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Let's say someone is playing a computer game in which the chance for some item to drop is 1 out of 100 each time he kills a monster.

The player concludes that if he kills 100 monsters then it is guaranteed for him to have found at least 1 of that item.

Is there a name for the fallacy above?

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    Appeal to probability: "...the logical fallacy of taking something for granted because it would probably be the case.." en.wikipedia.org/wiki/Appeal_to_probability – Bread Oct 28 '18 at 15:04
  • That would fit under the fallacy you mentioned but there might be a more precise fallacy for the case described. The core is the p to 1 / p relationship. – Răzvan Flavius Panda Oct 28 '18 at 15:10
  • Yes, that's why I didn't post it as an answer. Just a suggestion. I think the key point of the problem is in the logical leap of taking something for granted based solely on a mathematical probability. – Bread Oct 28 '18 at 15:14
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    Sounds like the gambler's fallacy: en.m.wikipedia.org/wiki/Gambler%27s_fallacy - over a long period of losses, it's a fallacy to think you're "due for a win". It's possible you just keep losing, or that your payout doesn't exactly average out. – Adam Oct 28 '18 at 15:47
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    I would say this is a mathematical error. Say, what is the probability of having a success (a single try has probability P) in N tries? P * N. Definitely wrong. – rus9384 Oct 29 '18 at 6:35
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This is definitely closest to the gambler's fallacy. An example of this fallacy demonstrated in your example would be that the player, having killed 80 monsters without a coin, thinks that the next 20 monsters will have a higher probability of finding a coin with each kill. This is fallacious, as no matter how many monsters the player has killed, even after killing 10,000 with no coin, he will still have a probability of 1/100 or 0.01 or 1% for receiving a coin after each kill.

On the flip side, there is the notion of "quitting while you are ahead," which is also the gambler's fallacy. This example would be that the player thinks that, after getting 2 coins but only killing 100 monsters, thinks they will have a lower probability of receiving a coin in the future and thus should quit playing or do something else since the probability of receiving a coin will be lower for future 100-kills. This is also fallacious. Related, gambler's conceit is where someone thinks they are on a "lucky streak" or that their skills are actually what is causing victory rather than probability, and so will continue gambling and almost inevitably lose all their earnings.

The probability of getting 1 coin per kill is 0.01 as a given that cannot change. However, the more times the player has killed groups of 100 monsters, the closer the resulting empirical probability will support this data. Monte Carlo simulations would be how you can simulate doing thousands of 100-kills in order to get the average probability of interest. Given an infinite number of attempts at killing 100 monsters, the probability of getting 1 coin in 1 kill will be 1%, though you may have instances where 2 coins are received in 100 kills, which makes it appear as though the probability of getting 1 coin in 1 kill is 2%. The best way to look at this and the probability of getting at least one coin in 100 kills is a Poisson distribution.

A Poisson distribution gives the probability of k number of events happening given lambda average number of events happening (assuming independent trials). In this case, lambda = 1 because on average, 1 coin will be given per 100 kills. P(k number of coins dropped) = e^(-lamda)*(lamda^k)/k! The result is P(0 coins dropped in 100 kills) = 37%, P(1 coin dropped) = 37%, P(>1 coin dropped) = 26%. So, P(at least 1 coin dropped)= 63%.

These situations and the fallacy assumes independent trials of a random process. If the process is not random or if the trials are not truly independent, then the situation is made more complex. With something such as 3-pointers made in a basketball game, there are certainly psychological (and physical) factors at play affecting the outcome of any given shot, even though player's shot may have a historical probability. Confidence as a result of a "hot streak" may serve to perpetuate a hot streak or can give someone pressure that causes them to lose focus or otherwise make mistakes. So, we can conclude from the gambler's fallacy that it is not fallacious to pass someone repeatedly the ball who is on a "hot streak." There is some debatable statistical evidence on the subject, with some evidence offering support. See hot hand.

There is also the inverse gambler's fallacy, which is the fallacious conclusion that, upon seeing an unlikely event take place, assumes there were a large number of trials and outcomes that already happened, attempting to explain why the unlikely event was seen.

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    Actually, given individual trials of killing 100 monsters, the probability that a coin drops in each 100 goes to about 60%. The average number of coins dropped does go to 1. Which to consider, for a large group of players, is whether the coin is transferable in the game, so we either have 100 players, each of which must get the coin drop on their own, or 100 players cooperating, so a player with more than one coin will give it to one with none. – David Thornley Oct 29 '18 at 17:25
  • @David Thornley where are you getting the 60% from? I don't really follow the rest of your comment either. – Alex Strasser Oct 29 '18 at 18:33
  • @DavidThornley If the average number of coins dropped per 100 kills is 1, then the average number of coins dropped per kill is 0.01, which represents the probability of dropping 1 coin in 1 kill. Are you saying that there is a 60% probability of >=1 coin dropping per 100 kills? – Alex Strasser Oct 29 '18 at 18:42
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    Yes. There's a difference between average number of coins dropped in 100 kills (1) and the probability of at least one coin dropping (about 62%). Since these are independent trials (kills), there's no guarantee that one will drop (gambler's fallacy), because the 99% chance of no drop could come up all the time. This is pretty close to a Poisson distribution with expected number of wins = 1, and so there's about a 37% chance of none, 37% chance of one, and 26% chance of more than one. en.wikipedia.org/wiki/Poisson_distribution. – David Thornley Oct 30 '18 at 13:49
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    @DavidThornley Well, given that a probability of success is P, on average there would be needed 1/P trials for a success. In the worst case it is infinite (P < 1) though. – rus9384 Oct 30 '18 at 16:35

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