The odds of getting it right

While it is easy to point out when people are getting things wrong – IMHO or in your opinion – it may serve a greater purpose to examine why things may go so utterly wrong as they often do, especially when we’re speaking about software development.

Software development is mostly about communication. Whether it is communicating with a programmer to make what you want, or it is telling a project manager to get them to tell a programmer what you want – it is in any case a matter of communicating vision to understanding.

So let us try to map out the different possibilities when facing a decision – or what may seem clear to you, but isn’t for at least one of the links in the development chain.

binary tree

binary tree

I have chosen a binary tree to depict the decision “right” or “wrong”. While normal interpretation of such a tree is that it is a 50/50 split, let us not make such a hasty assumption – at least we – as developers – should be better than a 50% guess at understanding customer requirements.

In the binary tree above there are only 4 decisions  which has to be right. If we simplify the model to have a fixed probability, p, that we make the right decision, we can use Bernoulli’s binomial distribution to determine the odds of making s successes in as many trials. In this case the binomial distribution deteriorates into a simple power function, ps.

Given either p or s we can calculate the other if we want at least a 50% chance of ending with a right solution.

Let us try that with a 6-sigma probability – p = 0.9999966.

s log p = log (.50) <=> s = log(.50) / log(p)

s = log(.50) / log(.9999966) ~  203867

That is, if we have an almost unheard of quality for understanding customer communication, then at a bit more than 200,000 decisions, the solution has a 50/50 chance of hitting the anticipated solution.

If we want to be 90% sure, then we cannot make more than 30988 decisions with 6-sigma understanding.

So, let us try the other way around – we would like to know with a sufficiently high confidence that our project meets our expectations, let us say 90% sure. We have identified 10,000 key decisions. How good must the communication then be?

s log(p) = log(x) <=> p = exp(log(x)/s)

p = exp(log(.90)/10000)  = 0.999989

Which means we need 6-sigma communication to achieve this goal.

On top of all this, then the calculations are assuming that the customer knows and communicates exactly what he or she wants, and that all decision points are uncovered and communicated at the same high level.

The only immediate sane solution to improving the odds is to reduce the scope drastically. It may sound silly, but more having to fulfill 10 things right becomes daunting for most of us. In the binary tree above, we would need to have 2 (10+1) -1= 2047 nodes – the sheer size of such a tree should be sufficient to deter anyone wanting more than 10 decisions.

Reduce scope. Improve communication by shortening the feedback loop.

Naturally, we could reduce scope right down until a single decision – but that would quickly throw us off balance, as a single point makes it impossible to determine direction.

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