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!
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'll show you what I mean.
$100-90=10$
$10/3 = 3.\overline{3}$
Therefore, Cups C, D, and E have a theoretical probability of $3.\overline{3}$ of having the ball.
Therefore, Cup A is made up of $\frac{90}{3.\overline{3}}$ or $27$ separate cup A's.
Thus, there are currently 27 Cup A's which you consider to be your one cup A.
So, $27A+B+C+D=100$.
If you remove Cup D, which makes up $3.\overline3$ of the percentage, you have to add another cup A.
$28A+B+C=100$.
Follow this reasoning, and you will obtain your answer.