Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the final number is $15$.

Does the final number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

My thoughts

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

After some step, if we find ourselves in "open territory" (no previously visited hexagons nearby), then there is a sequence of five moves that would end the walk. This is illustrated by this image provided by @Stef in the comments:

(The first red line segment after the blue line segment could have been in any other direction, so we don't count that one.)

The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, but am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much deeper than the expected depth. The numbers in the triangle increase very quickly. Generally speaking, if the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example:

• Does the interior of Pascal's triangle contain three consecutive integers?
• I tried to kill the central binomial coefficients, but they cam back
• Conjecture: In Pascal's triangle with $n$ rows, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$..

Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the final number is $15$.

Does the final number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

My thoughts

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

After some step, if we find ourselves in "open territory" (no previously visited hexagons nearby), then there is a sequence of five moves that would end the walk. This is illustrated by this image provided by @Stef in the comments:

(The first red line segment after the blue line segment could have been in any other direction, so we don't count that one.)

The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, but am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much deeper than the expected depth. The numbers in the triangle increase very quickly. Generally speaking, if the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example:

• Does the interior of Pascal's triangle contain three consecutive integers?
• I tried to kill the central binomial coefficients, but they came back
• Conjecture: In Pascal's triangle with $n$ rows, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$..

Sometimes I am more interested in the methods used to answer the questions, than the answers themselves (but I am still curious what the answer is!). Many of my questions initially seemed almost unapproachable to me, but then turned out to have a rational explanation.

Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the final number is $15$.

Does the final number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

My thoughts

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

After some step, if we find ourselves in "open territory" (no previously visited hexagons nearby), then there is a sequence of five moves that would end the walk. This is illustrated by this image provided by @Stef in the comments:

(The first red line segment after the blue line segment could have been in any other direction, so we don't count that one.)

The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, but am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much deeper than the expected depth. The numbers in the triangle increase very quickly. Generally speaking, if the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example here, here and here.

Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the final number is $15$.

Does the final number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

My thoughts

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

After some step, if we find ourselves in "open territory" (no previously visited hexagons nearby), then there is a sequence of five moves that would end the walk. This is illustrated by this image provided by @Stef in the comments:

(The first red line segment after the blue line segment could have been in any other direction, so we don't count that one.)

The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, but am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much deeper than the expected depth. The numbers in the triangle increase very quickly. Generally speaking, if the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example:

• Does the interior of Pascal's triangle contain three consecutive integers?
• I tried to kill the central binomial coefficients, but they cam back
• Conjecture: In Pascal's triangle with $n$ rows, the proportion of numbers less than the centre number approaches $e^{-1/\pi}$ as $n\to\infty$..

Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the ending number is $15$.

Does the ending number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

My thoughts

After any step, there is usually a sequence of about five moves that would end the walk. For example, if the previous move was down-left, and we find ourselves in "open territory", then the following sequence would end the walk: down-right, right, up-right, up-left, down-left. The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Then using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, not am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much much deeper, where the values of the numbers increase very quickly. Generally speaking, if the values of the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example here, here and here.

Let's take a random walk on Pascal's triangle, starting at the top. Each number is in a regular hexagon. At each step, we can move to any adjacent hexagon with equal probability, but we cannot revisit a hexagon. The walk ends when we cannot move.

Here is an example in which the final number is $15$.

Does the final number have infinite expectation?

(Paradoxically, a random variable whose possible values are all finite, can have infinite expectation.)

My thoughts

I suspect the answer is yes, because as we move down Pascal's triangle, the numbers quickly become very large. But I don't know how this could be proved.

After some step, if we find ourselves in "open territory" (no previously visited hexagons nearby), then there is a sequence of five moves that would end the walk. This is illustrated by this image provided by @Stef in the comments:

(The first red line segment after the blue line segment could have been in any other direction, so we don't count that one.)

The probability of such a sequence would be $\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{5}\right)\left(\frac{1}{4}\right)=\frac{1}{2500}$. Using the expectation of a geometric distribution, the expected number of steps in a walk would be roughly $2500$. I believe, but am not sure, that the expected row number at the end would be on the order of $\sqrt{2500}=50$. (This rough analysis ignores several factors, for example the effect of the boundaries of the triangle.)

But sometimes our walk would go much deeper than the expected depth. The numbers in the triangle increase very quickly. Generally speaking, if the numbers increase faster than the probabilities of reaching them decrease, then the expectation of the final number could be infinite.

Context

This is one of several questions about Pascal's triangle that I've asked recently, for example here, here and here.