Calculating Probability

Theoretical probability is the likelihood of a specific outcome or set of outcomes resulting from a single trial. Given that the actual outcome is determined nigh purely by chance, it cannot be used as an accurate tool of prediction. That is to say, if a single trial has six equally possible outcomes, such as with an ordinary die, the first six permutations won’t necessarily feature all six possibilities. In fact, it’s less probable for this to happen than not. But, after six hundred permutations, it’s reasonable to expect about a hundred of each possibility. In other words, actuality reflects the science more clearly the larger the sample. Professional gamblers use theoretical probabilities throughout their careers, determining the likelihood of victory and if the pot is worth pushing their luck.

So, how do we go about calculating a theoretical probability? Actually, despite the fact that some people go through several years of training to become statisticians, the foundation is actually quite simple. To start, count the number of possible outcomes. Let’s continue to consider an ordinary die with six possible outcomes. Since each number from one through six appears once, the theoretical probability of rolling any specific number is one in six. The probability of rolling an odd number is three in six, since there are three of them, which can be simplified to one in two. The same goes for even numbers. This can be expressed as one in two, one out of two, 1/2, 1:2, 0.5 or 50%. So, the theoretical probability is the number of outcomes that fit the criteria divided by the total number of possible outcomes.

For multiple permutations, there are two general situations. One is where the an outcome – or a permutation in general – affects subsequent ones. The other is where each is wholly independent. The die fits the latter; a deck of cards the former.

Let’s find the odds of rolling a one followed by a five. The odds of a one is 1/6, as are the odds of rolling a five. We multiply the two fractions. 1/6*1/6 = 1/36. Each probability is still independent, but the odds of rolling those two numbers in that order on two consecutive rolls would be 1/36. If we were looking to find the odds of rolling those two numbers in any order, we would multiply the outcome by the number of possible orders. In this case, it’s two (1,5 and 5,1). So, the probability of rolling two specific numbers on consecutive rolls in either order is 1/18. When the two desired outcomes are identical, the odds for any order are, of course, the same as for a specific order. Order is moot.

Now, let’s look at a deck of cards. The dynamic nature of its probabilities is the foundation for counting cards at the blackjack table. (Note: Counting cards in a live casino is only recommended for players who can do so surreptitiously.) An ordinary deck of 52 cards has 4 suits each with 13 cards. Thus, the probability of drawing a specific suit, let’s say a club, is 13/52, since there are 13 clubs and 52 cards, or 1/4, since there are four suits, one of them clubs. The probability of drawing a specific number, let’s say a 9, is 4/52, since there are four 9s and 52 cards, or 1/13 to simplify. Thus, the odds of drawing the nine of clubs would be 1/52 since there is only one such card. Simple enough when you think about it, right? But after drawing a card and discarding it, everything changes, and unfortunately, the fractions can no longer be simplified.

Let’s suppose that first card was a Jack of clubs. We now have twelve clubs and three jacks in our deck of 51. At this point, the odds of drawing a club are 12/51, twelve for the number of clubs, fifty-one for the total number of cards. The odds of a jack are 3/51. What this means is that calculating the odds of a specific set of draws requires changing the base.

Thus, the odds of drawing a Jack followed by a nine would be 4/52, or 1/13, multiplied by 4/51. As with the die, the odds of a Jack and a nine in either order would be found by multiplying the outcome by two. The odds of two consecutive draws of equal value, let’s say two Jacks, would be 4/52 multiplied by 3/51, since only three Jacks remain assuming the first draw is successful.

In situations where the desired possibilities of one permutation encompass those of any subsequent, calculating the exact probability requires us to incorporate the probability that one of those subsequent desired possibilities is eliminated in advance. In simpler terms, if we want to find the possibility of drawing an numeric card followed by a seven, we have to account for the odds of drawing a seven first. However, this is the realm of more complex probabilities. Besides, disregarding that possibility and simply multiplying the odds of a numeric card (36/52 or 9/13) by the odds of a subsequent seven (4/51) will get you a close enough answer. Close enough to weigh the pot against it in a casino, at the very least.