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
Report

Answer

Answers can only be viewed under the following conditions:
  1. The questioner was satisfied with and accepted the answer, or
  2. The answer was evaluated as being 100% correct by the judge.
View the answer
Schwartstack Schwartstack
277
  • 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!

The answer is accepted.
Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.