Pumpkin carving is a popular activity that is closely associated with Halloween. One of the most common designs that people choose for their carved pumpkin is a witch's face. This classic design features a spooky and witch-like face with exaggerated features and a wicked expression. To create a witch face pattern for pumpkin carving, there are a few key features to keep in mind. First, the eyes should be large and slightly slanted, giving the face an intense and piercing look. Adding wrinkles and bags under the eyes can enhance the witchy appearance.

Magic 8 Ball Problem

I imagine the following kind of sequence occurring: NNNNONNOONOONOOONOOOONONNOONOOONNOONOOOOOONOOONOOOOOOONOOOOOOOONOOOOOOOO.

Now imagine we don't know there are twenty sides to a Magic 8 Ball. Or imagine that you have a Magic 8 Ball with 1000 sides. As the number of "new" trials approaches the actual number of sides, we'll get increasingly more "old" trials showing up in the mix. Once we've seen all of the possible results, we'll always get "old" results. But we're never 100% sure we've seen every single possible result.

So here are the questions I'm interested in:

• As we proceed with trials, can we estimate the total number of "sides" of the Magic 8 ball based on the number of "new" and "old" trials up to this point?
• Can we calculate a probability that our current estimate is correct, or a probability that some bounded estimate is correct?

$\begingroup$ Would you say that without even seeing a result the "magic 8 ball" has an equal chance of having any positive integer number of sides? There is no maximum possible number sides, and no preference amongst them? (I am looking for the a priori probabilities). $\endgroup$

$\begingroup$ These are not the same problems; just to give some idea what sort of approaches are used for such problems: en.wikipedia.org/wiki/German_tank_problem, en.wikipedia.org/wiki/Mark_and_recapture. $\endgroup$

$\begingroup$ @muaddib I would say that it could have any positive integer number of sides, there is no maximum, and no preference among them, yes. Oh, and that all results are equally likely to occur (uniform distribution). $\endgroup$

Adding wrinkles and bags under the eyes can enhance the witchy appearance. The nose should be long and pointed, resembling the classic depiction of a witch's nose. It can be curved or straight, depending on personal preference.

1 Answer 1

This problem is similar to the German tank problem. You can use the same approaches used for that problem, with the number of distinct sides seen and its conditional distribution taking the place of the highest serial number seen and its conditional distribution. (Note that the only information we have that tells us anything about the number of sides of the magic ball is the number of distinct sides we've seen; all the details of which side we saw when, both the ones you recorded in your sequence and the ones you didn't record, are irrelevant, since their conditional distribution given the number of sides seen is the same irrespective of the total number of sides, i.e. they carry no information on the total number of sides.)

I'll work out the Bayesian approach in analogy to the treatment in the Wikipedia article linked to above; I won't go into much detail for the parts that are common to the two calculations.

So let $n$ be the number of sides of the magic ball, $k$ the number of shakes performed so far, and $m$ the number of distinct sides seen. (All the notation is in analogy to the article.) Then the desired credibility $(n\mid k,m)$ is given by

We can take the required conditional probability $(m\mid n,k)$ of seeing $m$ distinct sides on $k$ shakes of a ball with $n$ sides from this answer:

where the $\left\$ are Stirling numbers of the second kind. As in the article, we can eliminate $(m\mid k)$ by marginalizing over all possible $n$:

Now comes the shady part, $(n\mid k)$, your prior credibility that there are $n$ sides after having performed $k$ shakes. In contrast to the tank problem, where observing a tank changes not only the highest serial number seen but also the minimum number of tanks, in our case $(n\mid k)$ doesn't depend on $k$. We can use the same trick of introducing a maximum number $\Omega$ of sides up to which the prior credibility is uniform, and hoping that $\Omega$ will magically drop out of the result. Then $(n\mid k)=1/\Omega$ for $1\le n\le\Omega$. Putting it all together, we have

Now, interestingly, if we want to let the sum in the denominator run to infinity to get rid of $\Omega$, we find that it converges if and only if $m\le k-2$. That is, if you had no prior idea how many sides the magic ball might have and you've never seen the same side twice, then you still have no idea how many sides it might have. That's perhaps to be expected; what's perhaps not quite as expected is that this is still true (though "only" with a logarithmic divergence in the denominator) if you've seen a single reoccurrence of a side. Only with at least two reoccurrences can you overcome your prior ignorance as to the possible scale of the number of sides.

Let's work out some examples. The first convergent case is $m=1$, $k=3$, i.e. you shook three times and got the same side every time. Then

So at this point you're already $6/\pi^2\approx61\%$ sure that the thing is a lemon and only ever shows this one side. More generally, we have

so after seeing the same side four times in a row your inclination to go to the store to get your money back has risen to $1/\zeta(3)\approx83\%$. On the other hand, if you do see $m=2$ different sides in $k=4$ shakes, you have

so now only about $28\%$ are concentrated on the possibility that these are the only two sides.

Now, you'd asked about an estimate of the number of sides. The expected value of $(n\mid m,k)$ is

So it turns out that even after two reoccurences you can't get a finite estimate of the number of sides if you didn't have any prior clue how many there might be – you need three reoccurrences for the sum in the numerator to converge. In the case of always observing a single side, the expected value is

so after seeing the same four sides and nothing else, your estimate of the number of sides is $\zeta(2)/\zeta(3)\approx1.368$.

To give a more realistic example, let's say you shake the ball with $n=20$ sides $k=20$ times. Then the mode of the distribution of the number of distinct sides you see is at $m=13$ (at a probability of about $28\%$). So let's say you observe that most probable value and try to estimate the number of sides. Then

Here's a plot, with a nice mode at $n=20$. The estimate from the expectation value is

which perhaps shows that you might be better off using the mode as the estimator rather than the mean, at least if you don't expect huge numbers of sides.

Another question we might ask is: If we've seen all $m=n$ sides, what number $k$ of trials do we need before we're, say, $99\%$ sure that that's all there is? That is, for which $k$ do we first have

Some trial and error on the borders of Wolfram|Alpha's time limits shows that for $m=n=20$ this requires a full $k=157$ trials. (If the calculation times out, just try again.)

Regarding the approximation considered in the comments: If the term for $n=m$ is dominant, we can compare it to the next term, for $n=m+1$:

We can already see that this is going to get very large for large $k$ (no approximation so far); to see more clearly how large, we can use $\log(1+x)\simeq x$ to get

Thus, for $k\gg m$ this ratio grows exponentially. The third term is again smaller by a similar factor, so we can ignore it and all remaining terms and approximate the credibility for $n=m$ as

So our doubts whether indeed $n=m$ decay exponentially with $k/m$.

Magic 8-Ball

A Magic 8-Ball is, in general, any tool that is used in order to gain advice or try to predict the future. It must be a prop or other physical object, since any characters would fall under Mentors or some other advice dispenser instead. It must be able to accept questions, and respond in turn.

In reality, a magic 8 ball is a classic toy consisting of a hollow plastic sphere containing blue liquid and a 20-sided die with different phrases printed on the sides. You would ask it a question, then flip it over and a random response would come up in the window built into it. As a result, there are various uses of the possible responses involved, as well as of the 8 ball itself. The details and phrases involved can be found on Wikipedia.

Adding a wart or a mole to the nose can add an extra element of realism. The mouth is another important feature of the witch face pattern. It can be wide open in a sinister grin or closed in a tight and menacing frown. Adding fangs or sharp teeth can make the design even scarier. Lastly, the hair is a crucial aspect of the witch face pattern. It should be wild and unruly, flowing in all directions. Long strands or a traditional pointy witch's hat can be added to enhance the effect. When it comes to actually carving the pumpkin, it is important to take your time and be careful. Start by cutting off the top of the pumpkin and cleaning out the insides. Then, transfer the witch face pattern onto the pumpkin by either drawing or using a stencil. Use a small knife or pumpkin carving tools to slowly and carefully carve out the features, being mindful of the thickness of the pumpkin. Once the carving is complete, place a small candle inside the pumpkin and light it to bring the witch face to life. The glow from the candle will highlight the intricacies of the carving and create a spooky ambiance. Overall, creating a witch face pattern for pumpkin carving is a fun and festive way to celebrate Halloween. With attention to detail and careful carving, you can create a unique and impressive design that will impress your friends and neighbors..

Reviews for "Add a Touch of Magic to Your Pumpkin Carving with these Witch Face Patterns"

1. John Doe - 1/5

I was very disappointed with the witch face pattern for pumpkin carving. The design seemed uninspired and lacked creativity. It was just a basic outline of a witch's face, with no extra details or unique features. I was expecting something more intricate and impressive, but this pattern fell flat. Additionally, the instructions were not clear and it was difficult to transfer the design onto the pumpkin. Overall, I would not recommend this pattern, as there are much better options available.

2. Jane Smith - 2/5

I was not impressed with the witch face pattern for pumpkin carving. The design was very simplistic and didn't have the same level of detail and intricacy that other patterns offer. It was just a generic representation of a witch's face, and it didn't stand out in any way. The carving process was also quite challenging, as the pattern didn't have clear guidelines to follow. I ended up with a messy and uneven carving, which was very disappointing. I would suggest looking for a different pattern that offers more creativity and easier instructions.

3. Sarah Johnson - 2/5

I found the witch face pattern for pumpkin carving to be quite underwhelming. The design lacked creativity and appeared to be a basic outline of a witch's face without any unique or interesting elements. Additionally, the instructions provided were not very clear and it was difficult to accurately transfer the pattern onto the pumpkin. The final result was disappointing, as the carving didn't have the same level of detail and depth as I had hoped. I would not recommend this pattern to others, as there are better options available for pumpkin carving designs.

Level Up Your Pumpkin Carving Skills with these Witch Face Designs