Determining Probability Based on Prior Events
With my other probability articles I demonstrated how to determine the likelihood of independent and dependent events. In those examples, I used no prior knowledge of events, but in many scenarios when testing for probability on dependent events, one event has already occurred for which the probability has been determined, and you may use that knowledge to test the second, or subsequent events.
In this example I’ll demonstrate finding the probability of an event related to playing the card game 21. This game has one clear goal – to end with a total value of either 21 or value closer to it than the dealer, without going over 21. In this example game, I have one card and the dealer one card. My card is an Ace, meaning I currently have 11 (need 10 to make 21), and the dealer’s card which is face-up is a Jack (10, they need an Ace). Also, it was previously determined the probability me the player=11 after the dealer=11 is 1/32.
For this example game, I want to find the probability the dealer will “hit” 21.
For this to occur, the dealer will need to draw as their next card an Ace.
This formula determines the probability of future event(s) based on the occurrence of a previous event.
Using our example 21 game, here is the formula I’ll use.
Regarding the formula’s terminology, here is a brief overview:
- The probability of event2, given that event1 is…
- The probability of event1 given event2 multiplied probability of event2…
- Divided by…
- The probability of event1.
Now I’ll plug into the formula the example 21 game.
- The probability of dealer=11 (ace), given that player=11…
- The probability of player=11 given dealer=11 multiplied probability of dealer=11…
- Divided by…
- The probability of player=11.
Beginning with the numerator (top of the fraction), I will start completing the formula: P(P11|D11) * P(D11).
First, I’ll fill-in the first portion on the left using the probability previously determined from the beginning of this article = P(P11|D11), or the probability of player=11 given that (already occurred) dealer=11.
Next, I’ll determine the next (right) portion of the formula in the numerator: P(D11).
Therefore, I’ve determined the probability of the dealer=11 is 4/52 because 4 aces exist (matching criteria) out of 52 possible cards in the desk (All Possibilities).
Here is the original formula with the numerator updated:
Now, I will complete the formula for the denominator.
Since the probability in the numerator for P(D11) has already been determined (4/52) and results in the same probability as P(P11) – the player has the same probability of receiving 11 as the dealer, I’ll just copy that probability into the denominator.
Now I’m ready to complete the formula:
Therefore, using Bayes’ Theorem I’ve determined the likelihood of the dealer drawing 11 after I received an 11 is .7%.