I am trying to develop an understanding of Degree of Belief (Probability) and Expectation of a Bet. I have two related questions.
- Let’s consider P(A) = 2/3 and P(B) = 1/3. If I am given an odd of 2:1 on occurrence of A, my expected value of win is $$2*2/3 - 1*1/3 = 1$$ However if I am given an odd of 4:3 on occurrence of A, my expected value is $$4*2/3 - 3*1/3 = 5/3$$ which is greater than my expectation from previous bet. How can I have higher expectation value from a lesser odd (considering 4:3 < 2:1).
- In the 1926 paper “Truth and Probability” Ramsey defines “Degree of Belief in proposition
pby the odds at which the subject would bet onp. I am finding it hard to build an intuitive understanding of this. Is “minimum” bet based on this degree of belief related to the “fair” bet based on the degree of belief?
Thanks for your help.
Firstly, it is simpler to consider the British system of quoting odds against, eg odds of $3:1$ means a net gain of $3$ against a stake of $1$. Thus if P(A wins) = $\frac23$, the fair odds will be $1:2$ against, meaning that for a stake of $2$, you get back $1+2$ (net win +stake)
You can check that net winnings $E[X] = 1*\frac23 - 2*\frac13 = 0$
If you are given an odds of $3:4$ against, while P(win) $= \frac23,$
then $E[X] = 3*\frac23 - 4*\frac13 = -\frac23$
less as it should be.
Re degree of belief, that would depend on whether you are are risk seeking, risk averse, or risk neutral e.g risk averse persons would only bet if the odds were nearly fair, risk neutral people might be a little more adventurous, and risk seeking people might bet on the dark horse with low probability of winning but very high profit if it won.
You could also read up various strategies in game theory such as mininmax, maximax, minimax regret etc