Chapter 11 Conditional Probabilities

Probabilities change as circumstances change. The probability of drawing an Ace from a full deck of cards is 4/52. If you draw an Ace out and don’t replace it, the probability of drawing an ace changes. Now, there are only 51 total possible outcomes, only three of which are favorable. So, the conditional probability of drawing an ace given that one ace has been removed is 3/51

Now, a bit more symbolization: \(\Pr(A \textrm{ given } B) \textrm{ is written as } \Pr(A \vert B)\)

11.1 Calculating Conditional Probabilities

The probability of A given B is the probability of the conjunction of A and B, divided by the probability of B, provided \(\Pr(B)\) is not 0. This is called the Definition of Conditional Probability. Symbolized, this looks like:

\[ \Pr(A \vert B)= \frac{\Pr(A \& B)}{\Pr(B)} \]

This formula is rarely used. It’s usually only necessary to think in terms of favorable outcomes over total possible. Conditional probabilities change one or more of these outcomes. For example, you draw a king from a standard deck, keep it, and draw a second card. What is the probability of getting a king on your second draw given that you got a king on the first draw? That is, what is \(\Pr(K2 \vert K1)\)? Now there are 51 total cards, three of which are kings. So, it is 3/51.

You roll a die and it goes off table. Your friends tells you that it’s an even number, but you can’t see. Given this extra information, what’s the probability that you got a 2? Now, you can eleimate all of the odd numbers. So, your only possibilities are 2, 4, and 6. So, the probability of rolling a two given that you rolled an even number is 1/3.

Now, we can add one more conjunction rule:

\[ \textrm{If A and B are dependent, } \Pr(A \& B) = \Pr(A) \times \Pr(B \vert A) \]

Draw two cards from a full deck and don’t replace the first card before drawing the second. What is the probability of drawing two Kings? \(\Pr(K1 \& K2) = \Pr(K1) \times \Pr(K2 \vert K1) = 1/13 \times 3/51 = 1/221\)

A and B are independent events just in case \(\Pr(A) = \Pr(A \vert B)\). This is just what you would expect. In this case, the extra information that B is true did not change the probability of A. This can only be the case if B makes no difference, in other words, they are independent.

11.2 The Rules

Here are all the rules up to this point:

11.2.1 Simple Events

\[ \Pr(A)=\frac{\textrm{ favorable outcomes }}{\textrm{ total possible outcomes }} \]

11.2.2 Negations

\[ \Pr(\neg A)=1-\Pr(A) \]

11.2.3 Disjunctions

\[ \Pr(A \, \textrm{ or } \, B)=\Pr(A)+ \Pr(B) \quad \textrm{(Incompatible Events)} \]

\[ \Pr(A \, \textrm{ or } \, B)=\Pr(A)+ \Pr(B)- \Pr(A \,\textrm{\&} \, B) \quad \textrm{(Compatible Events)} \]

11.2.4 Conjunctions

\[ \Pr(A \, \textrm{\&} \, B)= \Pr(A) \times \Pr(B) \quad \textrm{(Independent Events)} \]

\[ \Pr(A \, \textrm{\&} B)= \Pr(A) \times \Pr(B \vert A) \quad \textrm{(Dependent Events)} \]

11.2.5 Definition of Conditional Probability

\[ \Pr(A \vert B)= \frac{\Pr(A \& B)}{\Pr(B)} \]

11.3 Odds

Odds and probabilities are different, but closely related. Probabilities are numbers between 0 and 1. Odds are not numbers, but ratios between two numbers. Odds are defined as the ratio of favorable outcomes to unfavorable outcomes. The top number will always be the same in probabiliteis and odds. The bottom number will be smaller for the odds than for the corresponding probability.

\[ \textrm{If } Pr(X) = m/n, \textrm{ the odds in favor of X} = m \textrm{ to } (n - m) \] The odds against something occurring are the same as the odds for it occurring, except reversed. So, if the odds for A are 2/3, the odds against A are 3/2.

Converting odds to probabilities are just as easy. The top number stays the same, and you add the two numbers together to get the bottom. If odds for S are m to n, probability of S is m/(m + n)

If the odds of OBU’s beating SNU are 1:5, then the probability that OBU wins is 1/6. The proability that SNU wins is 5/6.

Why are odds useful? One reason is that they can be used to easily determine what a fair bet would be. – To make a fair bet a two will come up when you roll a die, you bet 1 dollar you will get a two and your opponent bets 5 dollars you won’t. If you both bet these amounts, over the long run you both tend to break even. Since the probability of rolling a two is 1/6, you should expect to win one out of every six times. On that time that you win, you’ll win five dollars. Of course, you’ll lose the other five times, and will lose a dollar each time. So, in six turns, you should expect to win five dollars and lose five dollars. Notice how this follows naturally from the odds. The odds of rolling a two are 1/5. If the person betting that the event will occur bets the top number and the other person bets the bottom, it’s a fair bet. Gamblers call such a bet an even-up proposition.

Casinos couldn’t turn a profit with even-up bets. So, you will never get a fair bet in a casino. The house essentially takes a percentage of the winnings. They do this by paying the winners less than the actual odds would require. A good example is roulette. A roulette wheel has 38 compartments: 1-36, 0 and 00. You bet the ball will land on number 10. The probability that the ball lands on number 14 is 1/38. So, the odds of the ball landing on 14 are 1/37. If you win, the house should be willing to pay you 37 dollars on a one dollar bet. At these odds, the true odds, you break even. The house, though doesn’t pay according to the true odds, they pay according to house odds. The house odds pretend that the 0 and 00 squares aren’t there. So, as far as the house is concerned, the odds of your winning are 1/35. The pay 35 dollars when, in fairness, the winner deserves 37 dollars. Another way of stating this is that they are taking two dollars of your 37 dollar winnings, a percentage of 2/37 or 5.4%. Now, since they are taking two dollars of your winnings every time, you will lose in the long run.