Say I pull $n$ balls from a box while blinfolded. The balls can be either red or blue. I do not know the distribution of the balls.

After that I receive a list predicting the color of every ball I pulled out. I am also told that every prediction has a $70\%$ chance of being right and $30\%$ chance of claiming the wrong color (if blue it says red and vice versa).

Is there any way I could estimate within a range of error the true distribution of the balls? Or estimate the most likely distribution of the colors of the balls I pulled after the predictions I have been given?

  • $\begingroup$ Is this a question from a course or textbook? If so, please add the self-study tag & read its wiki. $\endgroup$ Commented May 30, 2022 at 13:03
  • $\begingroup$ Not really. I trained a semi-supervised model that predicts whether a variable is "yes" or "no" and after comparing the predictions it gives to the true value I noticed that it assigns the correct value a 70% of the time. So I wondered about the posed question. Training the model was part of a course yes, but the question itself is not. I was just curious. $\endgroup$
    – st30
    Commented May 30, 2022 at 13:50

1 Answer 1


Try to find a likelihood function, and treat it as a likelihood estimation problem. Lets say you have an urn with $N$ balls, of which you draw $n$, say without replacement. That makes this a hypergeometric problem, with some extras ... But for now, assume $n << N$ and treat it as a binomial problem first. The extension I leave as an exercise. Define some random variables, and let $\theta$ be the fraction of red balls in the urn:

First some unobserved random variables, the true colors of the balls: $$ \newcommand{\red}{\color{red}{\text{red}}} \newcommand{\blue}{\color{blue}{\text{blue}}} \DeclareMathOperator{\P}{\mathbb{P}} \P(\tilde{X}_i=\red)=\theta \\ \P(\tilde{X}_i=\blue)=1-\theta $$ Then the variables we actually observe: $$ \P(X_i=\tilde{X}_i) = 0.7, \quad \P(X_i \not=\tilde{X}_i) = 0.3 $$ And by conditioning we find the distribution of $X_i$ in terms of $\theta$: $$ \P_\theta(X_i=\red)= \\ \P_\theta(X_i=\red \mid \tilde{X}_i=\red) \P(\tilde{X}_i=\red) + \P_\theta(X_i=\red \mid \tilde{X}_i=\blue) \P(\tilde{X}_i=\blue) = \\ 0.7 \theta + 0.3 (1-\theta) $$ and likewise (or by subtraction) $$ \P_\theta(X_i=\blue)= 0.7 (1-\theta) + 0.3 \theta $$ Then the likelihood function becomes $$\mathcal{L}(\theta)=\left( 0.7 \theta + 0.3 (1-\theta)\right)^{n_{\red}}\cdot \left( 0.7 (1-\theta) + 0.3 \theta \right)^{n_{\blue}} $$ From now, usual likelihood methods. You can then do the same within the hypergeometric upset.


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