Two friends were playing a friendly game of chess. The both have young children who they recently taught the rules and they agreed both children could help if they wished.

In the position below (black to move) the two friends took a short brake for dinner. The kids took a few extra minutes to follow for dinner and when the first of them arrived, he surprised both adults

Noah, the child of the father playing black said "I took their piece. I saw Jane (the child of the father playing white) make a move as I left the room but I have no idea what it was.

Both friends knew that by piece Noah did not mean a pawn and that him capturing one piece was no more likely than capturing another.

They also knew that if Noah played Bxd4 Jane surely responded Qxd4.

If Noah played Bxf4 there was an 80% chance, Jane would have moved her knight and 20% she would have done something else. If Noah played Bxf4 and Jane moved her knight the likelyhood of her moving it finding the best square would be (2/n) with n being the number of legal knight moves. All other sub optimal knight moves would share the same probability.

What is the likelihood (in percentage terms) that white is winning the game?

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1 Answer 1


Only two pieces are capable of being captured, not counting pawns

(50% chance) Bxd4 was played and then white played Qxd4 and is winning

(50% chance) Bxf4 was played and with an 80% chance white moved the knight:

  • Nc6 with (2/n) or (2/5) probability and is winning (40%)
  • Another knight move with 3/5 probability and is losing (60%)
  • n = 5 because there are 5 possible knight moves

If Bxf4 was not followed by a knight move, white is losing

The answer is:

50% + 50%(80%)(40%) = 50% +16% = 66% chance white is winning

  • $\begingroup$ Correct! Can anyone tell me if the unknown actions of the children were a net positive to the probability that white will win the game with best play compared to the starting position posted above? $\endgroup$ Commented Jun 9, 2016 at 8:29
  • $\begingroup$ The kids polarized the result as now either black or white is convincingly winning. Since with best play the previous position would end in a draw by perpetual a 66% chance of winning is an improvement for white. $\endgroup$
    – Luke
    Commented Jun 9, 2016 at 8:36
  • $\begingroup$ Necessary precision: 1...Bxf4 2.Nc6 Be5 3.Nxe5 (3.Qe3? Bxc6 4.Qh6 Re8 loses for White) dxe5 4.Qxe5 f6 5.Qh2! wins (much less clear is 5.Qd5 e6 6.fe6 Bc8!), e.g. 5....Kf7 6.Qh7 Ke8 7.Qxg6 Rf7 8.Rh8# $\endgroup$
    – Evargalo
    Commented Oct 4, 2018 at 16:46

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