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- This article is about the mathematical background of Daily Tasks. For general information, see Daily Tasks.
The stochastic process Daily Tasks is a composed experiment in Asphalt 8. It consists of 3 trials that can be conducted once per day. The timer is reset at 0:00 player's local time. The trials are connected with the gameplay element Daily Tasks: When a player taps on the Daily Tasks panel of the Play screen, the process assigns the rewards to the 3 tasks Vehicles, Events and Multiplayer.
History
- As of July 19, 2019, the process grants Optimal Shuffle Boxes, Optimal Split Boxes and Credits. The Vehicles task can now grant boxes, too. As long as the sample size is too small to make predictions for this period, the bar chart at the top of this page shows the complete data since the 2019 Spring Update.
- From the start of the Showdown Update until July 18, 2019, Optimal Boxes had been removed as possible rewards, so all trials were deterministic as there were no other rewards than Credits.
- Before the Showdown Update, the process granted Credits, Optimal Shuffle and Split Boxes. The Vehicles task granted only Credits while Events and Multiplayer could grant all three prizes. This period is examined in the following analysis.
Sample space
As of the Fast Lane Update, the possible rewards are
- 1 Optimal Shuffle Box,
- 1 Optimal Split Box or
- 1,500,
so the sample space of all possible outcomes is $ \Omega = \{\operatorname{Credits}, \operatorname{Shuffle}, \operatorname{Split}\} $.
Trials
After more than 200 trials (tasks), it has been observed that the Vehicles task never granted a box while the others granted both Shuffle and Split Boxes. As there where cases when both Events and Multiplayer granted a box, there seems to be no restriction that only one box per day is granted. However, it is possible that there cannot be two boxes of the same kind on one day as there was no such event of 2 Shuffle or 2 Split Boxes. This leads to the following conclusions:
- Events and Multiplayer have the same sample space of $ \{\operatorname{Credits}, \operatorname{Shuffle}, \operatorname{Split}\} $, but it cannot be said if they are independent.
- Vehicles is a separate deterministic experiment with only one outcome: $ \{\operatorname{Credits}\} $.
Statistics
There are no offical drop rates for Daily Tasks.
But as the law of large numbers states that in the long run, the average relative frequency of an event converges almost surely to its expected value (= its drop rate), conclusions can be drawn from statistical data if the sample size is big enough.
After reaching an acceptable sample size of 200 tasks, the average frequencies converged to the following drop rates:
- ~ 5 % Optimal Shuffle Box
- ~ 5 % Optimal Split Box
- ~ 90 % Credits
Probabilities
Borel's law of large numbers asserts that for independent and identically distributed (i. i. d.) repetitions of a Bernoulli trial, the drop rate of an event almost surely converges to its probability on any particular trial.
As the Vehicles tasks has only one outcome, it is not identical to the other tasks and has to be separated. It is already possible to say that the probability of getting Credits from this task is 100 %, whereas the probability of getting a box is 0 %.
- The Vehicles task makes up $ \frac13 = 33.\bar3 % $ of all tasks. Separating it will leave the remaining $ \frac23 = 66.\bar6 % $ tasks.
- Separating a task also requires separating its outcomes, so we have to subtract the guaranteed $ 33.\bar3 % $ Credits from the drop rate of $ 90 % $, which gives $ 56.\bar6 % $. This means that $ 56.\bar6 % $ of the remaining $ 66.\bar6 % $ of tasks will be Credits. $ \tfrac{56.\bar6}{66.\bar6} = \tfrac{85}{100} $, so the drop rate of Credits for the remaining tasks is 85 %.
- The box drop rates of 5 % out of 3 tasks deliver $ \tfrac{5}{66.\bar6} = \tfrac{7.5}{100} $ out of 2 tasks, so the drop rates of Optimal Shuffle and Split Boxes from the remaining tasks are 7.5 % each.
If we assign random variables to obtaining each item (Credits, Shuffle Box, Split Box), they can only take the values 1 (getting the item) or 0 (not getting it), so the respective trials are Bernoulli trials. As the Events and Multiplayer tasks approximately grant the same amounts of each item, they presumably have the same probability distribution, so they are identically distributed.
But, as mentioned above, we cannot be sure if they are independent: If it is possible that both tasks grant the same box on one day, they are independent. If obtaining a box from the first task means that the same box cannot be obtained from the second task on the same day, they are not independent, as the first task influences the outcome of the second.
Independent events
If the tasks are independent, all conditions for Borel's law of large numbers are fulfilled. This means that the calculated drop rates equal the probabilities of getting the item from one of the two remaining tasks Events or Multiplayer:
- $ \mathrm P(\{\operatorname{Credits}\}) = 85 % $
- $ \mathrm P(\{\operatorname{Shuffle}\}) = 7.5 % $
- $ \mathrm P(\{\operatorname{Split}\}) = 7.5 % $
The probabilities for one day can be calculated as follows (the indices $ _V $, $ _E $ and $ _M $ denote the event of getting the item from Vehicles, Events and Multiplayer, respectively):
The probability of getting only Credits on one day is
- $ \mathrm P(\{\operatorname{Credits}_V, \operatorname{Credits}_E, \operatorname{Credits}_M\}) = 100 % \cdot 85 % \cdot 85 % = 72.25 % $.
As the probability of $ \{\operatorname{Credits}_V\} $ is always 100 %, it can be seen that it will always be a factor 1 and can therefore be omitted.
The probability of getting a Shuffle Box from Events and a Split Box from Multiplayer is
- $ \mathrm P(\{\operatorname{Shuffle}_E, \operatorname{Split}_M\}) = 7.5 % \cdot 7.5 % = 0.5625 % $.
To obtain the probability of getting 1 Shuffle Box and 1 Credits reward, all possible cases have to be taken into account: The Shuffle box can either be granted by Vehicles or Multiplayer, with the other reward being Credits:
- $ \mathrm P(\{\operatorname{Shuffle}_E, \operatorname{Credits}_M\} \cup \{\operatorname{Credits}_E, \operatorname{Shuffle}_M\}) = 7.5 % \cdot 85 % + 85 % \cdot 7.5 % = 12,75 % $.
The same goes, for example, if one wants to know the probability of getting exactly 1 box (Shuffle or Split) on one day. In this case, there are 4 possibilities: $ \{\operatorname{Shuffle}_E, \operatorname{Credits}_M\} $, $ \{\operatorname{Split}_E, \operatorname{Credits}_M\} $, $ \{\operatorname{Credits}_E, \operatorname{Shuffle}_M\} $ and $ \{\operatorname{Credits}_E, \operatorname{Split}_M\} $. The probability for each of them is $ 7.5 % \cdot 85 % = 6.375 % $, so the total probability is
- $ 4 \cdot 6.35 % = 25.5 % $.
Analogously, there are 4 ways of getting exactly 2 boxes: $ \{\operatorname{Shuffle}_E, \operatorname{Split}_M\} $, $ \{\operatorname{Split}_E, \operatorname{Shuffle}_M\} $, $ \{\operatorname{Shuffle}_E, \operatorname{Shuffle}_M\} $ and $ \{\operatorname{Split}_E, \operatorname{Split}_M\} $, each of them having a probability of $ 7.5 % \cdot 7.5 % = 0.5625 % $. The probability of getting exactly 2 boxes is
- $ 4 \cdot 0.5625 % = 2.25 % $.
Dependent events
If obtaining a box from one task means it cannot be obtained from the other, there are less possibilities. The above example of getting exactly 2 boxes would only have $ \{\operatorname{Shuffle}_E, \operatorname{Split}_M\} $, $ \{\operatorname{Split}_E, \operatorname{Shuffle}_M\} $, resulting in a probability of
- $ 2 \cdot 0.5625 % = 1.125 % $.
As both probabilities are very small (2.25 % for independent and 1.125 % for dependent tasks), more experiments have to be conducted until the event of 2 equal boxes occurs. If it does, the tasks are independent. If not, the probability that the tasks are independent gets smaller the more experiments are conducted.
See also
The statistical data on this page is part of WikiProject Statistics. It contains original research which may be incomplete, incorrect or biased. |