How exactly does 2^n work?

n is the number of dependencies an item carries. 2^n approximates the odds of finishing without delay, assuming each dependency has an even chance of slipping independently. Five dependencies is about 1-in-32. Ten is 1-in-1,024. Fifteen is 1-in-32,768.

Is the answer just to have fewer dependencies?

Fewer, yes — but also to strengthen the ones that remain. Challenge whether each dependency is real, necessary, or avoidable. Clarify ambiguous problems. Stabilise unstable decisions. Simplify the path. These five moves raise finishing odds without cutting scope.

The Problem

Dependencies Keep Delaying Delivery

Your teams are capable. Your plans look achievable on paper. And yet dependencies keep hijacking the finish line. The maths is less forgiving than most leaders realise.

Does this sound familiar?

Work items bundled with many dependent children
Cross-team handoffs stretching finish dates
Portfolio initiatives looking safe on paper, slipping in reality
Repeated rework because integration surfaces blockers late
Leaders asking for certainty when the maths says 1-in-1,000

You're not alone

Each dependency doubles the odds of delay. Five dependencies is 1-in-32. Ten is 1-in-1,024. Fifteen is 1-in-32,768. Teams routinely carry items with 10–20 dependencies without realising they are betting against themselves. This is the silent killer of delivery confidence — and it is maths, not attitude.

The IMIRT response

How the plays address this

Stack the Odds reframes delivery as managing probability, not just managing plans. Teams stop asking "Can we do it?" and start asking "What are the finishing odds?" Five moves strengthen fragile work: resize the work, remove or reduce dependencies, clarify the problem (ambiguity is a hidden dependency), stabilise the decision, and simplify the path. Improving finishing odds isn't about doing less — it's about making the work finishable.

Tease and gate

Get the complete plays with step-by-step guided exercises.

FAQ

Common questions

How exactly does 2^n work?: n is the number of dependencies an item carries. 2^n approximates the odds of finishing without delay, assuming each dependency has an even chance of slipping independently. Five dependencies is about 1-in-32. Ten is 1-in-1,024. Fifteen is 1-in-32,768.
Is the answer just to have fewer dependencies?: Fewer, yes — but also to strengthen the ones that remain. Challenge whether each dependency is real, necessary, or avoidable. Clarify ambiguous problems. Stabilise unstable decisions. Simplify the path. These five moves raise finishing odds without cutting scope.

Plays that address this problem

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