xyz Posted January 1, 2017 Report Share Posted January 1, 2017 Hello, I will try to explain the topic title to the best of my ability . We shall start by placing three different values in a row . Numbered 1,2 and 3 . 123 We shall then hide the values and randomly mix up the positions, leaving 3 values but now in unknown position. XXX I shall now offer you the first on the left unknown value . P1=1/3 P2=1/3 P3=1/3 Before I continue, are we all in agreement with the probability stated, that the 1st value on the left has an equal 1/3 chance of being either 1,2 or 3? Quote Link to comment Share on other sites More sharing options...
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