#### Counting Successes and Failures

So what is the probability of winning from ten tickets? Let's start with the marginal cases: not a single win and not a single failure.

Each ticket is independent of the previous one (at least if there is large number of tickets, otherwise one must strive for the so-called hypergeometric distribution). The chance of winning at e.g. ticket number ten is thus exactly the same as for tickets one to nine. The probabilities simply multiply:

- P (10 wins) = 10 percent x 10 percent x ... x 10 percent = 0.00000001 percent [we have a factor of 0.1 ten times]

- P (10 failures) = 90 percent x 90 percent x ... x 90 percent = 34.86 percent [we have a factor of 0.9 ten times]

What is the chance of (exactly) one winning ticket? To answer this, we need ten factors again. One of them is a 10 percent probability of winning, the remaining nine are the failure probability of 90 percent:

10 percent x 90 percent x ... x 90 percent = 3.87 percent

This is not yet the probability you are looking for. It is now necessary to take into account that there are ten positions on which the winning ticket appears. Therefore:

P (exactly one winning ticket) = 10 x 3.87 percent = 38.7 percent

In order to calculate, for example, »exactly two winning tickets« some combinatorics is necessary. You need to know how many ways you can position two winning tickets in ten tickets. This is then done by the binomial distribution for one. Fortunately, we can take a shortcut here. If we are not interested in exactly one, but at least one profit, this is exactly the opposite of »no profit«:

P (at least one profit) = 100 percent - P (no profit) = 100 percent - 34.86 percent = 65.14 percent

Let us sum up briefly. Only profits almost never happen. In about 35 percent of cases we only have failures, and in 39 percent of cases exactly one winning ticket. In the remaining 26 percent of cases, there are two, three or more wins. At least one win is made in 65.14 percent of cases.

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