MAS113 Introduction to Probability and Statistics –
Supplementary Notes
More on subjective probability
Informally, we’ve said that you can choose your subjective probability by
considering your fair betting odds for an event. Here, we give a formal
definition for the fair price of a bet, and show setting probabilities based on
fair betting prices implies that you must follow the ‘usual’ rules of probability,
under the assumption that you are not willing to accept a guaranteed loss.
Suppose England are to play Australia at cricket. How would you choose
your value of P (England win)? One method is to consider a hypothetical
bet. Consider a ticket that pays £1 if England win, and £0 if they do not.
The bet proceeds as follows.
(a). You choose a price £p1 for this ticket.
(b). I can then choose to buy a ticket from you for £p1 , or sell you one
ticket for £p1 .
(c). You should choose p1 such that you don’t have a preference for whether
you buy or sell the ticket.
Depending on your price, whether I choose to buy or sell, and the outcome
of the match, your profit will be as follows.
I buy a ticket from you for £p1
Your profit
I sell a ticket to you for £p1
Your profit
England win
England do not win
You pay me £1
You pay me £0
£(p1 − 1)
£p1
I pay you £1
I pay you £0
£(1 − p1 )
£(−p1 )
What would you choose p1 to be? We make the assumption that you want to
avoid a guaranteed loss. Given this assumption, you must have 0 ≤ p1 ≤ 1.
1
• If you choose p1 < 0, I will buy the ticket from you, and your profit
will definitely be negative.
• If you choose p1 > 1, I will sell the ticket to you, and your profit will
definitely be negative.
As long as 0 ≤ p1 ≤ 1, there is no restriction on what you choose p1 to
be, but you can see how the choice relates to your “degree of belief” that
England will win. For example:
• If you thought the events “England win” and “England do not win”
were equally, the sensible choice for p1 would be 0.5. If you instead
chose p1 = 0.55, selling the ticket to me would be more desirable than
buying the ticket from me.
• If you thought the event “England win” was very unlikely, a small value
of p1 would be sensible, as you wouldn’t want to sell the ticket for p1
close to 1.
Now suppose we repeat the exercise for a second ticket, that pays £1 if Australia win, and £0 if they do not. Suppose you choose a price p2 for this
ticket. Finally, consider a third ticket, that pays £1 if either England or
Australia win, or 0 if neither side wins. What price would you choose for
this ticket?
If you what to avoid a guaranteed loss, your price must be p1 + p2 . For
any other price, you will definitely lose money:
Your price £x for
{England or Australia}
x < p 1 + p2
x > p 1 + p2
My betting strategy
Your guaranteed loss (£)
I buy {England or Australia} for £x
I sell {England} for £p1
I sell {Australia} for £p2
I sell {England or Australia} for £x
I buy {England} for £p1
I buy {Australia} for £p2
x − p1 − p2 < 0
2
−x + p1 + p2 < 0
The losses in the third column do not depend on the outcome of the match.
If the match if a draw, neither of us have a winning ticket. If either England
or Australia win, we have one winning ticket each, so the only money that
can change hands is due to the buying and selling if tickets.
Finally, suppose the event we are betting on is the whole sample space
{England win, Australia win, Draw}. The only way you can avoid a certain
loss in this case is set the price at £1.
In summary, if we write pr(E) as your price of an event E we have shown
(a). 0 ≤ pr(E) ≤ 1 for any event E.
(b). pr(S) = 1 for a sample space S.
(c). If A ∩ B = ∅, then pr(A ∪ B) = pr(A) + pr(B).
If we now define your probability P (A) to be your price pr(A), and you
wish to avoid guaranteed losses, your probabilities must be consistent with
a probability measure. Interestingly, we haven’t assumed the three axioms
of probability here, we’ve proved them, under a different axiom of avoiding
guaranteed loss1
1
Almost. We can only prove “finitePadditivity” pr(A ∪ B) = pr(A) + pr(B) rather than
∞
“countable additivity” pr(∪∞
i=1 Ei ) =
i=1 pr(Ei ) for disjoint sets E1 , E2 , . . ..
3