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Geometric Distribution Calculator

Find the probability of the first success on trial k.

How it is calculated

Here p is the success probability of one trial, written as a decimal between 0 and 1, and k is the trial number. The tool counts trials starting at 1 and returns three values with five decimals: P(X = k) = (1 − p)k−1 × p, P(X ≤ k) = 1 − (1 − p)k and P(X ≥ k) = (1 − p)k−1.

The value of p must be above 0 and at most 1; a percentage such as 20 is rejected, so enter 0.2. The trial number is rounded to the nearest whole number, so 3.6 is treated as 4, and anything below 1 gives no result. The model assumes independent trials with the same p every time, and X counts the trial on which the first success happens, not the failures before it.

Worked example

Each attempt succeeds with probability p = 0.2. What about the 4th attempt?

The failure probability is 1 − 0.2 = 0.8. Exactly on trial 4: 0.83 × 0.2 = 0.512 × 0.2 = 0.10240. Success by trial 4: 1 − 0.84 = 1 − 0.4096 = 0.59040. Needing at least 4 trials: 0.83 = 0.51200.

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Common mistakes

  • Entering the probability as a percent. Use 0.2, not 20; values above 1 return no result.
  • Mixing up the two cumulative figures. P(X ≤ k) is the chance of succeeding within k trials; P(X ≥ k) is the chance that the first k − 1 trials all fail.
  • Using the version of the distribution that counts failures before the first success. This tool counts the trial number, so k = 1 means success on the first try.
  • Assuming a streak of failures makes success more likely. The trials are independent, so the chance on the next trial is still p.
  • Entering a fractional k and expecting it to be used as is. It is rounded to a whole number first.

About this calculator

The geometric distribution models the number of independent Bernoulli trials needed to get the first success, where each trial succeeds with probability p. The probability of the first success landing exactly on trial k is P(X=k) = (1−p)^(k−1)·p; the cumulative probability of succeeding within k trials (≤k) is 1−(1−p)^k, and the probability of needing more than k trials follows directly from the survival function (1−p)^k.

The distribution is memoryless — the probability of needing m more trials doesn't depend on how many failures have already occurred — which makes it a natural model for time-to-first-event problems. Quality control engineers use it to model the number of items inspected before finding the first defect; reliability engineers use it for time-to-first-failure in discrete trials; and it appears in sales (calls until first conversion) and in probability courses as the discrete counterpart of the exponential distribution.

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Comments (1)

  • Editorial team

    To ask how many tries you need to be fairly sure, use the cumulative figure and step k up. With p = 0.2, the chance of succeeding within 10 trials is 0.89263 and within 11 it is 0.91410, so 11 attempts are needed to pass 90%. Notice how far that is above 5, the average number of tries at this p. What success chance and what confidence level are you working with in your own problem?

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