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down in a military mission. It seems quite natural to conclude that in 50 missions
we can guess that he is almost sure to die. But this is completely wrong. In fact, let
us apply the rules above, then the probability of surviving is 0
98
50
, assuming that
each mission is independent from the others. Therefore the probability of dying is
1
.
98
50
−
.
=
.
64, which is markedly different from 1. The same argument shows
that the probability of obtaining one head, by tossing a coin twice, is not 1.
This kind of error was in the problem posed by the
Chevalier de Mere
to Pascal,
about a popular dice game:
Why the probability of one ace, rolling one die 4 times,
is greater than that of both aces, rolling two dice 24 times?
In fact, one could led
to reason in the following way. The probability of one ace is 1
0
0
/
6, and 4
/
6
=
2
/
3,
analogously the probability of 2 aces is 1
3, by concluding
that in principle the two events: “1 ace in 4 rolls”, and “2 aces in 24 double rolls”
are equiprobable. But the empirical evidence reported by Chevalier de Mer´ewas
against this wrong conclusion. Namely, Pascal (interacting with Fermat on this puz-
zle) solved the apparent paradox by the following computation. In the first game,
P(no-ace-in-4-tolls) =
/
36, and 24
/
36
=
2
/
4
, therefore P(ace-in-4-rolls)= 1
4
(
5
/
6
)
−
(
5
/
6
)
=
0
.
5177. In
24
, therefore P(2-aces-
the second game, P(no-2-aces-in-24-double-rolls)=
(
35
/
36
)
24
in-24-double-rolls) = 1
4914. A detailed analysis shows that the
simple mistake in the argument suggesting the same probability for the two events
is due to the violation of rule 5 given above.
Sometimes difficulties arise in choosing the right event space. The problem which
was posed to Galileo by the archduke of Tuscany was the following:
Why rolling
three dice the value 10 is more frequent than 9?
. His question was motivated by the
fact that there are 6 ways for decomposing 10 into three parts: (6+3+1), (6+2+2),
((5+4+1), (5+3+2), (4+4+2), (4+3+3), but even 6 ways of decomposing 9: (6+2+1),
(5+3+1), ((5+2+2), (4+4+1), (4+3+2), (3+3+3). The answer Galileo found (in mod-
ern language) is that the space of events is not given by the partitions of integers in
three parts, but by the possible outcomes of the three dice, and it is easy to realize
that there are 27 ways of obtaining 10 with three dice, while the 25 ways of obtain-
ing 9 with three tolls (for example (3+3+3) is realized in only one way, but (6+3+1)
can be realized in a number of ways corresponding to the 6 permutations of three
objects.
Another interesting case of confusion is related with conditional probability.
Bayes' theorem, in a simplified form, asserts the following equation:
−
(
35
/
36
)
=
0
.
P
(
A
|
B
)=
P
(
A
)
P
(
B
|
A
)
/
P
(
B
)
.
Its proof is very simple. In fact, by the definition of conditional probability we have:
P
(
A
|
B
)=
P
(
A
∧
B
)
/
P
(
B
)
but
P
(
A
∧
B
)=
P
(
B
∧
A
)=
P
(
B
|
A
)
P
(
A
)
