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Hello,

I would like to ask, if I am right with my computation.

Let´s have a set of integers from 0 to 12. We start at 0 and we can go to 1 with the probability 1. From 1 we can go to 1 or back to 0, both with probability 0.5. When we start at zero, how many steps (exp. number of them) do we need to go until 12?

My way of solving:

We start at 0 ... there is only one possibity - to go to the 1 with probability 1

From 1, we can go to the 0 (P=$\frac{1}{2}$) or to the 2 (P=$\frac{1}{2}$)

From 2, we can go back to the 1(P=$\frac{1}{2}$), or forwards to the 3(P=$\frac{1}{2}$) ... but when we start at 0, there is the probability to reach 3(P=${\frac{1}{2}}^{2}$).

etc.

So, I created a binomial distribution:

From 0 to 1: 1 step; prob. 1 ... 1*1

From 0 to 2: 2 steps; prob. ${12 \choose 1}*\frac{1}{2}^{1}*{\frac{1}{2}}^{11}$

From 0 to 3: 3 steps; prob. ${12 \choose 2}*{\frac{1}{2}}^{2}*{\frac{1}{2}}^{10}$

...

From 0 to 11: 11 steps; prob. ${12 \choose 10}*{\frac{1}{2}}^{10}*{\frac{1}{2}}^{2}$

From 0 to 12: 12 steps; prob. ${12 \choose 11}*{\frac{1}{2}}^{11}*{\frac{1}{2}}^{1}$

And now, to get the whole expected value, I need to sum all the expected values:

$1*1+12*2(steps)*\frac{1}{2048}+66*3\frac{1}{2048}+220*4\frac{1}{2048}+495*5\frac{1}{2048}+792*6*\frac{1}{2048}+924*7*\frac{1}{2048}+792*8*\frac{1}{2048}+495*9*\frac{1}{2048}+220*10*\frac{1}{2048}+66*11*\frac{1}{2048}+12*12*\frac{1}{2048}$.

The sum is 15353/1024 = 14.9932 steps. I think it is too low number. Can I ask, where I am not doing right?

Thank you a lot.

Have a nice day.

I would like to ask, if I am right with my computation.

Let´s have a set of integers from 0 to 12. We start at 0 and we can go to 1 with the probability 1. From 1 we can go to 1 or back to 0, both with probability 0.5. When we start at zero, how many steps (exp. number of them) do we need to go until 12?

My way of solving:

We start at 0 ... there is only one possibity - to go to the 1 with probability 1

From 1, we can go to the 0 (P=$\frac{1}{2}$) or to the 2 (P=$\frac{1}{2}$)

From 2, we can go back to the 1(P=$\frac{1}{2}$), or forwards to the 3(P=$\frac{1}{2}$) ... but when we start at 0, there is the probability to reach 3(P=${\frac{1}{2}}^{2}$).

etc.

So, I created a binomial distribution:

From 0 to 1: 1 step; prob. 1 ... 1*1

From 0 to 2: 2 steps; prob. ${12 \choose 1}*\frac{1}{2}^{1}*{\frac{1}{2}}^{11}$

From 0 to 3: 3 steps; prob. ${12 \choose 2}*{\frac{1}{2}}^{2}*{\frac{1}{2}}^{10}$

...

From 0 to 11: 11 steps; prob. ${12 \choose 10}*{\frac{1}{2}}^{10}*{\frac{1}{2}}^{2}$

From 0 to 12: 12 steps; prob. ${12 \choose 11}*{\frac{1}{2}}^{11}*{\frac{1}{2}}^{1}$

And now, to get the whole expected value, I need to sum all the expected values:

$1*1+12*2(steps)*\frac{1}{2048}+66*3\frac{1}{2048}+220*4\frac{1}{2048}+495*5\frac{1}{2048}+792*6*\frac{1}{2048}+924*7*\frac{1}{2048}+792*8*\frac{1}{2048}+495*9*\frac{1}{2048}+220*10*\frac{1}{2048}+66*11*\frac{1}{2048}+12*12*\frac{1}{2048}$.

The sum is 15353/1024 = 14.9932 steps. I think it is too low number. Can I ask, where I am not doing right?

Thank you a lot.

Have a nice day.

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