No House Advantage in Craps?

Here is a situation I was hoping someone could explain or give me the percentages to: In Craps the odds bet has no house advantage. Everyone (or at least a lot of people) know this. Also, betting the pass and don't pass on the Come Out roll at the same time doesn't even out because of the 12 (takes the pass and pushes the don't pass). So here's the question/situation: What if I bet the pass and don't pass and put a $1 bet on the 30-1 sixes ONLY on a Come Out roll? I don't consider this hedging in the traditional sense since it is ONLY for a Come Out roll. The way I'm seeing it you only need to hit the sixes once in every 30 Come Out Rolls to break even. If the average shooter sevens out in 6 rolls then there's a Come Out roll every 7 rolls (on average). Of course this doesn't account for a shooter hitting a seven or crapping out on a Come Out which would lose the $1 and cause you to put it up again, but then again it doesn't account for a shooter going for 20 rolls after a point either. Anyway, you would only need to hit the 30-1 once in every 210 rolls to break even. If you could do that then the pass/don't pass bets even out on all other numbers and the odds after a point is established has no house advantage. If my logic is off can someone explain why and/or provide the percentages of betting a 12 ONLY on a Come Out roll? I have been at a Craps table for five hours each of the last four days and have analyzed that a 12 comes on a Come Out roll once in about every 21 Come Out rolls. This easily covers the 30-1. I know this sampling is not nearly enough for a long-term hypothesis but that's what I have so far and I am somewhat new to the game! This is just something I noticed and thought about while playing and hoped someone could explain it to me. Thanks!

Two Things:

A) You're forgetting the buck you'll lose for every come out roll that's not a 12
B) A 12 will come up, on average, every 36 rolls.

Well, I won't delve into the perfect math of it, because somebody else will do so for you, and will do it better than I can. I will only suggest you think of the following. Assume you make 36 come out rolls, in which each number is rolled one time. As you already know, if you have bet $5 on the pass, and $5 on the don't pass, then you will break even 35 times, and lose $5 one time, on the doubles 6's.

Now let's take those same 36 come out rolls, but you're betting $1 on the 6's on each come out roll. Now, there are 35 times that you lose $1, and 1 time that you win $30. It's easy to see that you have lost money.

You are correct in that if you get double 6's at least once every 30 come out rolls, then you will break even. Unfortunately, the chance of rolling double 6's is 1 in 36, thus making it a losing proposition.

Quote: nyyankee0101

I have been at a Craps table for five hours each of the last four days and have analyzed that a 12 comes on a Come Out roll once in about every 21 Come Out rolls. This easily covers the 30-1. I know this sampling is not nearly enough for a long-term hypothesis but that's what I have so far and I am somewhat new to the game! This is just something I noticed and thought about while playing and hoped someone could explain it to me. Thanks!

Your sampling isn't nearly enough, because the 12 will come out once every 36 rolls. You are actually giving the house a greater advantage by putting the dollar on the 12 on the Come Out Roll. Think about it like this: You are fretting over the very small edge that's already on the Pass and Don't Pass (~1.37%) by losing a dollar on 35 of every 36 Come Out Rolls. Then on the roll that you do get the 12, you are still losing half of your combined Pass/Don't Pass. You may say, "Well, what's a dollar?" But the house edge on $10 Pass Line bet is a lot less than a dollar! Don't take a very good bet and think you can make it better by adding a bad bet!

Quote: cclub79

Don't take a very good bet and think you can make it better by adding a bad bet!

I agree. Its better to just roll the dice and win or lose than to try to nickle and dime Lady Luck all night long. Oh sure, you can always put your entire bankroll on one bet, but that kinda smarts if it loses. So stretch things out a bit and get some free drinks but don't hedge. The PRICE of gambling is the house edge. Pay the price and have fun.

Yes A 12 will come up about 36 rolls, but that doesn't account for 36 Come Out rolls. What I'm saying is let the rolls with an established point do the work of increasing your odds and get paid betting the Come Out. I just think that the odds on betting the Horn 12 are different when only betting it on the Come Out.

Quote: nyyankee0101

Yes A 12 will come up about 36 rolls, but that doesn't account for 36 Come Out rolls. What I'm saying is let the rolls with an established point do the work of increasing your odds and get paid betting the Come Out. I just think that the odds on betting the Horn 12 are different when only betting it on the Come Out.

The dice don't know it's a Come Out Roll or not, so the odds aren't different. It's a one roll bet, so the horrible odds are the same no matter when you place it, whether it's a hedge for you or not. I understand you aren't losing a dollar every roll, but on every Come Out that's a 2/3/7/11, you are losing it and automatically rebetting it. That's 11 out of every 36 Come Outs, or almost 1/3 of them. Trust us, adding a high house edge bet as a hedge is only INCREASING the amount you lose...it's not helping you.

One thing to realize is that a million players have tried to figure out some way to beat the House Edge. If it could be done, someone would have figured it out long ago [and how its done would now be banned]

Quote: nyyankee0101

Yes A 12 will come up about 36 rolls, but that doesn't account for 36 Come Out rolls. What I'm saying is let the rolls with an established point do the work of increasing your odds and get paid betting the Come Out. I just think that the odds on betting the Horn 12 are different when only betting it on the Come Out.

Uhh...no, no, no. Why would the odds be different on the 12 on a come out roll compared with any other roll? You getting your mind stuck in what's known as Gambler's Fallacy.

The odds on ANY roll, come-out or not, being a 12 is 1 out of 36. But you only get paid 30-1. Some anectdotal evidence that you observed at a craps table does not change the odds. This is a BAD method, and will just make you lose money faster in the LONG RUN.

Now in the short run, it could work great. See a trip report of mine from a few weeks ago, when my buddy hit something like 5 or 6 randomly placed $5 12s in an evening.

Alright Alright Alright! So, if it doesn't matter if it's a Come Out roll or not, then here's another question: What is the average number of rolls between Come Out rolls?

Quote: nyyankee0101

Alright Alright Alright! So, if it doesn't matter if it's a Come Out roll or not, then here's another question: What is the average number of rolls between Come Out rolls?

3.375758

Quote: Wizard of Odds

First, if the probability of an event is p then the expected number of trials for it to occur is 1/p. Let's call x the expected number of rolls per shooter. The probability that any given round will end in one roll (with a 2, 3, 7, 11, or 12) is 1/3. If the player rolls a 4 or 10 on the come out roll the expected number of additional rolls is 4, because the probability of rolling a 4 or 7 is (6+3)/36 = ¼. . Likewise If the player rolls a 5 or 9 on the come out roll the expected number of additional rolls is 3.6 and for a 6 or 8 is 36/11. Assuming a point was thrown the probability of it being a 4 or 10 is 3/12, a 5 or 9 is 4/12, and a 6 or 8 is 5/12. So the expected number of throws per round is 1+(2/3)*((3/12)*4 + (4/12)*3.6 + (5/12)*(36/11)) = 3.375758.


The average number of rolls per turn for a dice shooter is 8.52551

Explanation
Next, the probability that the player will seven out is (2/3)*((3/12)*(2/3) + (4/12)*(3/5) + (5/12)*(6/11)) = 0.39596. The probability that player will not seven out is 1 - 0.39596 = 0.60404. So...

x = 3.375758 + 0.60404*x
0.39596*x = 3.375758
x = 8.52551

Interesting area...
There is indeed a way to make a "no house edge" crap game - perhaps useful for a player-to-player tourny....
1. PASS LINE: same, except 12 loses only one-half of bet amount;
2. Don't pass: same, except 12 wins only one-half of bet amount;
3. COME: same except 12 = one-half loss;
4. DC: same except 12 = one-half win;
5. NUMBERS 4..6,8..10 pay at true odds
6. Hard six & eight = 10 TO 1, not FOR.
7. Hard 4 & 10 = 8 TO 1
8. 2, 12 = 36:1
9. 3,11 = 18:1
10. Horn bet: one unit each from (8 & 9) above.

Now, every bet offered would be 0% edge, either for the house or the player.

Quote: Paigowdan

Interesting area...
There is indeed a way to make a "no house edge" crap game - perhaps useful for a player-to-player tourny....
1. PASS LINE: same, except 12 loses only one-half of bet amount;
2. Don't pass: same, except 12 wins only one-half of bet amount;
3. COME: same except 12 = one-half loss;
4. DC: same except 12 = one-half win;
5. NUMBERS 4..6,8..10 pay at true odds
6. Hard six & eight = 10 TO 1, not FOR.
7. Hard 4 & 10 = 8 TO 1
8. 2, 12 = 36:1
9. 3,11 = 18:1
10. Horn bet: one unit each from (8 & 9) above.

Now, every bet offered would be 0% edge, either for the house or the player.

You did forget the FIELD bet, where you would pay triple for both 2 or 12.

See, I think that locals would pay to play a craps game with those kind of payouts. I would probably keep the house edge for the PASS/DP, COME/DC plays and require a minimum bet on PASS/DP so that the "pay to play" charges won't have to be as high. I would think it would make a nice alternative for smaller earning locals casinos.

The 11 craps tables in North Las Vegas made $4.474 million in the last year. That is about $100K per year per table. I think you could get almost that much in fees.

The 37 craps tables in Downtown Las Vegas made $28.256 million in the last year. That is close to $1 million per year per table.

The major casinos on the strip had 157 craps tables that made $210.545 million in the last year. That is about $1.3 million per table.

Frank,

Hmmm... At a $10 craps table, at 60 rolls per hour, at a per roll advantage of .463% on a $12 6 or 8, you're looking at $.0463 x 60 = $3.33/hour take.

There are two ways to collect a fee: $1 for every shooter, or $10/hour. The thing is with a $10/hour charge, as dealers, you would slow up the game as much as possible to collect the profit, whereas with the $1 / shooter, you would speed up the game. But for this game, you would have a max bet of $100.

Even better, make the $1 a mandatory "jackpot" bet with a return of about $.20 on the dollar to make it seem like it's worthwhile to play... kind of like the Grand Victoria free promotion with $1,000 for making all six points (payback $.162/dollar) except charge a $1.