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Suppose I have a coin and a die. When the coin is heads that is a success. When the die is 6 that is a success. Now, I have a dataset that looks like the following:

Trial # | Coin | Die     
   1    |   0  |  1
   2    |   1  |  0
   3    |   0  |  0
   4    |   1  |  1
   5    |   1  |  1
   6    |   1  |  1

So I want to estimate the probability of a coin and the die being successful - P(Coin = 1, Die = 1). I want to estimate this every 3 trials from the data.

In spite of using a coin/die model, I don't assume there is independence in their outcomes: I know that the the Coin being successful is more likely if the Die is successful and vice versa, but I do not know exactly how much more likely that is.

Given the problem set up, I know the distribution of each individual probability, P(Coin = 1) and P(Die = 1) is a binomial distribution.

Could someone shed some light on how I would go about estimating the joint probability of P(Coin = 1, Die = 1) from the data?

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  • $\begingroup$ Are you assuming independence? (It's a pretty reasonable assumption, but if you're abstracting some other problem you may not want to) $\endgroup$
    – Glen_b
    Nov 8, 2017 at 8:48
  • $\begingroup$ @Geln_b No, I am not assuming independence given my statement of "Also, I know that the the Coin being successful is more likely if the Die is successful and vice versa, but I do not know exactly how much more likely that is." $\endgroup$ Nov 8, 2017 at 18:23
  • $\begingroup$ Apologies, I managed to miss that bit. I am going to make a small edit (it should not begin with 'also' because it's making quite a different point there). (edit:) I have now moved the assumption of dependence up and clarified it. You can roll it back if you wish. $\endgroup$
    – Glen_b
    Nov 9, 2017 at 1:30
  • $\begingroup$ "I want to estimate this every 3 trials from the data." Is there a reason you only want to look at three trials at a time? As this will make your estimates incredibly inaccurate. Are you hoping to see this probability change over time, or something? $\endgroup$
    – Patty
    Nov 9, 2017 at 2:10

1 Answer 1

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I might be missing something here, but the fact that the die and coin results are not independent doesn't really change how you estimate the joint probability.

If you have $n$ trials, and the trials themselves are independent (which seems to be the case given your wording), then you simply estimate $P(A)$ by

$$P(A) = \frac{N(A)}{n}$$

Where $N(A)$ is the number of trials where the event $A$ occurs. In the context of your question, this means simply counting the number of trials where you get $1$ for both trials simultaneously, and dividing by the number of trials.

The fact that the coin and die results aren't independent simply means that

$$P(A,B) \neq P(A).P(B)$$

It does not change how you estimate $P(A,B)$.

However, you mention only using $3$ trials at a time, which will give incredibly inaccurate results as far as estimating this probability. So I feel like providing some context to your question might help.

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