I do understand that you may be disappointed by this solution of the problem, but I’m afraid you will be unable to solve this problem without a deep knowledge of the theory of probabilities.

Let's try to solve this problem in the way it must be solve using theory of probabilities. First, we must set a model. Our stochastic experiment is a series of games (one experiment stops when a series of wins ends with loss or vice versa). Therefore, the elementary events are the streaks of wings or losses of a certain length. Now we are ready to define the set a model of the set of elementary events. The elements of will be streaks of -1 or +1 where for example (11111) denotes a streak of 5 wins and (-1-1-1) – a streak of 3 losses.

Now let’s define a probability. As is discrete, it is enough to define probability only for the elements of . The probability of an elementary event is approximately equal to the number of experiments in which this event happened divided by the number of all experiments. The number of experiments is equal to the total number of streaks that is 720. So we get the following probabilities:

event probability

1 0.286111

11 0.123611

111 0.068056

1111 0.011111

11111 0.005556

111111 0.004167

1111111 0.001389

-1 0.248611

-1-1 0.1125

-1-1-1 0.072222

-1-1-1-1 0.026389

-1-1-1-1-1 0.025

-1-1-1-1-1-1 0.009722

-1-1-1-1-1-1-1 0.005556

Now, what do you want to find? You want to find the probability of a win if it is known that there has already been a streak of one win (we take length of one for example). This is a conditional probability of the event A that the streak is longer then one under the condition B that it is not less then one. The formula to calculate the conditional probability is

P(A|B)=P(A∩B)/P(B)

It is evident that A is a subset of B. So P(A∩B)= P(A)= 0.123611+0.123611+0.068056+0.011111+0.005556+0.004 167+0.001389=0.213889,

P(B)= 0.286111+0.123611+0.123611+0.068056+0.011111+0.005 556+0.004167+0.001389=0,5

P(A|B)= 0.213889/0,5=0,427778.

The probabilities for other length are calculated in the same way.