Quote:
Originally Posted by **Triggerfish** No! not after the event but actually before the event.
Box B 70% is a made up number!....an assumption which isn't true before the event.
ladbrokes.....sorry, am not a gambler nor do I play dice! |
The first point is that if one is thinking in terms of the probability of a favourable outcome, then there is no reason to prefer a ball to be drawn from one pot rather than the other in either of the two variants of Pat`s game. In the case of Pot A, we know that the probability of drawing a red ball (or a black ball) is 1 in 2 (50%). This is also the case for Pot B, even though the precise number of red balls (and black balls) it contains is unknown.
To understand why it is necessary to consider what follows from the fact that all combinations of red and black balls are equally likely. This means that the probability that Pot B will contain (50 - n) red balls is identical to the probability that it will contain (50 + n) red balls (where n is any number between 0 and 50). The n’s here cancel each other out, leaving an overall probability of 50 out of 100 (or 1 in 2).
If that’s not clear, think in terms of particular numbers. For example, the probability that the pot will contain 49 red balls (and 51 black balls) is identical to the probability that it will contain 51 red balls (and 49 black balls); the probability of 48 red balls (and 52 black balls) is the same as it is for 52 red balls (and 48 black balls) and and so on.
This means that the average probability across all possible combinations of red and black balls that a red ball (or black ball) will be drawn (given that the pot contains 100 balls in total, and all combinations of red and black balls are equally likely) is 50 out of 100 - that is, 1 in 2 (50%).
This means that whichever pot was selected made no difference to the chance of a red ball being drawn, |