How does Michigan secure double bye in BTT?

Um, we split with Wisconsin, but whatever. 

Here is page I view for this.  Good information.  Updated as of 2/25.  UM's chances of being the 5 or 6 are likely lower now due to the Wisconsin loss last night.  Still most likely scenario is the 3 seed.  https://www.btpowerhouse.com/2019/2/25/18239577/225-big-ten-tournament-seeding-simulation-standings-projections-purdue-michigan-iowa-rutgers-indiana

"

IV. In case of a tie for any place finish in the regular-season standings, the following tie-breaking procedure shall be followed in order to seed teams in the tournament bracket:

A. Two-team tie:

1. Results of head-to-head competition during the regular-season.

2. Each team’s record vs. the team occupying the highest position in the final regular-season standings (or in the case of a tie for the championship, the next highest position in the regular-season standings), continuing down through the standings until one team gains an advantage.

a. When arriving at another pair of tied teams while comparing records, use each team’s record against the collective tied teams as a group (prior to their own tie-breaking procedures), rather than the performance against the individual tied teams.

b. When comparing records against a single team or a group of teams, the higher winning percentage shall prevail, even if the number of games played against the team or group are unequal (i.e., 2-0 is better than 3-1, but 2-0 is not better than 1-0).

3. Won-loss percentage of all Division I opponents.

4. Coin toss conducted by the Commissioner or designee.

B. Multiple team tie:

1. Results of head-to-head competition during the regular-season.

a. When comparing records against the tied teams, the team with the higher winning percentage shall prevail, even if the number of games played against the team or group are unequal (i.e., 2-0 is better than 3-1, but 2-0 is not better than 1-0).

b. After the top team among the tied teams is determined, the second team is ranked by its record among the original tied teams, not the head-to-head record vs. the remaining team(s).

2. If the remaining teams are still tied, then each tied team’s record shall be compared to the team occupying the highest position in the final regular-season standings, continuing down through the standings until one team gains an advantage.

a. When arriving at another pair of tied teams while comparing records, use each team’s record against the collective tied teams as a group (prior to their own tie-breaking procedures), rather than the performance against the individual tied teams.

b. When comparing records against a single team or group of teams, the higher winning percentage shall prevail, even if the number of games played against the team or group are unequal (i.e., 2-0 is better than 3-1, but 2-0 is not better than 1-0).

3. Won-loss percentage of Division I opponents.

4. Coin toss conducted by Commissioner or designee."

The correct answer is- whatever Delany can do to screw over Michigan is how the tiebreaker will be calculated. I think it's written somewhere in the small font.

WIN

Beat penn state, or beat MSU on Sunday. We made our bed. Hopefully we get a reasonable draw come end of March. 

Actually, that does look to be correct -- with a win tomorrow, Michigan secures a record of at least 14-6.  That's guaranteed to be no worse than a tie for fourth with Wisconsin, and Michigan appears to win every tiebreaker against Wisconsin, as they've been brutal against the top of the Big Ten: 0-1 vs MSU, 0-1 vs Purdue, and 1-1 vs Maryland.

Just win, baby

stop the mental errors and move around without the ball. 

Per the sight below, win one of their remaining three games.

http://bball.notnothing.net/big10.php?sport=mbb

Denoting the real ranking of B1G team order of A by S p e r A {\displaystyle \mathrm {Sper} A}, the separating ideal of α and β in S p e r A {\displaystyle \mathrm {Sper} A} is the ideal of A generated by all polynomials g ∈ A {\displaystyle g\in A} that change sign on α and β, ie. g ( α ) ≥ 0 {\displaystyle g(\alpha )\geq 0} and g ( β ) ≤ 0 {\displaystyle g(\beta )\leq 0}. Any finite covering R n = ∪ i P i {\displaystyle R^{n}=\cup _{i}P_{i}} of closed, semi-algebraic sets induces a corresponding covering S p e r A = ∪ i P ~ i {\displaystyle \mathrm {Sper} A=\cup _{i}{\tilde {P}}_{i}}, so, in particular, when f is piecewise polynomial, there is a polynomial f i f_{i} for every α ∈ S p e r A {\displaystyle \alpha \in \mathrm {Sper} A} such that f | P i = f i | P i {\displaystyle f|_{P_{i}}=f_{i}|_{P_{i}}} and α ∈ P ~ i {\displaystyle \alpha \in {\tilde {P}}_{i}}. This f i f_{i} is termed the local polynomial representative of f at α. Let α, β be in S p e r A {\displaystyle \mathrm {Sper} A} and f be piecewise-polynomial. It is conjectured that for every local representative of f at α, f α f_\alpha, and local representative of f at β, f β {\displaystyle f_{\beta }}, f α − f β {\displaystyle f_{\alpha }-f_{\beta }} is in the separating ideal of α and β. 

That oughtta do it. (This formula is applicable for all Big Ten athletics). Just plug in the teams and you’re good to go.

Kill everyone else... Isn't that the solution to most of life's problems?

Disappointed it took thirty minutes for the obvious answer- WIN