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07-18-2019, 09:54 AM   #2
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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 scratch-off lottery tickets. The chance of winning a prize is greater on scratch-off tickets, but the chance of reclaiming your money on them is worse than standard print-off numbered lottery tickets (usually).
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Quote:
 Originally Posted by YoshL butts.

 07-18-2019, 12:58 PM #3 Andrew WCY terrible nerves     Join Date: May 2014 Location: Kwai Chung, Hong Kong Posts: 73 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 "All-or-none". 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 n-round 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 n-round game. This translates to paying \${[2^(n - 1) - 1] * (n + 1)}, which is an average of \$3 for a 2-round game, \$12 for a 3-round game, \$35 for a 4-round game and \$90 for a 5-round 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. __________________ Last edited by Andrew WCY; 07-18-2019 at 07:00 PM..
07-18-2019, 01:19 PM   #4
Andrew WCY
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Join Date: May 2014
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Re: VSauce2's Infinite Money Paradox Reaction

Quote:
 Originally Posted by TheSaxRunner05 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?
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.

Quote:
 Originally Posted by TheSaxRunner05 Wouldn't an expected median value be more telling? It seems more practical and less theoretical than an expected value.
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.
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Last edited by Andrew WCY; 07-18-2019 at 01:24 PM..

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