07182019, 12:27 AM  #1 
The Doctor
Join Date: Apr 2006
Age: 30
Posts: 5,917

VSauce2's Infinite Money Paradox Reaction
Original Video Here 
https://youtu.be/RBf1s4TassI I watched VSauce2 today and their video about an infinite money paradox. Basically that if any given risk/reward system had a potential infinite outcome, no matter how unlikely, the expected value would also go to infinity. While this is true, we certainly wouldn't spent a near infinite amount of money to play a game that would mostly likely gain us just a few dollars. The video a bit overdramatozes this and plays it off as "humans as not just followers of the strict numbers" idea. I think it's a bit misleading.I think we as humans just have misconceptions about probabilities than run deep. Despite the expected value going off to infinity, could you make an argument for the Median result being something modest like $4? Since half of all games end after the first flip, would it be safe to say the median value is either 2 or 4, without even running a sample size, or does the median require a list of results to exist? This ties in with the TCG a bit with the drop weight of Tier 12 cards  despite being in a wholly noninfinite expected value. I've calculated expected values left and right and I don't think they tell the full story, considering the absolute rarity of the top prize. Wouldn't an expected median value be more telling? It seems more practical and less theoretical than an expected value. It's like calculating a return on investment for lottery winnings when the chances of YOU winning are astronomical. This whole misconception between median expected value and average expected value is probably behind why people even play lotteries to begin with. I'm sure psychology plays a role here, but maybe we've just not been generally trained to look at these situations through the right lens, and has applications in practical decision making. If there's something I'm missing here, please enlighten me, I'd love to know more about this kind of stuff  it facinates me. Something about his video though struck me as a misconception and it inspired me to make this thread. 
07182019, 09:54 AM  #2 
Signature Extraordinare~~

Re: VSauce2's Infinite Money Paradox Reaction
The point of the video was to demonstrate a situation where you could make infinite money from a game, permitted that you had near infinite money, and the person willing to dish out money from the game, also has infinite money.
It's definitely a reason why people play lotteries and such, because the median expected value & average expected value is so misconstrued by the top prize. Heck, it's why people play scratchoff lottery tickets. The chance of winning a prize is greater on scratchoff tickets, but the chance of reclaiming your money on them is worse than standard printoff numbered lottery tickets (usually). 
07182019, 12:58 PM  #3 
terrible nerves
Join Date: May 2014
Location: Kwai Chung, Hong Kong
Posts: 87

Re: VSauce2's Infinite Money Paradox Reaction
The following is a rather lengthy discussion on the significance of expected gains.
Assume that everything we're talking about is (emphasis intended) ON AVERAGE.  We could begin examining the subject matter by creating a simple hypothetical scratchcard called "Allornone". Each scratchcard is valued at $1. If the expected gain were $1, one might naturally assume they'll have no net gain or loss. But let's define a few variations of the game: Variation 1: You either get $10 with a probability of 0.1 or you get nothing. Variation 2: You either get $100 with a probability of 0.01 or you get nothing. Variation 3: You either get $1000000 with a probability of 0.000001 or you get nothing. Notice how the expected gain in all 3 cases is $1 but the chance of winning continues to diminish. The expected gain doesn't hold much significance if we don't consider the chance we lose. To put this in another way, let us first determine the number of scratchcards we need to scratch before we earn the grand prize. Variation 1 requires 10 scratchcards before you win $10. Variation 2 requires 100, while the third one requires a whopping 1000000. Then, let's determine how much we would lose RIGHT BEFORE we reclaim our money. We would lose $9 right before we earn back what we gave for the first game. It's a $99 loss for the second game and a $999999 loss for the third. The average loss immediately before reclaiming money can be seen as a perceived 'risk' to winning. This is a bit similar to placing investments on stocks that have a high volatility (beta): The larger the volatility, the higher the risk; we could potentially earn more, but as a subsequence of bigger risk, we also have a larger chance of losing.  What if the Infinite Money game were to stop at a certain round? If the game has only 2 rounds, you'll have a 50% chance of winning $2 and a 50% chance of winning $4. The expected gain is $3. If the game has only 3 rounds, you'll have a 50% chance of winning $2 and a 50% chance of winning $4 or more. You'll have a 25% chance of winning $4 and a 25% chance of winning $8. The expected gain is $4. If the game has 4 rounds, you'll win $2 (50%), $4 (25%), $8 or $16 (12.5% for both). The expected gain is $5. If the game has 5 rounds, you'll win $2 (50%), $4 (25%), $8 (12.5%), $16 or $32 (6.25% for both). The expected gain is $6. In other words, you should pay $(n + 1) to have no gain or loss for an nround game. Doesn't seem like much, does it? But how many tries do you need before you earn back what you paid for? For a game with 2 rounds, you need 2 tries on average. In these 2 games, we would have 1 game with a $2 win and 1 game with a $4 win. For a game with 3 rounds, 4 tries. In 4 games, we would have 2 games with a $2 win, 1 game with a $4 win and 1 game with an $8 win. For a game with 4 rounds, 8 tries. For a game with 5 rounds, 16 tries. So, we need 2^(n  1) tries to have no net profit or loss for a game with n rounds. To calculate the average loss immediately before money reclamation, we need an average of [2^(n  1)  1] tries for an nround game. This translates to paying ${[2^(n  1)  1] * (n + 1)}, which is an average of $3 for a 2round game, $12 for a 3round game, $35 for a 4round game and $90 for a 5round game! The perceived risk factor increases exponentially the more we want to win.  In conclusion, we should look deeper than the surface of things: Not only should we consider the expected gain, we should also evaluate the risk. TL;DR: The expected gain alone doesn't mean much if we don't know how much we would risk losing.
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Last edited by Andrew WCY; 07182019 at 07:00 PM.. 
07182019, 01:19 PM  #4  
terrible nerves
Join Date: May 2014
Location: Kwai Chung, Hong Kong
Posts: 87

Re: VSauce2's Infinite Money Paradox Reaction
Quote:
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. Quote:
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.
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Last edited by Andrew WCY; 07182019 at 01:24 PM.. 

08242019, 12:21 PM  #5  
FFR Player
Join Date: May 2005
Posts: 19

Re: VSauce2's Infinite Money Paradox Reaction
Quote:
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 recalculate 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. 

08242019, 12:55 PM  #6 
FFR Player
Join Date: May 2005
Posts: 19

Re: VSauce2's Infinite Money Paradox Reaction
Here is a nice little python program that you can run to verify that this game sucks:
Please note you will have to fix the indentation. ``` import random import statistics def play_game(): value = 0 while True: flip = random.SystemRandom().randint(0, 1) if flip == 1: if value == 0: value += 2 else: value = value*2 else: print("Won: ", value) return value winnings = [] for x in range(0, 20000): print("Playing game ", x+1) winnings.append(play_game()) mean = statistics.mean(winnings) harmonic_mean = statistics.harmonic_mean(winnings) median = statistics.median(winnings) mode = statistics.mode(winnings) variance = statistics.variance(winnings) standard_deviation = statistics.stdev(winnings) full_range = max(winnings)  min(winnings) print("Range: ", full_range) print("Mean: ", mean) print("Harmonic Mean: ", harmonic_mean) print("Median: ", median) print("Mode: ", mode) print("Variance: ", variance) print("Standard Dev: ", standard_deviation) ``` Last edited by DoroFuyutsuki; 08242019 at 12:56 PM.. Reason: Clarification on indentation. 
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