You wrote an algorithm that implements what's called Markov localization.

Let me write this.

You might say OK, that's the probably of getting exactly 1 times the probability of getting 2 out of 3 plus the probability of getting 3 out of 3.

Then I just normalize.

So let's write this down.

Or let me write it this way.

And we're going to apply the exact same mechanics as we did before.

Then the value of the state for the action go up would be obtained as follows.

However, if I now divide, that is, I normalize those non normalized probabilities over here by this factor over here,

And it makes much more sense to talk about the probability or random variable equaling a value, or the probability that it is less than or greater than something or the probability that is has some property

OK.

So this equals 15 64.

We are again given 5 grid cells.

In a deterministic action, it obviously succeeds, unless of course we run into a wall.

What is the probability of randomly guessing the correct answer on both problems?

The reason why that is the case is it relates to the 0.9 probability of speaking the truth.

So in order to figure out the probability that I picked a fair coin, given that I got four out of six heads, I have to know the probability of getting four out of six heads given that I picked the fair coin, times the probability of picking out a fair coin, divided by the probability of getting four out of six heads, in general.

Probability

Phil, the odds against

This has nothing to do with total probability and all with Bayes Rules, because I'm talking about observations.

So the probability of guessing on both of them so that means that the probability of being correct on guessing correct on 1 and number 2 is going to be equal to the product of these probabilities.

Let's use the Laplacian smoother with K 1 to calculate the few interesting probabilities

And the answer is 0.45.

How they got there turns out not to be terribly critical in predicting.

That means X equals 1 over 1.5, which is 2 3.

So probability of correct on number 2 is 1 3.

And these are mutually exclusive events, you can't have both of them

Let's talk about inaccurate robot motion.

What are the odds?

So if you get a positive test result you're going to raise the probability of having cancer relative to the prior probability.

So how do I think about that?

And using the prior, we know that P of not C is 0.99.

If I pick a random T value, if I take a random T statistic

There's going to be a 10 percent chance you get a pretty good item.

Now let's have something a little bit more interesting.

Then you wrote a piece of code that used the measurement to turn this prior into a posterior, in which the probability of the 2 red cells was a factor of 3 larger than the posterior of the green cells.

This form is easily transformed into this expression over here, the probability of the message given spam times the prior probability of spam over the normalizer over here.

There is usually actually four of those.

For those of you that know what a Gaussian random variable is or for those of you that know what a normal random variable is, you can also set W equals Rand N, one by three.

So the probability of getting less than minus 5 is exactly 50 percent.

We now apply Bayes rule.

So, this probability is equal to the product over all i of the probability of words of i given all the subsequent words. So that would be from word 1 up to word i 1.

At this point, our robot has localized itself.

(Laughter)

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