Quote:
Originally Posted by Andrew WCY
The Infinite Money game in the video can be viewed as a discrete probability distribution. From https://en.wikipedia.org/wiki/Median..._distributions, the median (M) can be found where the probability of having a value equal to or less than the median AND the probability of having a value equal to or more than the median is at least 0.5.
Winning $2 has a 50% chance, winning $4 has a 25% chance and so on. Notice the gap between $2 and $4: This implies we don't have a value for the median; rather, the median is in the range of $2 <= M <= $4. It's a bit hard to interpret this.
I think it really depends on the context in which the median or expected value is used. For lotteries (discrete probability distributions), medians are reliant on the range of winnings you can earn, the number of possible winnings and the probabilities of which they occur; the expected value depends on the values of the winnings and their associated probabilities.
The above factors influence the choice of an appropriate statistical measure. Since I'm not familiar with the application of statistics in these fields, maybe someone with relevant experience could discuss about this. |
Sure thing.
When we talk about the "Median" of a thing in statistics, we are talking about recording all of the values observed in a distribution. For example, let's say that we have stated odds of something happening such as in the VSauce money game...
In order for those odds to be "true" we have to have an infinite sample size. As a reductive example, take a coin and flip it three times and record the outcomes. Even though you know that there is an even 50% chance for either outcome, in your sample set the odds are going to be .66r:.33r. The more you flip the coin, the closer to .5:.5 you get... but you never actually get quite to 50/50, usually you will end up with a very near split such as .4981:.5019. In practice, this is close enough to the probability statement of "even chance at A or B" that we have verified that the coin is a 50/50 proposition.
How does this apply to the concept of "Median Expected Value?"
Well, for starters, it should immediately jump out to you as a bunk term. "EV" is a concept that takes the weights of odds against the return on investment down to a simple proposition: "Should I play the game?" If you were going to construct a "median expected value" you would be sampling one value (THE expected value of the game) over and over again. The median would be the sample element and the range would be 0.
A better term would be "Median of Sampled Winnings" in which you perform a random game with the proposed odds and payoffs for some large value of iterations (let's say 20,000) and record the winnings and losings of each iteration. Once this is done you can then perform statistical analysis on the data set that was generated and produce a median value.
If you wanted to be really efficient, you could do a box and line diagram breaking your winnings into Quartiles. To be even more efficient, toss any significant outliers and re-calculate your analysis.
For me, this random sampling of events with outlier purging is the best way to get a feel for the median winnings on any single proposition (which isn't the same thing as expected value), which is more or less what you requested.