Basic Probability, part 6

August 24, 2009

The another type of presentation the probability is a table with occasions and their frequencies. For example, we have been watching basketball game and the team is stable with 5 players, and we calculated that players scores in a row are following:
table1

If we are watching a random scene of the games and our team scored. The probability that this score has done by player 3 is P(E)=\frac{number~ of~all~scores~by~ player~3}{number~of ~total ~scores}=\frac{300}{100}=\frac{3}{10}.

Expectation.

Consider an example, there is a one-armed bandit machine in a casino and it works on following idea: on every 400 attempts to win machine gives to win 2 times with $25 and 1 time with $45. If each attempt costs $1, what we can expect as average winning in long run?

To answer on this question we need to know the notion “expectation”. Expectation or expected value is a long run average and is calculated as:

E(X)=X_1 *P(X_1)+X_2*P(X_2)+...+X_n*P(X_n),
where X_1 is the first occasion in numbers and P(X_1) is the probability for the first occasion, and so on for the rest occasions.

Now let us look at our example in a casino. Denote the first occasion as lose, so X_1 = -1 doll. and P(X1)=\frac{397}{400}, the second occasion is win $25, so X_2 =25 doll. and P(X1)=\frac{2}{400}, the third occasion is win $45, so X_3 =45 doll. and P(X1)=\frac{1}{400}. Put all information together in the table:
table2

Calculate expected value or expectation:
E(X)=-1*\frac{397}{400}+25*\frac{2}{400}+45*\frac{1}{400}=-\frac{302}{400}
Round the value to -\frac{3}{4}.
And what does it mean? It means that if we are going to play with the machine for a long time every arm dropped will cost to us $\frac{3}{4}. It is the average value that we loose.

Pretend the following case.
We has sat all the night in a casino and played in one-armed bandit. We dropped the arm 400 times, on 25th and 142nd attempt we won $25, on 177th drop we won $45, the rest attempts were lose.

You can see that sometimes we lose, sometimes we won, however it looks like we came to the machine and gave $ \frac{3}{4} for every attempt without losing $1 or winning $25 and $45.

Now let us see the view from the casino owners. When one is going to play in our casino in one-armed bandit and saying “I am going to win”, he(she) is wrong. We are going to win! And at an average on every drop we win $ \frac{3}{4} or approximately 66 cents.

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