Originally Posted by
Phoenix Ignition
Out of curiosity I tested this theory, using a 52 card deck with 2 lands still in it (taiga + bayou), as well as a 52 card deck with 1 Bayou left in it. It's not nearly 1/3, it's actually closer to 1/2 that the belcher player will win. Your assumption was that he played double Bayou (I don't think any belcher list does), and in that scenario there is a ~37% chance of him winning. With a single Bayou left in the deck, there is a ~53% chance of him winning. With both a Bayou and a Taiga there is a (if I'm doing my math correctly) ~45% chance of him winning. Nearly a coin flip. This takes into account hitting no lands in the top 10, then hitting a taiga in the next 10 = win, and hitting a Bayou in the second 10 = lose unless the Taiga comes before the Bayou. Not 1/3 chances of winning, much closer to 50%.
Feel free to correct my math, I was using Hypergeometric Distributions and took a couple day break in the middle of the problem.
Sorry, sort of off topic in terms of deck design, but the question of the math was bugging me for a while and I wanted to clear that up, and it wouldn't hurt to see a break from people advising you to run non-jitte equipment (hah).
Rest of your post for reference:
I'm confused as to why you think it's both 1/3 and 1/2 chance to kill though. You say both the wrong and the right answer.
EDIT: After reading your post a few times you might be right, but I'll leave this here for math reference anyway. I think you mean the belcher player loses 1/3 of the time, and that something happening 1/3 of the time is not that far from normal, in which case you would be right.