And they all tell you, Yes, we have won the lottery.

And here is my answer. You can really read off the formula that I just gave you.

Well, it's going to be the probability that you initially picked wrong.

If min plays 2, so strategy 1 is the dominant strategy, and that should have probability 1.

She then proceeds to the next most probable number.

The motions don't move at all, move right, move down, move down, and move right again.

Another way to think about it is, when you first make your initial pick, there's a one third chance that it's there, and there's a two thirds chance that it's in one of the other two doors.

OK.

So this equals 15 64.

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

And I want you to tell me what the strategy is, whether it's pure or mixed.

And we solve that, and that works out to P 1 4.

Her election is in the bag.

So you actually have a two thirds chance of winning.

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

So five in fifteen that's the same thing as 1 3 that I pick a fair coin.

Probability

Phil, the odds against

So here, there's a 1 3 chance of each of these.

Right? What's the probability that I'd get someone with a temperature of 103?

The same argument going on this side.

Well what's the probability of you picking right the first time?

What are the odds?

You might nail it. You might crash and burn.

So to answer the question, the probability of Marcia being

I'm also going to ask you for the opposite.

In the else clause, the truth and my guess are not the same.

Who would have thought?

For the chance nodes, all the possibilities are equally probably.

So the only thing we need to compute now is P of H given R.

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

Probability of success

Probability of failure

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.

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

The smallness of that probability is what we mean by extremity.

In this case, remember we had a set of random variables X1 up to Xn.

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.

This might be a coin where heads is on both sides.

And I should note that this is just the translation part of the model.

The first coin flip might also come up tails, in which case I pick a very different coin.

I didn't even draw this to scale, this area is actually much smaller when you do it to scale...

Over here Min would choose 0, the minimum of 0 and 10.

But if you remember, my most extreme example was, what's the probability that

We now apply Bayes rule.