If a physicist were asked this question he might ask for more information, such as distance thrown, speed, spin, etc, and then attempt to calculate the result. Scientists believe that all events have a cause, and, in principle, we can determine the effect if we know all the causes and how they behave, even though, in practice, it might be too complicated to make the calculation. The point being that tossing a coin is not a causeless event -- we might not be able to determine the cause, but the event is nevertheless caused.

In practice and in probability we assume that we cannot predetermine how a fair coin will land, and we consider the tossing of the coin as a random experiment for which we cannot predetermine the outcome.

The result is uncertain.

Suppose then you ask me what outcomes are possible in tossing a coin. I might answer "Either heads or tails". That is, there are two possible outcomes, but only one of them can occur at a time. The set of all the possible outcomes is called a sample space. So the sample space for tossing a coin has two members, heads and tails.

Similarly, if we require the probability that a head will appear on the first coin, we count the number of events where this is so (H,H) and (H,T), and so the probability of a head on the first coin is 2/4, or ½.

If we rolled two dice, there are 6·6, or 36 possible events. The sample space is:

First die | ||||||
---|---|---|---|---|---|---|

Second Die | 1 | 2 | 3 | 4 | 5 | 6 |

1 | 1,1 | 1,2 | 1,3 | 1,4 | 1,5 | 1,6 |

2 | 2,1 | 2,2 | 2,3 | 2,4 | 2,5 | 2,6 |

3 | 3,1 | 3,2 | 3,3 | 3,4 | 3,5 | 3,6 |

4 | 4,1 | 4,2 | 4,3 | 4,4 | 4,5 | 4,6 |

5 | 5,1 | 5,2 | 5,3 | 5,4 | 5,5 | 5,6 |

6 | 6,1 | 6,2 | 6,3 | 6,4 | 6,5 | 6,6 |

If we required to know the probability of the spots summing to more than 7, we could add up all the scores and count those which exceeded 7. There are 15 such combinations of events that produce a number greater than 7. The required probability is therefore 15/36, or 5/12

When the sample space is unknown, or extremely large, we have to resort to caculation. However, when we can view a sample space, it helps us visualise the problem more clearly, and solve problems more directly (by simply counting).

The probability (P) of an outcomes or event is the ratio of the number of times a favourable outcome (F) can occur divided by the total number (N) of possible outcomes.

So the probability of heads when tossing a fair coin is 1/2, because heads can occur once and the total number of possible occurrences is 2 (heads and tails). The probability for tails is also 1/2.

If we toss a fair coin twice, we have the following possible outcomes, or events: {(H,H), (H,T),(T,H), (T,T). The total number of possible outcomes is therefore 4 and the number of outcomes where the result is two heads is 1. The probability of getting two heads in tossing a fair coin twice is therefore 1/4.

Probability, p, is a number such that 0≤p≤1 , or 0%≤p≤100%.

When, for instance, tossing a coin n times, the **average**
number of heads tends to 1/2 as the number of tosses, n, increase (to infinity).

The number of heads does not approach n/2. In fact, it becomes more and more
distinct from n/2. It is the **average** that approaches 1/2.

For example, by simulating tossing a coin 10 times, in 4 trials, we obtain the following (the next simulation may be quite different):

Trial |
Heads |
Tails |
Heads-5 |

1 | 6 | 4 | 1 |

2 | 5 | 5 | 0 |

3 | 7 | 3 | 2 |

4 | 3 | 7 | -2 |

The average number of heads is 5.25, and the maximum difference observed is -2 (tails were more frequent on this occasion). The probability of heads is 0.525.

When tossing a coin (in simulation) a million times, in 4 trials, we obtain (another 4 trials may be quite different):

Trial |
Heads |
Tails |
Heads-500000 |

1 | 499912 | 500088 | -88 |

2 | 500352 | 499648 | 352 |

3 | 499379 | 500621 | -621 |

4 | 499774 | 500226 | -226 |

The average number of heads is 499854.25, and the maximum difference observed
is -621. The probability of heads is 0.4999 (4 decimals).

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