The die has rolled 3 3 3 3 3 3 3 in the past. What are the chances of 1 2 4 5 6 being rolled next? The mathematician will say: P(k)=1/6 for each number, forget that short-term evidence. What will the probability expert say? And the statistician? And the philosopher?I have provided a detailed solution to this problem here.

In summary, it is based on a Bayesian network in which (except for the 'statistician') it all comes down to what priors they are assuming for the probability of each P(k).

- The mathematician's prior is that the probability of each P(k) is exactly 1/6.
- One type of probability expert (including certain types of Bayesians) will argue that, in the absence of any prior knowledge of the die, the probability distribution for each P(k) is uniform over the interval 0-1 (meaning any value is just as likely as any other).
- Another probability expert (including most Bayesians) will argue that the prior should be based on dice they have previously seen. They believe most dice are essentially 'fair' but there could be biases due to either imperfections or deliberate tampering. Such an expert might therefore specify the prior distribution for P(k) to be a narrow bell curve centred on 1/6.
- A philosopher might consider any of the above but might also reject the notion that 1,2,3,4,5,6 are the only outcomes possible.

The mathematician's posterior for each P(k) is unchanged: i.e. each P(k) is still 1/6.So there is still just a probability of 1/6 the next roll will be a 3.

For the probability expert with the bell curve priors, the posterior for P(3) is now a distribution with mean 0.33. The other probabilities are all reduced accordingly to distributions with mean about 0.13. So in this case the probability of rolling a 3 next time is about 0.33 whereas each of the other numbers each has a probability about 0.13.

And what about the statistician? Well a classical statistician cannot give any prior distributions so the above approach does not work for him. What he might do is propose a 'null' hypothesis that the die is 'fair' and use the observed data to accept or reject this hypothesis at some arbitrary 'p-value' (he would reject the null hypothesis in this case at the standard p=0.01 value). But that does not provide much help in answering the question. He could try a straight frequency approach in which case the probability of a three is 1 (since we observed 7 out of 7 threes) and the probability of any other number is 0.

Anyway the detailed solution showing the model and results is here. The model itself - which will run in AgenaRisk is here.