Sudoku. Nine columns across. Nine rows top to bottom. 81 small squares. This entire large frame of 81 squares is divided into nine equal sections, each with nine squares. The numbers 1 through 9 go in each of the small squares. Each number may only appear once in each column, row, and section. There are some numbers already filled in to start. The challenge is to fill in the rest of the numbers correctly. There will only be one correct solution.
I am faced with a situation. In one section, in one particular square, there can only be a 1 or a 9. All other possible numbers have been eliminated. There are no other ways left to get at the correct number; it’s time to guess. I could guess a 1 or a 9 and I’d have a 50% chance of being right with either number, correct? Seems simple.
Let me complicate it slightly. There are only two numbers that can go in that one square, but each of those two numbers might also go in other squares in that section. The 1 can go in that square or one other, all other possibilities have been eliminated. The number 9 can go in that square or four others. From this perspective, I have a 50% chance of being right if I guess a 1, but only a 20% chance of being right if I guess a 9. (Okay, Judy’s eyes just glazed over.)
So what is the probability of being correct if I guess a 9: 50% or 20%?
If the correct probability for a 9 is 20%, does that mean my probability of being correct if I guess a 1 goes up to 80%?
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