhaps not quite as highly as ourselves). This perceptual bias causes us to overestimate the risk that we stand just above our friends in the ranking and that our friends are dangerously close. This illusion is deceptive because we know only our own friends; there are other similar groups of friends elsewhere who think the same way about themselves. We should be careful not to let this natural misperception cause us to feel excessively threatened. We should also be careful not to overestimate the beneficial power of the help we give since such a misperception would further discourage us from helpful behaviour.
Although real life competitions might seem more complex than this calculation, we should realise that when one candidate approaches another, on average, the chance of the help being detrimental to the helper really is as low as that shown above. The ideal, cooperative solution is mutual exchange of help. If, however, you are still overwhelmed by the fear that their help to you may be less helpful than yours to them, and even the meagre probabilities listed in the table are too high, perhaps you are discovering that your personality is best suited not to the gentle art of medicine but to a more classically hard nosed calling that has cherished selfishness as the paramount goal.    It is never too late to make a change—the finance industry is hiring once again.Alexandra N Nowbar, fifth year medical student1, Darrel P Francis, professor of cardiology 2
1Imperial College London , 2International Centre for Circulatory Health, Imperial College London
Contributors: AN did the analysis and drafted and revised the manuscript. DF initiated the paper, did the analysis, and drafted and revised the manuscript. DF is guarantor.
Competing interests: None declared.
Provenance and peer review: Not commissioned; externally peer reviewed.
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Cite this as: Student BMJ 2014;23:g7345