Describing the Monty hall paradox following this context would have been amazing. I hope you do it in future. Specially making distinctions between the two premises.
Just what I was thinking as I was watching. The Monty Hall problem isn't just relevant to Let's Make a Deal. It shows up in disguise as the important Principle of Restricted Choice in the card game of bridge (as well as other places). An example of it in action is the following. Say you are missing the queen and king of some suit. If a player wins a trick with, say, the queen, odds are reduced that the same player has the king. Like Monty Hall opening a curtain, the player's choice of which card to reveal (in order to win the trick) may have been restricted. Hence the name.
I encourage you to make a commentary or analysis on the TED-ED frog riddle. It is quite interesting problem that involves sample space and probabilities and is quite confusing to many people (including myself) :D
I saw this paradox differently. The first assumption considers the fact that you've gained x and x is a positive number larger than 0. Therefore, losing half of x implies the difference must be less than what we can gain by doubling x, because dividing x by 2 is always a smaller difference than multiplying x by 2. So, I agree with guy 1...The odds of multiplying x for a double gain vs dividing x for a half gain make the implied odds worth the risk.
The values in the envelopes are fixed (this is crucial!). If you change the envelope, the gain/loss is always equal to the lower of the values in the envelopes. Let me demonstrate this. Let x equal the amount we've gained. Since the values in the envelopes are fixed, we can denote the lower value as y and the higher as 2y. If we change the envelope, there are two different cases: 1. We picked the lower value (x=y) and changing would net us 2y-y=y extra. 2. We picked the higher value (x=2y) and changing would lose us 2y-y=y. Either way by changing we always gain/lose y. And assuming the choice in the beginning was random (50-50) the expected value of gain/loss by changing is 0.5*y + 0.5*(-y) = 0.
One nice thing about this paradox is that you can test your ideas about it experimentally. If you know how to program, it's a pretty easy one to write. If not, just ask a friend to randomly put some value x in one envelope (or hand), 2x in another, and play the game. Do it over and over and see how you do with a strategy of always switching.
Your solution is more insightful than most, but in my opinion falls a little short. Part if that is that I would eplain the point you made a little differently. It all boils down to the fact that the amount of money in the envelopes is a random variable, but it is only _one_ independent random variable. You can use two, but then the second is dependent on the first. So, as you pick an envelope, you should use one random variable, T, to represent the _total_ amount in the two envelopes. The set of values t that T can take is part of the sample space.The problem is that we do not know what that space is, or what its probability distribution is. This turns out to not matter, since there is another random variable, R, for whether the envelope you pick is relatively the low value, or the high value. The set of values r that R can take is {1/3,2/3}, with probability distribution (50%,50%). Finally, the amount in your envelope is r*t. The expected value, with respect to R is (1/3)*t*(1/2)+(2/3)*t*(1/2) = t/2. It is the same as the expected value of the other envelope. We don't know what the probability distribution of T is, because t is used as a single value in this calculation. But that changes if we learn x, the value in our envelope. Then either t=3x, or t=3x/2. And now we need to include these two values of T in our expectation. For that, we need to know, at least, the relative probabilities of T=3 and T=3x/2. This is something we have no way of knowing. One conclusion where you are wrong, is around 3:40. Hobbes is not right that your chances are the same whether or not you switch. They are actually unknowable; much like asking how likely it is that a random child's favorite color is more likely to be green or blue. You might have a good guess, based on what colors you think a child might like, but you don't know. And in fact, there are distributions where it is (almost) always favorable to switch, _after_ you learn what is in your envelope. All you need is a probability distribution where the probability distribution increases as t increases. There are two problems with that: 1) To be true for all cases, there is no upper bound for the amount of money that could potentially be put in the envelopes; otherwise.... 2) If your envelope has 2*t_max/3, you will lose money by switching. A lot of money. Enough to make the expectation, if you go into the game already knowing you will switch, to be that there is no gain.
