One of my philosophies of life is that if you want to be good at anything I believe you should try to understand all facets of it, even parts you don’t necessarily have direct contact with. If you want to be a great plumber, for example, you should know how copper pipe is made, how solder is made, the mechanics of heat and glue bonding, the geometry of tube routing, the physics of water and atmospheric pressure, the history of plumbing, the bylaws and building codes, on and on and on.
It’s no different with the game of Craps. Before we just go out and play the game somewhere we should look into understanding some basic concepts and aspects of it first. This would help us play it more intelligently and at least give us a realistic expectation of the outcome of things.
Let's start with the 'tools of the game' --- Dice.
Two of them are called ‘dice’. A single one is called a ‘die’.
Dice are the most ancient gambling implements known to man, and the most universal, having been known in nearly all parts of the world since earliest times. You see them used a lot in board games and games of skill but their chief history and probably their greatest use today are in gambling games.
A modern die has a specific design. It’s a cube marked on its respective faces with 1, 2, 3, 4, 5 and 6 dots so that the dots on opposite faces always total 7. If the 2 side is vertical and facing you, with 4 on the top, 1 should be at your right and 6 on your left. In gambling places the dice provided there have been manufactured to be perfectly balanced at all faces and to be geometrically perfect cubes with sharp crisp corners and edges.
There are unique probability characteristics with dice, that I avoided getting into in one of my other little essays (about the 6/49 lotto), but of which I'll talk about now.
In most games (Craps included) two dice are rolled and the two numbers are added together. The result of the roll can be 2 (rolling a pair of 1’s) or anything up to and including 12 (rolling a pair of 6’s).
Is any combination of a roll as equally probable as any other?
No.
Let’s see why.
There are a total of 36 different pair-combinations.
I can list them in a chart by writing each possible combination you could roll in a (die#1,die#2) format:
Possible combinations of rolling 2 dice:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
This little grid shows all 36 combinations possible by rolling a pair of dice.
If you examine the grid you’ll notice that only 1 possible combination gives you a sum of 2, that's (1,1). So this means that the odds of getting 2 on a roll of the dice is just 1 chance in 36 or 1/36 or (if you divide 36 into 1) a 2.78% chance. On the other hand, look at all the combinations that add up to 7. They are: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). You can see that there are 6 ways to roll a 7 so this means the odds are 6 chances in 36 or 6/36 or a 16.7% chance you'll get a 7 on a typical roll.
By looking at just these two sums, 2 and 7, we already realize that different sums have different odds of showing up on a roll of a pair of dice. Let’s skip going through each remaining dice combination. Allow me to simply list all the probabilities and combinations for you:
Sum of Dice Roll |
# of Combinations |
Probability of Occurring |
% of Chance |
2 |
1 |
1/36 |
2.78% |
3 |
2 |
2/36 |
5.56% |
4 |
3 |
3/36 |
8.33% |
5 |
4 |
4/36 |
11.10% |
6 |
5 |
5/36 |
13.90% |
7 |
6 |
6/36 |
16.70% |
8 |
5 |
5/36 |
13.90% |
9 |
4 |
4/36 |
11.10% |
10 |
3 |
3/36 |
8.33% |
11 |
2 |
2/36 |
5.56% |
12 |
1 |
1/36 |
2.78% |
Do you notice the patterns in this chart?
Look at the top line first. To throw a 2 (column 1) there is only 1 possible combination (column 2) of the dice that will give it to you, that is (1,1). At the other end of the scale (the bottom line of the chart), 12 also has only 1 combination and that’s (6,6). Both of their ‘probabilities of occurring’ and ‘% of chance’ are equal (column 3 & 4). Next we look at 3 and 11 (column 1). Each have 2 possible dice combinations (1,2) (2,1) & (5,6) (6,5) and both have identical ‘probabilities of occurring’ (2/36), etc. And so it goes with all the possible dice combinations as you discover the pattern of an upper number matching the parameters of a lower number in the chart. Except for 7. It’s unique in its ‘probability of occurring’ and is also, as it turns out, the most favorable of all the possible sums that a person can get in a throw of the dice as indicated by its ‘% of chance’ (16.7%).
So if you sit at a table and throw a pair of dice, the most likely sum to come up will be 7 (about 16.7% of the time), next most likely are 6 or 8 (column 1), each of which occur 5 times out of 36 (or about 13.9% of the time). And so on.
You might ask: Why is 7 the most likely sum to show up?
Why does this happen?
Well, think of it this way:
To picture this, first of all pretend you’re rolling the dice one after the other rather than at the same time.
So you roll the first die.
No matter what comes up on the first die, the second die always has a 1 in 6 chance of rolling a number that will cause the pair to equal a sum of 7. For example, if you roll a 1 with the first die, the 1 in 6 chance of the second die roll may give you the 6 required and thus a total sum of 7 between the two dice. However, if you roll a 2 with the first die, the 1/6 chance of the second die may give you the 5 and once again, a total of 7. And so it goes for the other number combinations adding up to 7. In other words, it doesn’t matter what the first die rolls because the second die always has a 1 in 6 chance of rolling the number you need to give you a total sum of 7.
To put this probability-of-a-sum-of-7 thing mathematically you’d describe and write it this way:
Since it doesn’t matter what number you roll with the first die, any possible number will do, any number will be fine. So the probability of getting the number you need on the first die is 6/6. This means it does not matter what number comes up. Any one of the 6 choices out of 6 will work.
So we roll the first die and a number comes up and (I'll say it again) it doesn’t matter which one.
Now we roll the second die.
But the second die needs to compliment the first one with a specific number that will total to 7 when added with the first die. So just any number won’t do. You need a specific one to come up. The odds of that happening are 1 chance in 6 possible numbers on that die or 1/6.
So we’d write this down in an equation like this:
6/6 X 1/6 = ‘the-odds-of-a-seven-coming-up-on-a-roll-of-a-pair-of-dice’
6/6 X 1/6 = 6/36 = 16.7% chance-of-occurring. (Refer to chart above to confirm results!)
Another way of looking at all this is with the law of averages or commonly called the ‘Law of Large Numbers’. In probability, when you do something for a long time, with a great amount of repetition, the results always start to drift to an average.
If you toss a coin only 10 times you might get 3 ‘heads’ and 7 ‘tails’ even though the mathematical odds are 50% for ‘heads’ and 50% for ‘tails’. This is because the sample is just too small to arrive at an accurate display of the probabilities. But if you toss the coin a million times, you’re likely to see results closer to the average of 50% of the time ‘heads’ and 50% of the time ‘tails’ within the calculated margin of error. It always goes this way when dealing with a large number of trials. It’s a law of nature actually. Things always drift towards an average. You can count on and expect this always.
The average number on a single die is 3.5.
No, it’s not 3 as you might think.
You calculate the average like this:
Add all the faces, 1+2+3+4+5+6, then divide by 6 to get the average, you get 3.5.
So if you have a pair of dice, the Law of Large Numbers says the average these would drift to is:
3.5 X 2 = 7.
That’s why 7 is the most probable of the combinations --- because it’s the average of that particular probability system!
And that's all dice are; a mini probability system.
Now we’ve looked at the 7 combo let’s look at a different sum. What if you wanted to roll an 8 during your Craps game? If the first die rolled to a 1 then it would be impossible to get a total sum of 8 from any number that would turn up on the second die. Unlike the 7 sum, to get an 8 overall, the number the first die falls on IS important. What if you needed to roll a sum of 6 during your Craps game and the first die rolled to a 6? It would now be impossible for anything to turn up on the second die to give you a sum of 6 overall. This is why 7 is the most probable roll and all other numbers, greater and lesser, start decreasing in probability of occurrence.
The odds for these other combinations are easily calculated with math equations just as the 7 combination was and will conform exactly with the probabilities listed in my chart above. Let me show you.
To mathematically calculate the odds to get a roll of 9, for example, would go this way:
The first die needs to land on one of 4 different possible numbers, 6, 5, 4 or 3. It doesn’t matter which one. The odds of this happening are 4 chances in 6 or 4/6. If the first die rolls to a 2 or 1 it will be impossible to create a sum of 9 with a second die. For now assume the first die indeed hits one of the 4 numbers. The second die now has only 1 chance in 6 of hitting the corresponding exact right number to create a total sum of 9. So the odds for the second die to hit its exact right number is 1/6.
Now we simply multiply the probability of one die against the other:
4/6 X 1/6 = 4/36 = 11.1% = the-odds-of-rolling-a-sum-of-9!
What’s cool of course is that these probabilities are a function of a pair of dice regardless of what game they are being used in. For example, if you were playing a game of Monopoly and were in a position of buying a hotel for some properties that your opponent could possibly land on in his next move, you’d now have some valuable information about the odds of a pair of dice, to help you place your hotel wisely.
If your opponent was 2 spaces away from one particular square of your property, 3 hops away from another and 7 from another square of yours, it would be a smart move to put your hotel on the property 7 hops away from him. It’s more likely this will come up with his next roll of the dice than the 2 or 3 roll.
Are you starting to see that understanding the underlying principles of things gives you a slight advantage in life?
So let’s move on to the game of Craps now…
By the way I looked up the origin of the name Craps in a big Webster's Encyclopedia/Dictionary book I have. The two 1's on a roll of the dice, what we call 'snake-eyes' today used to be called 'crabs' by the English, a couple hundred years ago. This term was an offshoot from an earlier English card game in which two aces were also called 'crabs'. Maybe because the letter 'A' looked a bit like the body of a crab with it's two extending claws?
Anyway, when the French got a hold of the dice game they mispronounced 'crabs' by saying 'craps'. Since it became quite a popular game in the French section of the United States in the late 1700's and throughout the 1800's (in the New Orleans area and along the Mississippi), the French mispronunciation 'Craps' soon became strongly associated with the game and it eventually stuck.
Craps used to be played a lot in the back alleys, streets and on riverboats in the U.S. before the days of casinos and gambling halls. Roman soldiers are even known to have played it among themselves during their free time. In modern times it is a common popular game at any casino that you visit and is curious for its seemingly strange and complicated rules. At first the rules seem arbitrary and peculiar. As we soon will see they developed for a very good and exact reason.
The rules are actually pretty straightforward:
So, on a first roll of a round, a 2, 3 or 12 is bad while a 7 or 11 is good. Any other sum sets off a race of rolls to match that sum before rolling a 7. That's about it.
Of course betting is involved.
At the beginning of the round the player (shooter) bets a certain amount of money, let’s say $10. If he wins at any stage in the round he receives his $10 back plus he wins $10 from the House. If he loses, he naturally forfeits his bet of $10 to the House. That round is finished and a fresh round of play begins.
People will wonder where these strange rules came from. Well it’s our old friend --probability-- of course. The long-term odds of any gambling game should be weighed in the casino’s favor to guarantee they continue to make a profit from their operation. Their business relies on this principle. It isn’t too important for them to win big from their customers but rather to win consistently over a long time in exact proportion to the odds of the particular game. And make no mistake, the odds always favor the House. But it is also very important that the odds are never too heavily weighed against the customer otherwise it would discourage players and after a while no one would want to play.
After working on the odds of Craps at its various stages I realized some interesting things.
Craps is a cleverly designed game in that its odds vary for and against the player depending on the stage of the game they are in. In the beginning of a round the odds work out to be largely in the player’s favour and this no doubt is the allure of the game. It draws you in and pumps your confidence up so that losing isn’t noticed because it’s a gradual bleeding. It almost creeps up and grabs you when it’s too late to do anything about it but it can be so slight you may hardly notice it’s happening.
We know, from my little chart way back, that the odds of getting the preferred 7 or 11 on your first throw of the dice is 6/36 and 2/36 respectively or 8/36 collectively. That works out to about a 22.2% chance of winning immediately. Rolling a 2, 3 or 12, on the other hand, and losing immediately, carries with it a 4/36 chance collectively or 11.1%. Isn’t it slick how the chance of winning is designed to be almost exactly twice the chance of losing? You have odds of 2 to 1. You are twice as likely to win than lose on your first throw. This is a sweet deal for the player and it pulls them into the game.
But the House ain’t worried. It’s all about the big picture to them.
Deep pockets, plenty of customers and time.
So far so good.
The 22.2% chance of winning and the 11.1% chance of losing on the first throw account for exactly 33.3% of the chances of what will come up on the first toss of the dice or 1/3. That leaves all the remaining numbers that could come up on a first throw such as 4, 5, 6, 8, 9 and 10. You can do the math yourself and see, from my chart, that their combined individual probabilities equal 66.7% or the other 2/3 of the equation. So you have a 66.7% chance of not losing with these numbers, but not winning either. If you get one of them, it becomes your ‘point’ and you are now allowed to scoop up the dice for another throw to see if you can throw a match to that ‘point’ before you throw a 7.
Notice that now throwing a 7 has become a bad thing. If you throw one, you immediately lose and worst of all, 7 is of course the most likely number to come up. The tables have suddenly turned in favor of the House at this point. Hopefully your ‘point’ is a 6 or 8 since these are the next most favored numbers to come up in a throw, after 7. If your ‘point’ is a 10 or a 4, you’re up against hard odds indeed as you hope to roll them before you roll a 7. (Refer to my chart to see the odds).
Now this is where things get a bit complicated. If you roll a 4, 5, 6, 8, 9 or 10 on your first roll of a round, the probability of winning on the next throw depends on what your ‘point’ equals of course. Calculating the probability of winning now requires computing a complicated sum of different probabilities to obtain the various ‘point’ values, multiplied by the probabilities of winning once you know your ‘point’.
I undertook this task (I’ll spare you the page and a half of calculations) and when all the fractions have been multiplied and sums have been completed, it turns out that the overall probability of winning at Craps is equal to 244 chances in 495. This equals 49.2929%. Put another way, the chances of winning are just a smidge below a fair 50-50 bet.
In reality this means that if you play Craps over a long period of time on average you will win $10 exactly 49.2929% of the time and you will lose $10 exactly 50.7071% of the time (assuming you're making $10 bets each round). So your average winnings per bet, overall, could be figured out to be:
$10 X 49.2929% minus $10 X 50.7071%
This comes to -$0.141. (Notice! That’s minus!)
So as you play round after round at the Craps table you slowly lose about 14.1 cents for every 10 bucks you bet.
Wow! you say…
It's losing but it's not much! Right?
Ahhhhhh. Sorry but it’s something!
And to the casino bosses it’s everything.
They rely on the Law of Large Numbers and this constant trickle of cash dripping in their favour. Eventually, with their huge reserves of money and unending patience, in the long run, they will win all the cash and the vast collection of gamblers at that table will go home empty handed. It’s inevitable I’m afraid.
When the odds are so slim between winning and losing like that, the guy with the deeper pockets can endure all the ups and downs and variables and ultimately outlast the guy with less money in his jeans. And that’s exactly what happens, every day. You may get what seems to be a 'lucky' streak but if you continue playing it will always eventually turn the other way at some point.
To kind of wrap this up I just wanted to mention another, final quirk of the Craps game. It’s a little tougher to grasp but is a very important aspect of the game in regards to probability so I need to bring it up.
In the game of Craps, people NOT throwing the dice (i.e. they aren’t the current shooter or roller) have the opportunity to also place bets on the round, just like the shooter. This is why we often hear a big cheer rise up from the casino Craps table throughout the evening. The spectators aren't expressing great affection for the shooter, they're cheering because their own side bet has just paid off on a particular throw he's made. They can either place a bet in favor of the shooter winning the round or, more interestingly, place one against him, hoping he’ll lose! They can plunk $10 down on a special area of the table designated for anyone betting against the shooter and if he loses the round, these betters immediately win an amount equal to what they bet, from the House!
But hang on… something’s not right here!
If you think about it for just a second, the odds of the game are designed to always favor the House right?
Yes.
So in other words, the odds favor the shooter losing more often than winning?
Yes.
So you betting against him is the same as what the house does?
Yes.
O.K. So naturally it follows that in order to win money at Craps a person simply has to get a great big wad of cash and constantly bet against the shooter all the time and they’ll win 14.1 cents on every 10 bucks they bet, just like the House does!
Yes, this would be true but as it turns out there's no such luck. They didn’t overlook anything.
There’s just one more subtle little rule to the game that I haven't mentioned yet, that's all.
It eliminates this advantage and turns it around in the House's favor. It's a rule for bets placed specifically against the shooter.
You might think, great!
After all, when you look at my little chart, the shooter has only a 1 in 36 chance of ever rolling a 12 on the first toss anyway. And even when he does, you simply get your money back? You don’t actually lose it to the House?
Good!
No harm done, right?
Not quite.
It's tricky but you have to think of it this way:
Without the rule you'd have exactly the same odds as the House or a 50.7071% chance of winning overall. If that rule didn't exist, on average you too would eventually win 14.1 cents on every 10 bucks you bet, as long as you constantly bet that the shooter will lose. After years and years of playing this way, you'd be rich! Now, just by referring to my little chart up above, you can see that the odds of rolling a 12 is a 2.78% chance. When the shooter rolls a 12, the House wins and without the rule, you would too. But with the rule in place, this chance of winning is taken away from you. So now in order to calculate your actual rate-of-return per $10 bet in this game, you must first subtract that 2.78% chance from your hypothetical superior odds of winning.
To do that, it looks like this:
50.7071% - 2.78% = 47.93%
This percentage will help us calculate your real average winnings per $10 bet.
Next we need to work out the difference between your previous superior average winnings (which used to equal that of the House before we considered the rule) and the level this average was reduced to by eliminating your opportunity to win by 1 time in every 36 when betting against a shooter. (the odds of a 12 getting rolled)
That difference is calculated this way:
$10 X 50.7071% minus $10 X 47.93% = 27.8 cents
What this is telling us is that your previous rate of return of +14.1 cents on every $10 bet (before the rule was sprung on you) just got reduced by 27.8 cents with the rule.
So finally, your real actual rate of return, for every $10 bet that you make, works out to be:
14.1 - 27.8 = -13.7 cents ...(notice the minus)
This means that as you play the game, round after round, placing side bets against the shooter, you slowly lose on average about 13.7 cents for every 10 bucks you bet. It’s a little better odds than the shooter (remember, he's losing 14.1 cents on each of his 10 bucks bet) but it still means your money is slowly trickling away from you and ends up being a total loss over time.
In the end, in the game of Craps, you’ll go broke betting against the shooter or for him.
It's a little depressing isn't it?
The trick to gambling is knowing when to hold up.
Knowing when to fold up.
Knowing when to walk away and knowing when to run…
Always keep in mind two things:
The odds always favor the House and It's smart to quit while you're ahead.