If I am given two repeatable events -

  • event A - that I am told has a probability of 0.5% to produce result X
  • event B - that I am told has a probability of 6% to produce result X

and given that the events have a cost associated with them, can I approximate how many times I would expect to have to perform event A to get result X with the same probability as performing event B once?

ie. how many times would I expect to have to do event A to get a probability of 6% of producing result X, to find out if the cost of doing event A that many times works out more or less expensive than just doing event B once?

Apologies if this has been asked before, but I don't know much about probability and what to search for, the other questions of this form all seem to want to know how many times to repeat event A to get a probability of 100% with some confidence level.

  • $\begingroup$ Since each run of A has a chance of producing X, running A multiple times may produce multiple X. Is this a problem? $\endgroup$ Jun 26, 2019 at 13:35
  • $\begingroup$ @StephanKolassa no, that would not be a problem. I'm approaching it from a perspective of a value proposition, would it in theory be better value to attempt Event A however many times, or just to attempt Event B once. Or alternatively, what cost should I assign to Event B to make it theoretically a better value proposition than doing Event A however many times. If that makes any more sense? :) $\endgroup$
    – Klors
    Jun 26, 2019 at 14:18

2 Answers 2


@Dave's answer is right, but I think you can approach it in a simpler way. Think instead of the probability of NOT getting result $X$, which is $0.995$ if you try once, $0.995^2$ if you try twice.... and $0.995^n$ if you try $n$ times.

In short, you only have to solve $0.995^n = 1-0.06 = 0.94$ With $n=13$ you get under $0.94$ for the first time (93.6% chance)

Finally, there is no way to guarantee that event $A$ will produce result $X$ for any number of trials, since $0.995^n$ is always greater than zero (it actually makes sense if you think about it)

  • $\begingroup$ This was my first thought, but I like phrasing it in terms of expected value because that makes explicit that we're talking about having a 6% chance on average. It does concern me, however, that this methods gives 13 and mine gives 12...continuity vs discreteness issue? $\endgroup$
    – Dave
    Jun 26, 2019 at 14:10
  • $\begingroup$ The result was "12.something" with that something being small, but since $n$ must be an integer, I rounded up the result. The 6% chance is a 6% chance every single time you repeat the experiment, there is no average there $\endgroup$
    – David
    Jun 26, 2019 at 14:45
  • $\begingroup$ I agree with rounding up to 13. What I'm wondering is why my approach results in exactly 12, while the calculation you did results in 12.something. $\endgroup$
    – Dave
    Jun 26, 2019 at 15:24
  • $\begingroup$ Could you please provide further details about how you get 12 as the answer? $\endgroup$
    – David
    Jun 26, 2019 at 15:29
  • $\begingroup$ N = 0.06/P = 0.06/0.005 = 12 $\endgroup$
    – Dave
    Jun 26, 2019 at 15:50

Assuming that it's not a problem if X happens more than once, you can just reverse the probability and calculate the answer very simply. If event A has a 0.5% chance of producing result X, that means it has a 99.5% chance of not producing result X. How many times do you need to multiply 99.5% (i.e. 0.995) by itself, to get down to 94% (0.94) ? Let t be the number of times:

$0.995^t = 0.94$

$log(0.995)t = log(0.94)$

$t = \frac{log(0.94)}{log(0.995)}$

$t = \frac{-0.0268721464}{-0.00217691925}$

$t = 12.3441171957$

Since t needs to be a whole number, you set it equal to 13. Now you can check your answer:

$0.995^{13} = 0.93691469288$

That means if you repeat event A 13 times, you have a 6.3% chance ($1-0.93691469288$) of result X happening at least once.

  • $\begingroup$ Correct, having it happen more than once is not a problem. Thanks $\endgroup$
    – Klors
    Jun 26, 2019 at 14:26

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