Sports Betting w Brackets

Question

Each of the teams has a power ranking, and the probability they will win a game is based on their power ranking compared to their opponents. For example, if Blue has a power ranking of 75, and Green has a power ranking of 25, Blue's win probability is 3/4, and Green's is 1/4.

The bracket is as follows:

Round 1A: Wisconsin (200), Utah (200)

Round 1B: Ohio (300), Washington(100)

Round 2A: Wisconsin v. Utah winner

Round 2B: Ohio v Washington winner

Winner of Round 2A/B: Champion

What is the probability for each team to be champion?

My current method is to file through each team's odds 1 by 1, such that Wisconsin is shown below for round 1:

$$\frac{1}{2} * \frac{2}{5} + \frac{1}{2} * \frac{1}{3}$$

This is my first time encountering sports/ bracket betting odds, and any help is appreciated!

Answer 1

You aren't considering the probabilities of who Wisconsin plays in the second round. The correct calculation would be $$ \frac12\left(\frac34\cdot\frac25+\frac14\cdot\frac23\right) $$ If Wisconsin wins its first game (probability $\frac12$), there's a $\frac34$ probability that it plays Ohio in the second round, and a $\frac25$ probability that Wisconsin wins that game. Similarly, there's a $\frac14$ probability that Wisconsin's second game would be against Washington, and a $\frac23$ probability that Wisconsin would win that game. (I assume that the $\frac13$ in the question is a typo.)

Answer 2

Let us compute in general. We fix four teams, P, Q, R, S with ratings respectively $p,q,r,s$. The eliminatory tournament has the scheme:

P --
     > Winner P/Q -- 
Q --                 \
                       > Winner
R --                 /
     > Winner R/S --
S --

Which is the winning chance for P?

First of all, P must win against Q, this happens with probability $\Pi_{pq}:=p/(p+q)$.

In a second step, P must win against the Winner R/S.

P plays against R with probability $\Pi_{rs}:=r/(r+s)$ and then wins with probability $\Pi_{pr}:=p/(p+r)$.

P plays against S with probability $\Pi_{sr}:=s/(r+s)$ and then wins with probability $\Pi_{ps}:=p/(p+s)$.

So the probability $W(p;q;r,s)$ for the team P to win is: $$ \begin{aligned} W(p;q;r,s) &= \Pi_{pq}\cdot\Big( \Pi_{rs}\cdot \Pi_{pr} + \Pi_{rs}\cdot \Pi_{ps} \Big) \\ &= \frac p{p+q} \cdot \left( \frac r{r+s}\cdot \frac p{p+r} + \frac s{r+s}\cdot \frac p{p+s} \right) \\ &= \frac{p^2}{(p+q)(r+s)} \cdot \left( \frac r{p+r} + \frac s{p+s} \right) \\ &= \frac{p^2(pr+ps+2rs)}{(p+q)(p+r)(p+s)(r+s)} \ . \end{aligned} $$ The same formula applies also for the other teams in this generality, the other winning probabilities are $W(q;p;r,s)$ for $Q$, $W(r;s;p,q)$ for $R$, $W(r;s;p,q)$ for $S$. In our case the probabilities are $7/30$, $7/30$, $9/20$, $1/12$.

sage: def g(p, q, r, s): return p^2*(p*(r+s) + 2*r*s) / (p + q) / (p + r) / (p + s) / (r + s)
sage: p, q, r, s = 200, 200, 300, 100
sage: WP, WQ, WR, WS = g(p,q,r,s), g(q,p,r,s), g(r,s,p,q), g(s,r,p,q)
sage: WP, WQ, WR, WS
(7/30, 7/30, 9/20, 1/12)
sage: WP + WQ + WR + WS
1