Glad it was helpful!

Glad it was helpful!

Yes, exactly!

Of course!

that's a valid question. I think of it this way: Before sending the robot out, we have some prior probability of where the treasure will be. Then, we send the robot out and observe some additional evidence (the beep or no-beep). Using that additional evidence, we make a decision about where to dig. So, different states for the evidence (beep vs no-beep) can differently affect our conclusion.

It came to a different conclusion from the two different robot events because the robot was not always right when it beeped or not beeped. So, you have a decision point. If the robot comes back with a beep, then you evaluate the probabilities for that event. If the robot comes back with no beep then you evaluate the probabilities for that event. It would appear that chances are greater for finding treasure if the robot does not beep for a particular square. Which is what makes it weird because the purpose of the robot is to help find treasure but here lies the variability in its measurement capabilities. Sometimes the idea is good but the equipment fails to measure at a useful accuracy or precision necessary to be useful.

So the logic would be that if you send the robot out and it comes back and does not beep, the most probable place for the treasure is square 1. However if you send it out and it comes back and beeps, the most probable place is square 0.

@@ritvikmath but not beeping = treasure unlikely, right? So if we’re using “no beep” in the posterior probability calculation, shouldn’t the highest posterior probability in this case be the last place you’d want to look for treasure? Wouldn’t you want to look at the lowest posterior probability when you’re using the “no beep” probability in the calculation?

@@ohokaythen The posterior probability here calculates the probability of finding the treasure given the robot beeps or does not beep. Hence in both cases its actually calculating the same probability of finding the treasure based on two different given conditions i.e. robot beeping and not beeping.

You're welcome!

The 'get_robot_beep' provides a sample of a binary law based on the robot skills to find the treasure on this specific island (robot_likelihoods). This robot skills depend on the true location of the treasure (loc). This is the only time where the true location is used, in order to simulate experimental data. def get_robot_beep(loc, robot_likelihoods): i, j = loc//2, loc%2 # Cartesian conversion u = np.random.random() # Sample from the uniform law U([0, 1]) return u < robot_likelihoods[j, i] # Binary output: beep or no_beep It is important to note that there is a treasure, and a single one. And the robot can provide a single behaviour for the entire island: "beep" or "no beep". And this is a "random event" (not uniform, but random)

Thanks