multi armed Bandit Problem

Question

I am studying machine learning, I remember what are distributions, mean, median mode, from my university statistics studies, but the author, says that given five slot machines with these distributions, the number 5 is good, as it is left skewed and has as consequence favorable mean, mode and median. Also he states that also the 4th should be good. Of course for good he means the probabilities I would have to win money from them. Why 5 distribution is best and why the 4th is also acceptable.

With common sense I would say that bigger is the area "expecially on top" better the distribution is. Which basic concept I am missing?

EDIT: the author of the course says:

"Each one of the machines has a distribution out of which the machine picks results, randoms pulling the arm of the machine. The one on the right is the best machine because has the best outcome as the highest mean, median and mode."

In the comments is written: Then in the comments of the course I tooks people say: the reason you are chosing the orange(yellowish) one is these graphs are Probability Distributions and therefore by definition they range between 0 and 1

And the reason we want ORANGE is because the MEAN of this probability distribution is LEANING to the RIGHT and hence a higher number in the range of 0 to 1

here 0 is 0% and 1 is 100% in terms of probability distributions.

And the tutor said, yes you are right.

Answer 1

In fact the question 'which distribution is the best' (i.e. gives me the most money) strongly depends on your goal. Let us throughout assume that you have 'enough' pulls at hand, so you can pull as often as you want. Let us first assume that your goal is to make as much money as possible in average and we just compare the following two distributions:

Would you rather pull the left one or the right one? Well, it does not make a difference because if your goal is to maximize the average amount of money you get in every pull then you just get to the equal mean(s) of both distributions.

Well this is weird, the right distribution is clearly 'better' because it has a high mean and it has a low variance, i.e. you have more certainty that you get a high amount of money. How can that distribution be equally good as the one that has a high mean by 'coincidence' and much uncertainty whether or not you will get that amount of money...

Now let us change the goal to 'we want to have a high probability that we get at least 10 money'. Then we have a higher chance to reach our goal with the right distribution (because the red area of failure is smaller in the right one than in the left one):

So in short: There are at least two parameters: How high is the mean and how 'spiky' is the distribution towards the right. How much you can interchange one property for the other depends on the goal that you want to achieve. Examples we have seen: We want to optimize the mean --> the spikyness is completely unimportant, just the mean is important. If we make the goal so that it includes the uncertainty somehow then the second property becomes important.