Vodoo ploppies are right, we can shove our math and sims :)
- Thread starter Gamblor
- Start date
It is not 50/50
If there is a limited amount in the envelop, there are chances only pair of (0.5X, X) exists, and pair of (X, 2X) never exist.
with limited amount(L) in the envelop,
no matter where the X lays between 1 and L, 0.5X is always be there....100% chance of having (0.5X, X)
if X is greater than 0.5L, there is no 2X...50% chance of having (X,2X)
so for an unknown X, EV of switching = 100%/150%*-0.5X + 50%/150%*+X = 0
without limited amount(L) in the envelop,
for any X(X still less than half infinity anyway), EV of switching = 50%*-0.5X + 50%*+X = +0.25X
X=$100,
for a limit of $200 or higher, the pair of ($100, $200) is possible, switching EV=+$25
with limited amount less than $200, the pair of ($100, $200) is impossible, switching EV=-$50
If there is a limited amount in the envelop, there are chances only pair of (0.5X, X) exists, and pair of (X, 2X) never exist.
with limited amount(L) in the envelop,
no matter where the X lays between 1 and L, 0.5X is always be there....100% chance of having (0.5X, X)
if X is greater than 0.5L, there is no 2X...50% chance of having (X,2X)
so for an unknown X, EV of switching = 100%/150%*-0.5X + 50%/150%*+X = 0
without limited amount(L) in the envelop,
for any X(X still less than half infinity anyway), EV of switching = 50%*-0.5X + 50%*+X = +0.25X
X=$100,
for a limit of $200 or higher, the pair of ($100, $200) is possible, switching EV=+$25
with limited amount less than $200, the pair of ($100, $200) is impossible, switching EV=-$50
[why it wouldn't be 50/50?]Gamblor said:
If there is a limited $amount, there are chances only pair of (0.5X, X) exists, and pair of (X, 2X) never exist
Say envelop max = 100,
If X=1 to 100,
there are 100 pairs of (0.5X, X),
but only 50 pairs of (X, 2X)....(51,102) is impossible
So for an unknown X, number of pair(0.5X,X) is always double the number of pair(X,2X)
...so it is 2/3,1/3 ratio, and not 50,50 ratio
Infinite money and possibilities
I was contemplating the infinite problem today thinking it would make the likelihood the same for the X/2 and 2X as the other envelope when X is in the first. But as I analyzed the problem I decided it was actually the opposite.
The likelihood of picking an envelope with a specific X (say $100) is 1 divided by infinity which is zero or at least approaches 0. But some amount has to be in the envelope so let us say it is $100. The odds of ever opening another envelope with $100 is again zero. This would theoretically make it a 100% and 0% situation for the 2 possibilities.
One might argue that the theory of large numbers would average it out. But if the sample range is infinite you would never even reach a point where each possibility could possibly be represented so the theory of large numbers would never apply. Rather than being the extreme that allows you to make the 50% and 50% possibility assumption, it is much closer to the 0% and 100%.
Infinity is a very tricky concept.
I was contemplating the infinite problem today thinking it would make the likelihood the same for the X/2 and 2X as the other envelope when X is in the first. But as I analyzed the problem I decided it was actually the opposite.
The likelihood of picking an envelope with a specific X (say $100) is 1 divided by infinity which is zero or at least approaches 0. But some amount has to be in the envelope so let us say it is $100. The odds of ever opening another envelope with $100 is again zero. This would theoretically make it a 100% and 0% situation for the 2 possibilities.
One might argue that the theory of large numbers would average it out. But if the sample range is infinite you would never even reach a point where each possibility could possibly be represented so the theory of large numbers would never apply. Rather than being the extreme that allows you to make the 50% and 50% possibility assumption, it is much closer to the 0% and 100%.
Infinity is a very tricky concept.