The basis of all games of chance is the concept of probability.
What is probability? Probability refers to the chance that an event will occur. Probability is a value that ranges from 1, for an event that will definitely occur, to 0, for an event that will never occur.
We get the probability of an event occurring by dividing the number of possible ways an event can occur by the total possible number of outcomes for all events. Let’s illustrate:
For a 6-sided die, we know that the total number of sides and numbers is 6. So, for any single roll of the die, we could have any one of 6 possible outcomes. 6 is then the total possible number of outcomes for all events.
Thus, the chance of rolling any number from 1 – 6 on a roll of the die is 1/6, assuming that the die isn’t loaded (we hope!).
A variation of this, is the chance that an even number will be rolled. We know that a 6-sided die has 3 even numbers – 2, 4 and 6. So that’s 3 possible outcomes / 6 total outcomes. Thus, the probability of an even number being rolled is 3/6 or 1/2. In percentages, that’s 50%.
This changes a little when we move to games like roulette, where the probability of winning is affected by the number of numbers wagered on. Roulette has 37 numbers from 0 to 36 for a single-zero game and 38 numbers from 00 and 0 – 36 for a double-zero game.
The probability of winning on a single-number straight up wager is 1/37 for a single-zero and 1/38 for a double-zero game. However, the probability of winning at all is affected by the total number of wagers placed. We’ll use the single-zero variant for our examples.
Thus, if a player bets 15 different straight up wagers on a single game, the probability of winning on any one of those 15 numbers isn’t 1/37 anymore, but 15/37. The drawback of this method of wagering is that it is overall a poorer choice. We’ll get to that later.
Dependent and Independent Events
A dependent event is an event that is influenced by previous events that preceded it. An independent event is otherwise unaffected. What does that mean?
Let’s assume that your friend and you have a single 52-card deck each. You each take turns drawing a card to see who has the higher value, with the higher value winning, irrespective of suits. Each drawn card is then discarded.
At the start of the round, your friend draws a King.
Only an Ace beats that, but since you have 4 Aces in your deck, the probability of you winning by drawing an Ace is 4/52 or 0.076923 or 7.6923%.
You draw an Ace and win the round. Now, both of you discard your cards.
In round 2, your friend draws a King again. The probability of that happening is not 4/52 anymore since a King has already been taken out of his deck. So the probability of your friend drawing a King is really:
(4-1) / (52-1) = 3/51 or 0.058824 or 5.8824%.
Now, for the enthusiasts, the chances of your friend drawing 2 Kings in a row in the first 2 rounds is:
0.076923 x 0.058824 = 0.004525 or 0.4525% (incredibly small!)
In roulette, the probability of any single number occurring is 1/37. No matter how many spins have occurred, the probability doesn’t change as the numbers on the layout remain the same, nothing added or subtracted. This is an example of an independent event, that is unaffected by preceding or future events. The probability of any single number occurring is 1/37 now and forever.
A combination of events is a different story. For someone who has wagered on straight up 35 and also on split 35/36 only, the individual probability of winning is 1/37 for the straight up and 2/37 for the split.
The probability of them BOTH winning (assuming only 1 wager had been place on both the straight up and the split) is as follows:
(1/37) x (2/37) = 0.001461 or 0.1461%
Now, for the enthusiasts, the probability of someone winning on 35 twice in a row?
((2 x 1) / ((2 x 1)(0))) x (1/37)2 x (36/37)0 = 0.00073 or 0.073%
How about winning on 35 two out of three rounds?
((3 x 2 x 1) /(2 x 1)(1x 1)) x (1/37)2 x (36/37)1 = 0.002132 or 0.2132%
No discussion on gaming and probability can be complete without at least broaching expected value. Expected value is the mathematical probability of winning x winning wagers – probability of losing x losing wagers.
In roulette, a straight up wager pays 35, thus, the expected value for a straight up wager is =
(The probability of a straight up win x the amount won) – (The probability of losing a straight up wager x the amount lost) = (1/37 x 35) – (36/37 x 1) = 0.945946 – 0.972973 = -0.02703.
This implies that overall, in the LONG TERM, everyone who wagers on straight up is expected to lose 0.02703 dollars for every dollar wagered.
We mentioned why wagering on multiple areas is a poorer choice than just wagering on one. Let’s calculate the expected value of someone who wagers straight up on 15 numbers. The overall probability of winning on any one of the 15 numbers is 15/37 or 0.405405 or 40.5405%. Not too bad you’d think?
The probability of a win is 15/37, but the amount that would be won is no longer 35, as we have to subtract the amount of losing wagers, so that’d be 35-14. The losing probability would be 37-15/37 or 22/37 and the amount lost would be 15. The calculation would look like this:
(15/37 x (35-14)) – (22/37 x 15) = (15/37 x 21) – (22/37 x 15) = 8.513514 – 8.918919 = -0.40541
So, it looks like anyone wagering this way would over the long term end up losing 0.40541 dollars for every dollar wagered. Far worse than just wagering on a single number, it seems.
Designing Games based on Expectation
Knowing how to calculate expectation allows us now to design and calibrate payments and odds for game wagers. Let’s look at one of the payments we know of.
A pair bet on Baccarat pays out 11:1. What is the expectation of this sort of wager? Let’s assume a wager of $1. The probability of a pair occurring is 0.0747. Thus, the expectation would be:
(0.0747 x ($1 x 11) – (1-0.0747 x 1) = 0.8217 – 0.9253 = -0.1036.
This means that for every dollar, the house would win 10.36 cents. Imagine if the payments were lower, like 8:1?
(0.0747 x ($1 x 8) – (1-0.0747 x 1) = 0.5976 – 0.9253 = -0.3277.
This means that for every dollar, the house would win 32.77 cents. That is almost 200% more! Just by reducing the payment odds.
Something to ponder, yes?
The best way to prove something is to take it to extremes. Let’s try this with a single-die wager. A single-die is rolled and players get to wager on numbers 1 to 6. The Sic-Bo payment we know of pays 1:1. As the probability of any of the numbers appearing is 1/6, if a player wagered $5 on number 1, the expectation would be as follows:
(1/6 x 5) – (5/6 x 5) = 0.833 – 4.167 = -3.333, meaning our friend is expected to lose 3.33 dollars for every $5 wagered.
As the payment is 1:1, it does not make sense to wager on more than 1 number. What if the payment was raised to 3:1? For a similar $5 wager on number 1, the expectation would be as follows:
(1/6 x 15) – (5/6 x 5) = 2.5 – 4.167 = -1.667, which is still a safe gap for the house.
Let’s mix this up a little. Now that the payment is 3:1, our player thinks it gives him a better chance by wagering $5 each on two numbers, say numbers 1 and 3. The expectation would be as follows:
(2/6 x (15-5) ) – (4/6 – 10) = 3.333 – 6.667 = -3.333
By doing the maths behind your games, you’d be able to attract players with better odds, so to speak, while still maintaining an edge for the house.
So, now you are probably asking – if the expectation is that everyone who plays would lose over the long term, how does one account for all the winners? Something to think about, which we will discuss in a later post.