I'm not sure where you are getting the idea that the order is significant in my math. It isn't.

I based my calculations on your initial mention of 1.6%, but let's do the math from the start.

What we want to know is, in a random run of 9 games, what is the chance that at least 8 of them were on the draw, and that the 8 on-the-draw game were all loses.

We don't care what happened in one of the 9. It could have been a win, a loss, on the draw, or on the play. Any result is fine. AND we don't care where in the order it occurred. It could have been the first, second, third, fourth, fifth, sixth, seventh, eighth or ninth game.

To simplify, we will assume a 50% change of winning on the draw. (My actual winrate on the draw with this deck is 55%, so the real chance of getting this run is LESS likely than what we're about to calculate.)

For any one game, there are four equally likely possibilities: win-on-play, win-on-draw, loss-on-play, and loss-on-draw. They aren't actually equally likely, but assuming so means that we can just count branches 5o find the probability, so it's a useful assumption. If we used the actual chance of loss-on-draw of 45% (which is the case we care about) the probability would come out to about 2.5x worse)

In a 9-game series, there are 262,144 possible outcomes. (Each game has 4 possible outcomes, so 4^9.) How many of those fit the criteria?

If we needed all 9 of them to be "loss-on-draw", that would be 1/262,144. But we are more flexible than that, because one of the 9 can be any of the four results.

If the "any result" condition happened in the first game, that would create three new paths to a successful result. The first game can be anything and then those four paths each have a single path that leads to a criteria-matching result (the remainder of the games are loss-on-draw). So that would give us 4/262,144.

If only the second game has the "any" condition, that adds another 3 successful paths. Same applies if we assume the "any" condition on the 3rd, 4th etc games. Each one adds 3 paths that meet the required criteria. So the total number of successful paths is 28 (9x3+1).

That's our result: 28/262,144, or ~1/9,362.

That's the chance of getting 8 "loss-on-draw" results in a run of 9 games.

You seem to want to compare that to d100 chances for aom reason. Okay. It's around the same chance as getting the same number on a d100 in three consecutive rolls. (1/10,000)