# Probability: Expected Value/ .

You have the opportunity to throw a coin. If it ends head up, you get 100 Dollars. If it ends tail up, you get 50 Dollars. If it ends standing on its edge, you have to pay 20 Cents. Would you decline the offer to play since you may loose 20 Cents?

Very often, a numerical value (money, prize) is attached to each outcome of an experiment. This is usually called a random variable, abbreviated as any variable by a letter, say X. Assume all simple events (possible outcomes) of an experiment are A1, A2,... An. Assume the amounts attached to these simple events are X1, X2,... Xn, respectively. How much money would you expect? Probably the average of all these values, provided all outcomes are equally likely, i.e. if p(A1) = p(A2) = ... = p(An) = 1/n. Note that the average in this case equals the sum of the products p(Ai)*Xi of probabilities and amount attached. If the outcomes are not> equally likely, ........ Overall, the expected value for the random variable X is defined by

E(X) = p(A1)*X1 + p(A2)*X2 + ... + p(An)*Xn.

In the "100-500-2400-7000-0" version of the one-round version of "5 Envelopes", all 5 envelopes are equally likely with probability 0.2. Then the expected value for the money is 0.2*100+0.2*500+0.2*2400+0.2*7000+0.2*0 = 0.2*(100+500+2400+7000+0) = 0.2*10000 = 2000, the same as in the "1000-2000-3000-4000-0" version.

You have to draw a card from a deck of 52 cards. You win \$10 if you have chosen an ace, and \$3 for a king. What is the expected value?
We have 52 simple events. To four of them (the aces), \$10 is attached, to four of them (the kings) \$3 is attached, to the remaining 48 of them \$0 is attached. Each of the 52 simple events have the same probability of 1/52. Then E = 48 * 1/52 * 0 + 4 * 1/52 * 3 + 4 * 1/52 * 10 = 52/52 = 1.
In a European roulette, all numbers between 0 and 36 are equally likely. If you bet on "Even", you obtain \$1 if the result is one of the 18 even numbers larger than 0 (!), otherwise, if the result is odd or 0, you have to pay \$1. What is the expected value?
All 37 simple events are equally likely of probability 1/37. 18 of them have the number "1" attached, 19 have "-1" attached. Then the expected value is E = 18 * 1/37 * 1 + 19 * 1/37 * (-1) = -1/37.
Another example is rolling two dice and considering the sum of the two dice as the random variable.
There is only one possibility to get a sum of 2. But there are 5 possibilities to get a sum of 6: 1+5, 2+4, 3+3, 4+2, 5+1. Note that the cases 1+5 and 5+1 are different cases. Assume the first die is red and the second blue. The red die showing a "1" and the blue one showing a "5" is obviously a different outcome to the red one showing a "5" and the blue one a "1". There are 36 total possible outcomes, therefore P(2) = 1/36 and P(6) = 5/36. In the same way P(3)=2/36, P(4)=3/36, P(4)=3/36, P(5)=4/36, P(6)=5/36, P(7)=6/36, P(8)=5/36, P(9)=4/36, P(10)=3/36, P(11)=2/36, and P(12)=1/36. Test these results empirically! Now the expected value for our random variable is 2*1/36+3*2/36+4*3/36+5*4/36+6*5/36+7*6/36+8*5/36+9*4/36+10*3/36+11*2/36+12*1/36=7, maybe not too surprisingly.