Bayes trees and correlation in an election bet

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$P(\text{Ortega})=0.7$, $P(\text{tax hike})=0.6$, $P(\text{tax hike}\mid\text{Ortega})=0.8$. Also $P(\text{uprising}\mid\text{Ortega, tax hike})=0.6$, $P(\text{uprising}\mid\text{Gonzalez, tax hike})=0.8$; without a tax hike the population does not rise up.

(a) Build the election/tax hike/uprising tree and compute the total probability of an uprising.

(b) Draw a Venn diagram of the two events “Ortega” and “tax hike.”

(c) Compute $P(\text{Ortega}\mid\text{tax hike})$.

(d) Are the two events positively or negatively correlated?

(e) Ortega’s odds of winning are $1.2$: compute the average payoff for Debbie, who bets $200{,}000$ pesos on Ortega.

(f) How should the odds be adjusted so the bookmaker has a $2\%$ margin?

Solution

From the data: $P(\text{Ortega}\cap\text{tax hike})=0.7\cdot0.8=0.56$, so $P(\text{Gonzalez}\cap\text{tax hike})=0.6-0.56=0.04$.

(a) Since without a tax hike there is no uprising, the total probability is $P(\text{uprising}) = 0.56\cdot0.6 + 0.04\cdot0.8 = 0.368$

(b) The Venn diagram shows the two circles “Ortega” ($0.7$) and “tax hike” ($0.6$) with intersection $0.56$.

(c) $P(\text{Ortega}\mid\text{tax hike}) = \frac{0.56}{0.6} \approx 0.933$

(d) Comparing with independence: $P(\text{Ortega})\cdot P(\text{tax hike}) = 0.7\cdot0.6 = 0.42 < 0.56 = P(\text{Ortega}\cap\text{tax hike})$, so the two events are positively correlated.

(e) With odds $1.2$: if Ortega wins (probability $0.7$) Debbie’s net profit is $200{,}000\cdot(1.2-1)=+40{,}000$ pesos; if he loses (probability $0.3$) she loses the $200{,}000$ she bet. The average payoff is $0.7\cdot40{,}000 - 0.3\cdot200{,}000 = 28{,}000 - 60{,}000 = -32{,}000 \text{ pesos}$

(f) The fair odds (fair game, zero average payoff) are $q_e = \dfrac{1}{0.7} \approx 1.429$. To guarantee the bookmaker a $2\%$ margin, the odds must be reduced: $q = \frac{1.429}{1.02} \approx 1.40$

$\boxed{P(\text{uprising})=0.368;\ P(O\mid t)\approx0.933;\ \text{positive corr.};\ \overline{\text{payoff}}=-32{,}000;\ q\approx1.40}$