# Given an event with probability of 0.5%, how many times do I need to repeat it to have a probability of 6%?

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.

• Since each run of A has a chance of producing X, running A multiple times may produce multiple X. Is this a problem? – Stephan Kolassa Jun 26 '19 at 13:35
• @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? :) – Klors Jun 26 '19 at 14:18

@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)

• 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? – Dave Jun 26 '19 at 14:10
• 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 – David Jun 26 '19 at 14:45
• 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. – Dave Jun 26 '19 at 15:24
• Could you please provide further details about how you get 12 as the answer? – David Jun 26 '19 at 15:29
• N = 0.06/P = 0.06/0.005 = 12 – Dave Jun 26 '19 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.

• Correct, having it happen more than once is not a problem. Thanks – Klors Jun 26 '19 at 14:26