Quote: HeadlockThe house edge on the Fire Bet is calculated to be approximately 20%. I've been wondering though if the house edge can be reduced by laying the 5th and or 6th point to guarantee at least some win if the 5th or 6th point is not made.
The edge on the Fire Bet doesn't change when you hedge. They are actually separate activities from a House Edge perspective, with each of the lay bets carrying their own edge.
You can guarantee a small profit, but there is a cost, namely the full win if you take down the 6th point.
I'm trying to crunch some numbers and deciding to bet $2 on the Fire Bet, and Lay $24 (+ vig) on the 4th number, Lay $90 (+ vig) on the 5th new number, and Lay $360 (+ vig) on the 6th new number. Maybe don't lay the 4 or 10 if it's the 5th or 6th number.
Quote: DeMangoSo on a bet with a 20% house advantage, you are advocating three hedges with even more house advantage? My prediction is that you will lose money, lots of money.
Not to mention that hitting the six is a tax form.
If I'm gonna hit 20 $5,000 wins a year, maybe I'll reconsider the tax form hazard. #DI
What would be the odds of hitting all 6 points at least twice before sevening out?
Quote: Ace2The odds of hitting all 6 points at least once before sevening out are 1 in 6,156.
What would be the odds of hitting all 6 points at least twice before sevening out?
With or without dice influence? #yoleven
Depends on whether you count numbers made twice in first go around. Otherwise 6,156 squared. Unless you are ChumpChange.Quote: Ace2The odds of hitting all 6 points at least once before sevening out are 1 in 6,156.
What would be the odds of hitting all 6 points at least twice before sevening out?
I get 1 in 5,668,725.Quote: DeMangoDepends on whether you count numbers made twice in first go around. Otherwise 6,156 squared. Unless you are ChumpChange.
Using the integral from zero to infinity of:
((1-1/e^(x*165/3960)*((x*165/3960)+1))*(1-1/e^(x*264/3960)*((x*264/3960)+1))*(1-1/e^(x*375/3960)*((x*375/3960)+1)))^2*1/e^(x*2352/3960)*2352/3960 dx