Converse (complementary) probabilities
Converse - "Something that has been reversed; an opposite."
Often when you work out the probability of an event, you sometimes do not need to work out the probability of an event occurring you need the opposite. The probability that the event will not occur. For example, The probability of throwing a 1 on a die is 1/6 therefore the probability of a 'non-1' is (1-1/6) which equals 5/6.
Converse probabilities are used to work out such problems such as, "What is the probability of exactly one soccer match ending in a draw within a group of three separate matches?"
Let us assume the chance of a draw occurring in any match is 1/3 or 33.33%. To fulfill our target of only one match ending in a draw we would require the other matches to not end in a draw or (1-(1/3)) which equals 2/3 or 66.66%.
Therefore the probability of only one match out of three being drawn is 1/3x2/3x2/3 which equals = 4/27 or (.33*.67*.67) = 14.81%
In our group of three matches there are three ways for only one match to draw, DXX, XDX, XXD, therefore we need to add together all the probabilities, three in this case.
The final answer to the probability of one match drawing is (4/27)+(4/27)+(4/27) = 4/9 or (=.1481+.1481+.1481) = 44.44%.
Converse probabilities are used to work out the infamous birthday problem. Many people find the answer puzzling but it can be proved by either asking your personal manger for birthday dates or flicking through a the who’s who in your reference library.
The question is:-
"How many people should be gathered in a room together before it is more likely than not that two of them share the same birthday?"
Ignoring the issues of leap years the problem is solved as follows:-
When the first person enters the room and announces their birthday, the probability of the second person sharing the same birthday is 1/365. Conversely, the probability of the second birthday being different is the opposite of the first calculation, 364/365. When two birthdays are known, the probability of the third being different is 363/365, as there are now two 'favourable' outcomes among 365. The compound probability of birthday 2 being different from birthday 1, and of birthday 3 being different from the other two, these being independent outcomes, is:-
(364/365)*(363/365) = 0.991796 or 99.2% chance that two people will not share the same birthday.
Note the start of the sequence is (365/365). We have removed this as it does not affect the result of the calculation.
All that is necessary now is
to continue adding terms to the fraction until it equals less than
1/2 or 50%, since as soon as the probability is less than 1/2 that
all birthdays are different, the probability is clearly more than
1/2 that any two are the same. In other words it is more likely than
not that two people in the room share the same birthday. The
following chart shows the number of the people in the room and the
probability that they DO NOT share the same birthday.
|
People |
Chance % |
|
2 |
99.7 |
|
3 |
99.2 |
|
4 |
98.4 |
|
5 |
97.3 |
|
6 |
96.0 |
|
7 |
94.4 |
|
8 |
92.6 |
|
9 |
90.5 |
|
10 |
88.3 |
|
11 |
85.9 |
|
12 |
83.3 |
|
13 |
80.6 |
|
14 |
77.7 |
|
15 |
74.7 |
|
16 |
71.6 |
|
17 |
68.5 |
|
18 |
65.3 |
|
19 |
62.1 |
|
20 |
58.9 |
|
21 |
55.6 |
|
22 |
52.4 |
|
23 |
49.3 |
|
24 |
46.2 |
|
50 |
3.0 |
|
100 |
3,254,690 to 1 on |
The fraction drops to less than 1/2 with 23 iterations, so it is
more likely than not that in any gathering of 23 or more persons,
two of them will share a birthday.
Only 50 people need be present for the 'coincidence' of two of them having the same birthday to become, roughly, a 30-1 on chance.
In a company of 100 employees the odds are more than three million to one on that two share a birthday.
The birthdays proposition is one where a gambler who can estimate probabilities can make money from unsuspecting punters.