In this applet we are going to have a rigged (or loaded) die. One of the outcomes ​​may have more possibilities than the others. We do not know it. But it can only be one of the possible outcomes.[br]You have to make at least 20 rolls to see the solution, or choose more before viewing it if you still don't know what it is.[br]The sample space of a normal die would be [math]E=\left\{1,2,3,4,5,6\right\}[/math] and therefore the probabilities will be:[br]P(1)=1/6,P(2)=1/6,P(3)=1/6, P(4)=1/6,P5=1/6 and P(6)=1/6[br]If the die is loaded, [math]E=\left\{1,1,1,2,3,4,5,6\right\}[/math] , "1" would be 3 times more likely than the other outcomes. [br]In this case the probabilities will be:[br]P(1)=3/8, P(2)=1/8, P(3)=1/8, P(4)=1/8, P5=1/8 and P(6) =1/8[br]It is about observing the frequencies of the outcomes and deducing what number is more repeated and how many times it is repeated. Since it is not easy, the advice is that you make all the necessary extractions to be able to deduce it.

- There is already a random configuration of the die, start rolling the die[br]- Look at the frequencies, you should notice that one of the outcomes ​​might repeat more times.[br]- Try to see the relationship between the number of occurrences of the different outcomes and deduce what kind of die it is (fair or rigged).[br]- If you think you have the solution and you have reached the minimum number of extractions, look at the solution.[br]- Generate a new die and try again