# Conditional Probability is Not Commutative: Formulas and Examples

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## Conditional Probability and Given Outcomes

In the above case of the double coin toss, it is clear that one could describe the overall wager’s success as conditional on the first toss. If the first toss is heads, then the conditional probability of (heads, heads) is 0.5. If the first toss is tails, then the wager is already lost. More properly, the “conditional probability” for the successful wager, “given that the first toss is heads”, is 50%.

It is also possible to reverse the direction of the argument, to decide the probability of a component given the overall outcome. If you tell a friend that you won the above wager, your friend could correctly deduce that the first toss must have been heads. “Given the overall result of winning the wager”, then the “conditional probability that the first coin was heads” is 100% certain. If you lost the wager, then the conditional probability that the first coin was heads would be 50%.

The double-head coin toss wager is a trivial example of different probability calculations, but serves as a good example that one situation may have different conditional probability models and calculations.

## Conditional Probability Does Not Commute

If the order of the terms of a math operation does not change the outcome, then that operation “commutes”, or is “commutative”. Addition and multiplication are indeed commutative: “a + b = b + a”; “a * b = b * a”.

The intersection of two sets is also commutative: for sets ‘A’ and ‘B’, A∩B = B∩A. For example, in a standard deck of cards, the intersection of “red” cards and “Jacks” has two elements: the Jack of Diamonds and the Jack of Hearts. The order of writing “red cards ∩ Jacks” versus “Jacks ∩ red cards” is irrelevant.

However, division is not commutative. For example, “6 / 3 = 2” but “3 / 6 = 0.5”.

It is all too easy to assume that P(A|B) = P(B|A), or to mistakenly switch the values when making the calculation. Perhaps people simply get confused about what event was ‘A’ or ‘B’ before plugging the values into the formula.

The equations are indeed different: “P(A|B) = P(A∩B) / P(B)”, but “P(B|A) = P(A∩B) / P(A)”.

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