A-Level Probability Counting Problem
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This sounds great, thank you. One issue I have is that there wouldn’t be a total of 16! ways to arrange the pieces, since there are 8 pawns, 2 rooks etc. I believe the correction to that is 16!/(8!*2!*2!*2!), reducing the probability to (322560)^2/16! = 32/6435. Is this a fair argument?
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No. In order to use the axiom of "probability is favorable outcomes divided by total outcomes," each outcome must be equally likely.
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Ah I see. Relating it to a fair 6-sided die, the total N° outcomes is 6*6=36, not 21 different outcomes. I got it! Thank you!
The answer is accepted.
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