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Say for every natural number z $\ge$ 1, we choose a number $a_z$ from the set {${1,...,z}$} randomly with equal probability for each number 1 to z. For example, $a_3$ we can choose 1, 2, or 3 all with probability 1/3. If b $\ge$ 2 what is the probability that the sum of $a_1$ + ... + $a_b$ is even for any b.

I know that whenever b is an even number the probability that the number chosen is odd is 1/2. I tried to use the fact that $a_1$ has to be 1 to try to see when I can have the sum of two odds since that is even. However, I am stuck on figuring out the probability that the entire sum is even.

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  • $\begingroup$ Why did you delete all the text in the question? I have rolled it back to your original. $\endgroup$ – Henry Mar 10 at 17:47
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Let's call the sum $S_b = a_1+a_2+\cdots+a_{b-1}+a_b$.

If $b$ is even and $b\ge 2$ then the probability is $\frac12$ that $a_b$ is odd and so changes the parity of the sum, and $\frac12$ that $a_b$ is even and so does not change the parity of the sum, making the probability that $S_b$ is odd or even $\frac 12$ each.

If $b$ is odd then $b-1$ is even. So as long as $b-1\ge 2$, we can now say $S_{b-1}$ is equally likely to be odd or even. Meanwhile $a_b$ is odd with probability $\frac{b+1}{2b}$ and even with probability $\frac{b-1}{2b}$. So the probability $S_b$ is even given $b$ is odd and $b\ge 3$ is $\frac12 \frac{b+1}{2b}+\frac12 \frac{b-1}{2b}= \frac12$ too.

That leaves the cases $b=0$ and $b=1$, which are easy to handle.

  • When $b=0$ then $S_0=0$ and even (the empty sum) so $\mathbb P(S_0 \text{ is even})=1$.
  • When $b=1$ then $S_0=1$ and odd so $\mathbb P(S_1 \text{ is even})=0$.
  • When $b\ge 2$ then $\mathbb P(S_b \text{ is even})=\frac12$.
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