Change of Axiom of Probability

[jack1234]

The reference book I have used stating that:
Axiom 1 stating that 0<=P(E)<=1
Axiom 2 stating that P(S)=1
Axiom 3, the probability of union of mutually exclusive events is equal to the summation probability of of each of the events.

And the author says that, hopefully, the reader will agree that the axioms are natural and in accordance with our intuitive concept of probability as related to chance and randomness.

But what if axiom 1 and axiom 2 is changed to
Axiom 1 stating that 0.5<=P(E)<=1.5
Axiom 2 stating that P(S)=1.5
(Axiom 3 no change)

or

Axiom 1 stating that 1.1<=P(E)<=2
Axiom 2 stating that P(S)=2
(Axiom 3 no change)

and rebuild the probability model base on the new axiom? Will there be any problem in this new probability model?

If not can I say that the original Axiom 1 and Axiom 2 is just taking some reference value so everybody on the Earth can follow it?

 

[HallsofIvy]

No there is nothing wrong with that-and there is no real difference. If, in the first case, you subtract 0.5 from "your" probability, you get "regular" probability. In case two, since 2- 1.1= .9, you would have to divide by .9 after substracting 1.1 in order to get "regular" probability. The reason for the (mathematically arbitrary) choice of 0 and 1 is to relate it to the common idea of a probability as a percentage: Probability of 1.0 corresponds, in common parlance, to "100% certain".

 

[arildno]

However, as an addition to Halls' comment, addition and multiplication rules for your new probabilities would be rather more tricky than using 0 and 1 as your limits.

 

[arildno]

For example, let 0<=p(E)<=1, whereas a<=P(E)<=b, so that P(E)=a+(b-a)*p(E)
Now, for disjoint events u and v, we have:
p(u+v)=p(u)+p(v).

But in P-notation, we would have:
P(u+v)=a+(b-a)*p(u+v)=P(u)+P(v)-a

This is an unnice addition rule..

 

[NateTG]

But what if axiom 1 and axiom 2 is changed to
Axiom 1 stating that 0.5<=P(E)<=1.5
Axiom 2 stating that P(S)=1.5
(Axiom 3 no change)

or

Axiom 1 stating that 1.1<=P(E)<=2
Axiom 2 stating that P(S)=2
(Axiom 3 no change)

and rebuild the probability model base on the new axiom? Will there be any problem in this new probability model?

If not can I say that the original Axiom 1 and Axiom 2 is just taking some reference value so everybody on the Earth can follow it?

You can effectively 'shift' the numbers for the range, but you'll basically end up doing everything by shifting the numbers back to the [0,1] range in order to do any kind of operation, so it makes little sense to try to shift things that way.

This may be statistics rather than probability, but if you start trying to apply probability, it rapidly becomes clear that the 0 and 1 are natural notions for the limits. For example, if you want to calculate the expected number of heads in 100 coin flips, sixes in 100 die rolls, or 0s in 100 spins of a Roulette wheel with the 0 and 1 probability, you can simply multiply the number of trials by the probability.