SBMM seeks to set up a 50% chance of winning any given game. Players assume that this will make going Flawless much more difficult. I am going to prove that using math. [b][u]Flawless Chance On A Single Card[/u][/b] This section discusses the likelihood to go Flawless with no Mercies, one Mercy, and 2 Mercies.  [spoiler][u]Question:[/u] If you assume that every game's outcome is independent and you have a likelihood of 50% of winning any individual game, what is the likelihood of winning 7 of 9 games?  [u]Solution: [/u] P(k of n) = H*(p^k)*((1-p)^(n-k))  Where… P = probability of winning "k" of "n" total games p = probability of winning each game  H = number of different ways it is possible to win "k" of "n" games  In this case… p = 0.5 (we are assuming the "ideal" SBMM is working, so every game is 50/50) k = 7 n = 9 H = 36 (found using Pascal's Triangle instead of using binomial coefficient formula) P(7 of 9) = 36*(0.5^7)*((1-0.5)^(9-7)) = 0.070 [b]Therefore, the likelihood of going Flawless with 2 Mercies is around 7%. [/b] After going Flawless, your next card becomes more difficult, as you can only lose 1 game.  P(7 of 8) = 8*(0.5^7)*((1-0.5)*(8-7)) = 0.031  [b]The likelihood of going Flawless with only 1 Mercy is around 3%.[/b] Maybe you want to see how difficult it would be to win 7 in a row without losing at all.  P(7 of 7) = 1*(0.5^7)*((1-0.5)*(7-7)) = 0.008 [b]The likelihood of going Flawless without ANY Mercy is around 0.8%.[/b][/spoiler] [b][u]The Effect Of Multiple Cards[/u][/b] Many players try multiple times to go Flawless. This section uses the numbers we found from the previous section and establishes that players can still achieve the same success if they play enough games.  [spoiler][u]Question: [/u] If everyone plays 5 cards, what is the chance of players going Flawless at least once? [u]Solution: [/u] P(k of n) = H*(p^k)*((1-p)^(n-k))  Where… P = probability of going Flawless "k" times out of "n" total cards p = probability of going Flawless  H = number of different ways it is possible to go Flawless "k" times out of "n" attempts p = 0.07 (from previous section, assumes 2 Mercies) k = 1 n = 5 H = 5 (found using Pascal's Triangle instead of using binomial coefficient formula) P(1 of 5) = 5*(0.07^1)*((1-0.07)^(5-1)) = 0.261 [b]If everyone plays 5 cards, then around 26% of players will go Flawless. [/b][/spoiler] [b][u]Conclusion[/u][/b]  Currently, around 25% to 35% of players go Flawless on any given weekend. The average number of games played per player is around 4 (around 1 card). If SBMM is added to Trials, and everyone plays 1 card, then the percentage of players who go Flawless will drop to around 7%. To attain the same percentage of Flawless players per weekend with SBMM as CBMM, players will have to play at least 5 cards per weekend.  [b]TLDR: Unless players are willing to play 500% longer each weekend, the introduction of SBMM will reduce the number of Flawless players by a significant margin. [/b]