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Unformatted text preview: the probability that the nth ﬂip was tails; this is because if we had an even
number of heads on day n − 1, we must obtain tails on day n to maintain an even number of heads. By
¯
the same reasoning, Pr[En En−1 ] is the probability that the nth ﬂip was heads. Plugging in the values
and the inductive hypothesis we obtain:
1 (1 − 2 p)n−1
1 (1 − 2 p)n−1
Pr[En ] = (1 − p)( +
) + p(1 − ( +
)).
2
2
2
2
After simpliﬁcation, we ﬁnd that:
Pr[En ] = 1 (1 − 2 p)n
+
.
2
2 Here is another way to think about this problem. To get an even number of heads in n tosses, you
must have either an even number of heads in n − 1 tosses (event En−1 ) followed by a tails in the nth
¯
toss (event T ) or an odd number of heads in n − 1 tosses (event En−1 ) followed by a heads in the nth
toss (event H ). Note that these two possibilities are disjoint, so the total probability is the sum of the
probabilities of each of the two. It follows that:
¯
Pr[En ] = Pr[En−1 ] Pr[T ] + Pr[En−1 ] Pr[H ]
Here we used the fact that the outcome of the nth coin toss is independent of the outcomes of the ﬁrst
n − 1 tosses. We apply the induction hypothesis as above to get:
1 (1 − 2 p)n−1
1 (1 − 2 p)n−1
1 (1 − 2 p)n
Pr[En ] = (1 − p)( +
) + p(1 − ( +
)) = Pr[En ] = +
.
2
2
2
2
2
2 CS 70, Fall...
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This document was uploaded on 01/28/2014.
 Winter '09

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