Originally Posted by

**SumGuy** Imagine Runner 3 wasn't running, the probability ratio would stay the same wouldn't it? Meaning that R2has a 1/9 chance of beating R1. The same could be said about R3 if R2 were taken out. **Now, since they are mutually exclusive (hoping the terminology is correct there), doesn't that mean there is a (1/9) * (1/9) chance of both R2 AND R3 beating R1?** The event of them both beating R1 means that R1 must be 3rd. It would therefore happen 1/81 times (or 0.0123 in probability terms). So the probability of 2nd is 1 - 0.8 - 0.0123) = 0.1877

Now, that means the chances of R2 and R3 to come 2nd are each (1 - 0.1877) / 2 = 0.40615

and the probability of 3rd is (1 - 0.0123) / 2 = 0.49385

Making the table as follows:

R1: 0.8, 0.18770, 0.01230

R2: 0.1, 0.40165, 0.48395

R3: 0.1, 0.40615, 0.49385

That is, assuming you haven't corrected any of my logic in the time it took me to calculate and post that :-)