Suppose a fighter only fights opponents he is considerably better than. Let's say that he only fights opponents he would defeat 8 out of 10 times, or that his "dominance" is .80. Clearly this is sandbagging, he's a can crusher, etc. If he has 5 fights, what is the probability he will go undefeated? You'd think it would be pretty high, but to get the answer, we multiply .80 * .80 * .80 * .80 * .80 = (.80)^5, which is about .33. So the fighter only has about a 1 in 3 chance of going undefeated.

In real life, obviously, the fighter will have different win probabilities against different opponents. But if we hold the win probability constant, as we have done above, the math is much simpler. So, if we hold the win probability constant, what is the probability of a fighter going 49-0 like Mayweather?

Say Floyd beats all of his opponents 9 out of 10 times. Then the probability he goes 49-0 is .90^49, which is about .006. What about the probability he loses 1 fight?

The probability he loses his first fight is .10. Then the probability he wins his next 48 is .90^48, so the probability he loses his first fights, then wins his next 48, is .10*.90^48, which is about .0006.

This also applies to a 48-1 record where he loses only his second fight, where he loses only his third fight, etc. So the total probability of a 48-1 record, given dominance .90, is about .0006*49=.0294, or about 3 percent.

The formula for the probability of a given record where the loss occurs in a given fight (the first fight, the second fight, etc.) is

And the total probability of having a given record is given by

Given this formula, we can graph what the record of a fighter of a given dominance will probably be. For a fighter with dominance .75 and 30 fights, we have

We can also graph with respect to the dominance required to have a given record. Probable dominance for a fighter with a 25-5 record:

It seems that a long undefeated record requires fighting opponents who aren't competitive, or a lot of luck.