A-Level Probability Counting Problem

An idiot is setting up their side of a chessboard. The idiot knows where the pieces go (in some order on the last 2 rows) but has no idea in what order the pieces must be placed. What is the probability the idiot sets up the board correctly?

Probability Combinatorics
Sameljk Sameljk
42
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Schwartstack Schwartstack
269
  • Sameljk Sameljk
    0

    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?

    • Schwartstack Schwartstack
      0

      No. In order to use the axiom of "probability is favorable outcomes divided by total outcomes," each outcome must be equally likely.

  • Sameljk Sameljk
    0

    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!

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