Suppose there are 4 cups, and a ball is under one of these cups. Assuming it has equal chance of being at under any cup, the probability of it being under cup A is 25% correct?Now assume after 1000 observations, I notice 90% of the times, ball is under cup A.Now, if I have to predict where the ball is, I’d say it has 90% probability of being under A. Does that make sense?If cup D is removed, how should I change my estimate? I know it must increase, since if I remove B,C and D, probability raises to 100%.Thanks you all in advance!

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Dominique Holmes · Accepted Answer

After 1000 observations, you notice that 90% of the time, the ball is under Cup A. However, as Doug M mentioned, after 10,000 times, your probability would become lower.So, instead of thinking of Cup A as one cup, think of it as many different cups. I&#039;ll show you what I mean.  100  −  90  =  10  10      /    3  =  3.      3    ¯  Therefore, Cups C, D, and E have a theoretical probability of   3.      3    ¯   of having the ball.Therefore, Cup A is made up of       90          3.              3        ¯             or 27 separate cup A&#039;s.Thus, there are currently 27 Cup A&#039;s which you consider to be your one cup A.So,   27  A  +  B  +  C  +  D  =  100If you remove Cup D, which makes up   3.      3    ¯   of the percentage, you have to add another cup A.  28  A  +  B  +  C  =  100Follow this reasoning, and you will obtain your answer.

Suppose there are 4 cups, and a ball is under one of these cups. Assuming it has equal chance of bei

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