the memoryless property of the geometric distribution

Suppose you have been rolling a die for ten turns hoping for a six, and it still has not come. How much longer should you expect to wait now? The unsettling, true answer is: exactly as long as when you started. The die does not know it is 'overdue'. This is the memoryless property: past failures give you no information about how much longer the wait will be. The geometric distribution is the only discrete distribution with this property.

Stated precisely, for a geometric waiting time X, P(X > m + n given X > m) = P(X > n). In words: given that you have already survived m failures, the conditional probability of needing more than n further trials is the same as the unconditional probability of needing more than n trials from a fresh start. You can see why directly: surviving m + n failures requires m + n independent misses, P(X > m + n) = (1 - p)^(m + n); dividing by P(X > m) = (1 - p)^m cancels the (1 - p)^m and leaves (1 - p)^n = P(X > n). The history simply factors out, because each trial is independent and identically distributed.

This is the mathematical heart of the gambler's fallacy. People feel that after a run of failures a success is 'due', but a memoryless process has no such bookkeeping — the coin is never owed a head. Memorylessness is also why the geometric and its continuous twin, the exponential, model events that occur 'at random' with no aging: radioactive decay of an undisturbed atom, the time to the next phone call at a steady call centre. By contrast, a light bulb that wears out is NOT memoryless — an old bulb really is closer to failing, so its lifetime needs a distribution like the Weibull, not the exponential.