Confidence interval

Question

Mathematics

Gentlemen, help me figure it out.
There is some periodic event - let's say I buy a lottery ticket every day. And in 3 years I won 2 times. It turns out the probability of winning 2/(365*2) = 1/365
But these 2 wins of mine could be random. How to assess whether the estimate of 1/365 is correct?
Or another example - we distribute leaflets on the street, out of 50 handed out leaflets, someone called us and made a purchase. How accurate is the estimate that every 50th person who receives a leaflet buys?

Answer 1

It's hard without a priori probability. One observation (or any number of observations) can clarify the probability of a particular situation, but cannot determine it from scratch.
For example, in the case of the lottery. It's one thing when you consider that the probability of winning can be anything, say, from 1/100 to 1/1000, and the probabilities of these values ​​are approximately the same. Then, after your observations, you can say that the probability is most likely close to 1/550, and even determine the distribution of this probability. Another thing is when you know that the lottery can only have a 1/200, 1/500, or 1/1000 chance of winning, but you don’t know which lottery you are playing (although the chances that you slipped each of them are equal ). Then your observations will show that the probability is most likely 1/500 (and not 1/550 at all - since such an outcome was not on the list).
So you have to take or come up with a priori probabilities of the models, and use the Bayes formula.

Answer 2

With a lottery, if its rules haven't changed in the meantime, a score of 2/(365*3) will be fairly accurate.
With leaflets, everything is much more complicated - it is necessary to assess the representativeness of the sample, that is, how exactly the 50 people who received the leaflets correspond to the audience among which all leaflets will be distributed and the target audience for which they are designed.

Answer 3

You're trying to plot a distribution function from a single observation - that's not possible. You do see the expected value, but you don't see the variance. Type at least 5 observations.