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Am I right? (Casino Mathematics)?
I am trying to find the EV of this bet or the EV of the bet in its current situation, I feel tho I am very wrong in figuring this out..
Situation - All 2 dollar bets on Craps. There are 4 points(5,6,8,9) that have 2 dollar flat bets(bets from the come).
We are on the come out roll and there is 2 dollars on the pass line.(odds are off, not that it should make a difference)
What is the expected EV on the come out roll.
EV = (+2)(20/36) + (-6)(6/36) + (-2)(4/36) =
EV = (+2)(5/9) + (-6)(1/6) + (-2)(1/9) = 11.1%
-0.11111111111
20 ways to make 2 dollars(point numbers and the yo)
6 ways to lose 6 dollars(rolling a 7)
4 ways to lose 2 dollars(craps)
6 ways to stay even for that roll(4,10)
I understand all the bets have a 1.4 percent house edge but in the current situation they are in the odds are different.
So I get the EV of all the bets individually are 1.4 percent but the expected EV on the next roll is 11.1%??
So its 11.1 cents for this wager?
Ah, $8/.11 = 1.375% or the normal house edge!
2 Answers
- Anonymous7 years agoFavorite Answer
In this situation the house edge for each bet individually is not 1.4%, because your scope of interest is just one roll.
It depends on the point of view. If you calculate the house edge of the pass/come per bet (and wait until the bet gets resolved) it is indeed 1.4%. If you count per roll, it is just 0.4%.
However, in your particular situation it is a bit pointless to calculate the "House Edge" at all because you have "contract bets" - you can't take down your points. You can calculate the EV for this particular roll, but it tells you nothing about the House Edge for your system (it looks like you are trying to calculate the overall house edge for some kind of "max points" system).
"Ah, $8/.11 = 1.375% or the normal house edge" - This is completely misleading and wrong conclusion. It is just a coincidence that you got a number close to the house edge _per bet_, by calculating the EV _per roll_. The normal house edge per roll would be 0.4%. Actually, you don't have $8 but $10 on the table.
- Divide By ZeroLv 77 years ago
It looks like your calculations are right (but I don't play Craps). But your conclusion is wrong.
Your EV is not 11.1%, it's -11.1 cents.
Edge = EV / wager = (-1/9)/wager = -1/(9*wager) = some negative decimal (which you'll then convert to a percentage)
(I can't tell what your total wager is since I don't play craps.)
EV is the expected profit or loss in dollars. So,
EV = edge * (total wager)
If each individual bet contributes some negative EV then yes, making multiple bets will result in all their EV's combined, for a larger negative EV. Just like wagering twice as much on a single bet will result in twice the negative EV.
Edit -- yeah if your wager was $8 (which I guess it was), then
-.11111... / 8 = -1 / 72 = about -1.389%, so about -1.4% like you originally said.
Edit #2 -- I defer to Freeroller since it sounds like he actually knows how Craps works.
