## Saturday, February 7, 2015

### Unizor - Probability - Geometric Distribution - Problems 1

Problem A

Assume that the probability of winning on a particular number in roulette equals to P=1/38. The bet is 1 bitcoin. You have exactly 10 bitcoins.

What is the probability of losing 9 times in a row and winning on the last bitcoin left?

Solution

The probability of winning is P=1/38 and the probability of losing is 1−P=37/38.

All games are independent, so the probability of their combination equals to a product of their probabilities.

Therefore, a sequence of 9 losses and 1 winning has a probability of

[(1−P)^9]·P = [(37/38)^9]·(1/38) ≅ 0.02

Problem B

Let's expand the previous problem and assume that winning in the 1st or the 2nd or any other game up to 10th are also what we consider as an ultimate success because we end up with money, while losing the ten games in a row would bankrupt us and should be considered as an ultimate failure.

What is the probability of ultimate success?

Solution 1

The easy solution is to calculate the probability of a failure, that is the probability of 10 losses in a row.

This is equal to

(37/38)^10 ≅ 0.7659

Therefore, the probability of success is

1 − (37/38)^10 ≅ 0.2341

Solution 2

On the other hand, we can calculate 10 different probabilities of winning in the 1st game, in the 2nd game etc. up to 10th game and add them up.

This would be a sum of a geometric progression with the first member a=1/38 and the factor d=37/38.

The sum of the 10 members of this geometric progression equals to

S = 1/38+37/(38^2)+...+(37^8)/(38^9)+(37^9)/(38^10)

To summarize this geometric progression without using a formula, which only few people remember, let's multiply this sum by a factor:

S·(37/38) = 37/(38^2)+...+(37^9)/(38^10)+(37^10)/(38^11) =

= S − (1/38) + (37^10)/(38^11)

Now we can consider this as an equation for an unknown S that can be easily solved:

S·[1-(37/38)] =

= (1/38)·[1-(37/38)^10]

Hence,

S/38 = (1/38)·[1-(37/38)10]

Therefore,

S = 1-(37/38)^10 ≅ 0.2341

The same exactly answer as in the Solution 1

.

Problem C

Assume you are playing the following game.

Two dice are rolled and the resulting numbers are added together. If the sum is N, you win. Otherwise, you lose.

What should N be if you would like the average number of experiments until the first winning to be equal to 6?

Solution

Since we are talking about the number of experiments until the first success, this is an example of a geometric distribution.

The probability of success p is the probability of a sum of two numbers rolled on two dice to be N.

Obviously, probability p is a function of the number N. If, for example, N is less than 2 or greater than 12, the probability of success equals to 0 since any experiment results in the sum of two numbers to be between 2 and 12.

The average number of experiments until the first success for a geometrically distributed random variable with the probability of success p equals to 1/p, as we know from the lecture on properties of geometric distribution, and is given in this problem as being equal to 6.

Therefore, we can determine the probability of success. If 1/p=6, p must be equal to 1/6.

All we have to do is to determine N from a statement that the probability of a sum of two numbers to be equal to N is 1/6.

This is a combinatorial problem.

Let's research the number of combinations that lead to each value of a sum of two numbers on rolled dice.

N=2: 1+1 (1)

N=3: 1+2;2+1 (2)

N=4: 1+3;2+2;3+1 (3)

N=5: 1+4;2+3;3+2;4+1 (4)

N=6: 1+5;2+4;3+3;4+2;5+1 (5)

N=7: 1+6;2+5;3+4;4+3;5+2;6+1 (6)

N=8: 2+6;3+5;4+4;5+3;6+2 (5)

N=9: 3+6;4+5;5+4;6+3 (4)

N=10: 4+6;5+5;6+4 (3)

N=11: 5+6;6+5 (2)

N=12: 6+6 (1)

The total number of combinations of numbers on two dice is, obviously, 6·6=36. The only value of the sum of two numbers that gives the probability 1/6 is N=7 with 6 different combinations of two numbers since 6/36=1/6. All other values of N are rarer than this and, therefore, their probabilities are less than 1/6. Therefore, the solution of this problem is N=7.

If, for example, the problem stated that the average number of experiments until the first success must be equal to 9, the solution would be either N=5 with 4 combinations or N=9 with 4 combinations because 4/36=1/9.

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