You Can't Fool the Random Nature of Craps

25 April 2015

By Jerry Stickman

An email exchange from Bill:

Hello Stickman,

First, I really enjoy your writing. It is very informative.

I have been studying craps a lot and in my upcoming trip to Las Vegas, I am trying to guestimate my approximate win/lose amounts. If I make two $10 bets and have 9 as the point and 8 as the come bet number do I lose 28 cents (1.4% x $20) per roll which in 36 rolls would be a loss of $10.08 or do I estimate by knowing that when I get in that situation I have 9 ways to win $10 and 6 ways to lose $20 which would be a $30 loss? I am a little confused and would appreciate your insight.

Thanks,
Bill

Hi Bill,

The best way to look at this situation is by taking the 14 cents per bet that you will lose PER DECISION. Each bet as a separate game. They should not be combined. So, on the point of 9, over a long period of time you will lose 14 cents per bet. For example, if you bet $10 1,000 times that the point is 9 you will have bet a total of $10,000. You will lose very close to 1.4 percent of the total $10,000 bet or about $140.

Of course you will not lose that amount in a nice smooth pattern. You may lose 10 bets in a row ($100 loss) and then win 8 bets in a row ($80 win) then lose 6 bets ($60 loss) then win 7 bets ($70 win), etc. Your bankroll will move dramatically up and down through the period of 1,000 bets, but ultimately your total loss will be about 1.4 percent (14 cents per $10 wagered).

The main thing you overlook in your example is what happens during the come-out roll. That is where the player has the advantage over the casino. There are 8 ways to win on come-out (6 possible combinations that total 7, 2 possible combinations that total 11) and four ways to lose (two combinations totaling 3, one totaling 2, and one totaling 12) giving the player a 2-1 edge. After the point is established the edge swings dramatically toward the house. The 1.4 percent edge is calculated by including the 2-1 player edge on come-out.

I hope that clears it up. Please feel free to contact me again if you still are unclear.


From Bill:

Thank you, I have studied your answer and have one last question. My head is telling me that it would take several rolls for the casino to overcome my pass line edge. It seems that if I would win $40 on the first roll my edge per roll on the pass line is $40/36 or $1.11 per roll.

If the first roll is a number then I could make a come bet on the second roll and the my edge on the come bet would more than offset the 1.4% casino edge on the pass line bet giving me an edge on the second roll. If the come bet becomes a number then the casino edge on the third roll and following rolls is 28 cents per roll. That means it would take 4 rolls to offset the come out roll (4x.28=1.12).

If I bet like this and stick to it will I be even or nearly even with the casino? My head is also telling me not to replace any winning decision but to let the roll end before repeating this method of betting so do I have a decent chance against the casino?

Thanks,
Bill

Hi Bill,
I know betting and odds can be a bit confusing. There is a whole lot going on during the game of craps. I think it might help if think about the big picture and not “if I would win …”

There are two things to remember. First, if you bet $10 on the pass line and you win that bet, you only win $10 – not $20. You get your original bet back plus your $10 win. I am not sure where you are coming up with your $40 win. If you bet $10 on the pass line and win it on the come out roll, you cannot place a come bet. It will be another pass line bet. You can only place a come bet after a point has been established on the come out roll.

The second thing to remember is the 1.4 percent house edge includes the player advantage on the come out roll. You have a 2-1 edge on 12 of 36 possible outcomes. On the remaining 24 possible outcomes of the come-out roll the house has a 9.09 percent edge on the 6 and 8, 20 percent edge on the 5 and 9 and a whopping 33 1/3 percent edge on the 4 and 10.

On a random dice shooter (one who does not have any dice control skill) you will lose 14 cents for every $10 you wager on a pass / come bet over time and that time could include large swings in wins and losses. So the more money you bet, the more you will lose over time.

The best way to bet on a random shooter is one table-minimum pass or come bet with the maximum odds allowed (assuming you can afford it). That way, over time, you will only lose 1.4 percent of the base portion of you bet and break even on the odds portion. Each additional pass or come bet carries an additional house edge on the base portion of the bet.

May all your wins be swift and large and all your losses slow and tiny.

Jerry “Stickman”


Jerry “Stickman” is a regular contributor to top gaming magazines. He authored the video poker section of "Everything Casino Poker: Get the Edge at Video Poker, Texas Hold'em, Omaha Hi-Lo, and Pai Gow Poker!" You can contact Jerry “Stickman” at stickmanjerry@aol.com