House Edge Math Questions...

Hey all - posting these questions that a friend recently sent me that I cannot answer... I play cards, not dice, so please forgive me for any ignorance in the following questions..

Given: Placing a point of 6 or 8 has a house edge of 1.52%
Placing a point of 5 or 9 has a house edge of 4.00%
Placing a point of 4 or 10 has a house edge of 6.67%
1. If I bet on all the numbers and leave my money on the layout until the point is made or a seven is thrown, what is the true house edge that I face?
2. Given that craps is a series of independent trials, does your answer change if the shooter hits 1 or 3 or 15 numbers before he either makes his point or sevens out?
3. If the shooter has made several numbers and I begin increasing my bets, does the house edge become a factor of my initial bet or what I CURRENTLY have on the table?
4. If the shooter has made enough numbers for me to have recouped all of my initial bets, is the house edge now zero?
5. The house edge on the Pass line bet is 1.41%. The edge on the "odds" bet is zero. Depending on the casino, the amount of the odds bet allowed can be a function of which number becomes the "point". It could be 2,3,4,5,10 or even 100 times the line bet. How low can I reduce the house edge by taking the maximum odds?

The short answer is that you can go through the following steps to answer your questions:

Calculate the edge per roll for each bet you want.

Modify the number of rolls by the conditions that complete the new bet you want to consider (leaving it up until you seven out).

Come up with the new edge.

So for example, the edge on the 6 or the 8 place bet is 1.52%. But per roll it's .4644. The roll normally lasts 36/11 3.27 rolls, but if you wait for a seven it lasts 6 rolls.

So 1.52 * ( 11/36 ) * ( 36/6 ) = 1.52 * 11 / 6 = 2.78% -- the edge per roll is still 0.4644, but you get six rolls out of the bet instead of 3.27.

You can do this same thing for the other bets.

4.00 * ( 10/36 ) * ( 36/6 ) = 4.00 * 10/6 = 6.66% for the 5 or 9.
6.66 * ( 9/36 ) * ( 36/6 ) = 6.66 * 9 / 6 = 9.99% for the 4 or 10.

The short answer is that to truly analyze the game, you should consider each bet to be a one-roll bet and just look at the edge per roll and minimize the number of rolls and try to meet your goals one roll at a time. Learn the phrase "bring it all down" and make bold plays and hit your target and stop paying the house to keep gambling if you want to win.

If you just enjoy the process of gambling, make sure you do it on a table where the amount of cost to play the game is not too high. HAVE FUN.

Good luck!

Quote: nobeedsram

Hey all - posting these questions that a friend recently sent me that I cannot answer... I play cards, not dice, so please forgive me for any ignorance in the following questions..

Given: Placing a point of 6 or 8 has a house edge of 1.52%
Placing a point of 5 or 9 has a house edge of 4.00%
Placing a point of 4 or 10 has a house edge of 6.67%
1. If I bet on all the numbers and leave my money on the layout until the point is made or a seven is thrown, what is the true house edge that I face?

If all the bets are resolved, the HA is a weighted average of those numbers.

Quote: nobeedsram

2. Given that craps is a series of independent trials, does your answer change if the shooter hits 1 or 3 or 15 numbers before he either makes his point or sevens out?

The house edge is a theoretical number; the actual results do not change it. However, if the shooter makes his point right away, your place bets will not be resolved; OTOH, if he sevens out right away, you will lose them all at once. This is not an issue of the HA at all.

Quote: nobeedsram

3. If the shooter has made several numbers and I begin increasing my bets, does the house edge become a factor of my initial bet or what I CURRENTLY have on the table?

The house edge applies to the amount on the table when the bet is resolved, so the more money you have out there the higher the expected loss.

Quote: nobeedsram

4. If the shooter has made enough numbers for me to have recouped all of my initial bets, is the house edge now zero?

No, results and house edge are separate concepts.

Quote: nobeedsram

5. The house edge on the Pass line bet is 1.41%. The edge on the "odds" bet is zero. Depending on the casino, the amount of the odds bet allowed can be a function of which number becomes the "point". It could be 2,3,4,5,10 or even 100 times the line bet. How low can I reduce the house edge by taking the maximum odds?

At 100X odds, the house edge, figured using the amounts of the flat bet and the odds bet, is 0.02%; at double odds, it's 0.6%. However, this is misleading, in my view. The expected loss, which is the HA times the bet amount, or the expected value of the bet, is the same, regardless of taking odds, because only the flat bet carries a house edge; there is no house edge on the odds. So, if you make 60 $10 pass bets, the expected loss is $8.48, no matter whether you take no odds or any odds multiple, including 100X. The odds bets add substantial volatility, because you can win and lose a lot more on each decision. Be aware that the more odds you take, the quicker you can lose your stake, but the more you win if you're lucky.
Cheers,
Alan Shank

Quote: Ahigh

The short answer is that you can go through the following steps to answer your questions:

Calculate the edge per roll for each bet you want.

Modify the number of rolls by the conditions that complete the new bet you want to consider (leaving it up until you seven out).

Come up with the new edge.

So for example, the edge on the 6 or the 8 place bet is 1.52%. But per roll it's .4644. The roll normally lasts 36/11 3.27 rolls, but if you wait for a seven it lasts 6 rolls.

Huh? The 3.27 rolls is an average number before either a 6 or a 7 is rolled. What do you mean by "if you wait for a seven it lasts 6 rolls"?

Cheers,
Alan Shank

Your $32 base bet (all points) would have a 3.90% House Advantage.

($12 x 0.01515 + $10 x .04000 + $10 x 0.06667) / $32 = 0.039016 as House Advantage, thus negative to the Customer.