## Monday, September 8, 2014

### Unizor - Probability - Random Variables - Expectation Examples

Example 1
What is the expected value of a random variable ξ that equals to a sum of two numbers obtained by rolling two dice?
We have 36 combinations of numbers as a result of rolling two dice. Each pair has the same chances to occur as any other, therefore, the probability of each pair equals to 1/36.
The sum of two numbers takes values from 2 to 12 in the following manner:
2:1,1
3:1,2/2,1
4:1,3/2,2/3,1
5:1,4/2,3/3,2/4,1
6:1,5/2,4/3,3/4,2/5,1
7:1,6/2,5/3,4/4,3/5,2/6,1
8:2,6/3,5/4,4/5,3/6,2
9:3,6/4,5/5,4/6,3
10:4,6/5,5/6,4
11:5,6/6,5
12:6,6
Since each pair has a probability of 1/36 and, as we know, the probability is an additive measure, the probability of a combination of N pairs equals to N/36. Hence, the probabilities of different values for a sum of numbers on two dice equal to
P(ξ=2)=1/36
P(ξ=3)=2/36
P(ξ=4)=3/36
P(ξ=5)=4/36
P(ξ=6)=5/36
P(ξ=7)=6/36
P(ξ=8)=5/36
P(ξ=9)=4/36
P(ξ=10)=3/36
P(ξ=11)=2/36
P(ξ=12)=1/36
Using the formula for expected value
E(ξ) = p1·x1+p2·x2+...+pN·xN
We can calculate the expected value in this particular case as
E(ξ) = (1/36)·2 + (2/36)·3 +
+ (3/36)·4 + (4/36)·5 +
+ (5/36)·6 + (6/36)·7 +
+ (5/36)·8 + (4/36)·9 +
+ (3/36)·10 + (2/36)·11 +
+ (1/36)·12 =
= (2+6+12+20+30+42+
+40+36+30+22+12)/36 =
= 252/36 = 7
So, the expected value of a sum of two numbers obtained by rolling two dice equals to 7.
It means that, if we repeat our experiment with rolling of two dice a very large number of times, the average sum of these two numbers per experiment would be very close to 7.

Example 2
What is the average value of a card in the game of Blackjack (assume for simplicity that Aces are always valued at 11 points)?
The standard deck of cards contains 52 cards valued by numbers on the cards (from 2 to 10) or by 10 for pictures (Jack, Queen and King), or by 11 (Ace) with four suits per each. Therefore, we have
4 cards valued at 2 points
4 cards valued at 3 points
4 cards valued at 4 points
4 cards valued at 5 points
4 cards valued at 6 points
4 cards valued at 7 points
4 cards valued at 8 points
4 cards valued at 9 points
16 cards (10,J,Q,K) at 10 points
4 cards valued at 11 points
Since the probability to get each card is 1/52, the expected value of any card pulled randomly from a deck is
(4/52)·2+(4/52)·3+(4/52)·4+(4/52)·5+(4/52)·6+(4/52)·7+
+(4/52)·8+(4/52)·9+(16/52)·10+(4/52)·11 =
= (8+12+16+20+24+28+32+36+160+44)/52 = 380/52 ≅ 7.3
Of course, the situation above is a simplification of the real case. The rule that Ace can be counted as 1 or 11, as you wish, complicates the real picture.
Another complication is that usually a few cards have already been given to other players or yourself, so the probabilities must be based not on a full deck, but on whatever is left after the first few cards were distributed and depends on these cards.
However, Blackjack is usually played in a casino with more than one deck (like 5 or 6), which decreases the influence of previously dealt cards on the probabilities.

In conclusion, let's notice that the expected value of a random variable is not necessarily one of its real values. For instance, in Example 1 the expected value of 7 is the real value for two dice in positions 1+6, 2+5, 3+4, 4+3, 5+2 and 6+1. However, in Example 2 the expected value of 380/52≅7.3 cannot be a real value of the card.