Jackpot odds- am I not understanding?
Raven2318
Member Posts: 226 ✭✭✭✭✭
OK, I will be the first to admit math is not my strongest suit. However, the odds of getting an all silver jackpot are stated as 15%, and the odds of all gold as 5%. I have run approximately 84 challenge missions during this event, and had two gold jackpots and four silver. It would seem the odds are not in my favour, or maybe I am misunderstanding the probabilities?
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well I've had one and this is it and I'm at 37.2 rsl
So the odds suck for everybody lol -
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> @Raven2318 said:
> OK, I will be the first to admit math is not my strongest suit. However, the odds of getting an all silver jackpot are stated as 15%, and the odds of all gold as 5%. I have run approximately 84 challenge missions during this event, and had two gold jackpots and four silver. It would seem the odds are not in my favour, or maybe I am misunderstanding the probabilities?
But there's only a 20% chance of that lol -
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I'm so lucky I got 2 cans of blue sardines and 7 experience boxes on a gear run.
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Got 6 green weapons in one of my gold jackpots. Had to pay the 15 gold for the next 3 to try get something descent.
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@Jenng would have told you the same thing I'm writing below:
Since the probability of getting a gold or silver jackpot completing one mission is 2/10, there is no guarantee you will ever get either no matter how many times you play (getting either jackpot the first time does not affect the probability of getting a jackpot the next time, that is the events are independent). Thus every time you complete a mission the probability of a jackpot is 2/10 and you can never be sure that you will win. You can, however, determine the number of missions you would need to play to come within some acceptable limit of certainty of getting a jackpot, such as 99% sure you would get a gold or silver jackpot.
This may be calculated as follows:
Let n be the number of missions you need to play
The probability of no jackpot is 1 - 2/10 = 8/10.
The probability of no jackpot in n attempts is equal to the probability of no jackpot in the first attempt AND no jackpot in the second attempt AND ... AND no jackpot in the nth attempt which is (8/10)^n since the probability of the intersection of n independent events is the product of the probabilities.
So the probability of getting a jackpot at least once in n attempts is 1 - (8/10)^n
Since we want to be 99% sure of a jackpot, we require 1 - (8/10)^n to be greater than or equal to 0.99.
Rearranging this we find we need (8/10)^n to be less than or equal to 0.01. Trying some values for n you can see that (8/10)^20 = 0.01153 and (8/10)^21 = 0.00922 and so you should play 21 missions in order to be 99% sure of a jackpot at least once.
If you got 2 golds and 4 silver the odds per above were in your favor.
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Finally some recognition for the f’n mathematical genius I am ............ deep, practically unnoticeable......... inside. Brains and rage my friends. -
@Cronus
In 100 attempts, it would be expected (on average) that 5 attempts yielded gold jackpots and 15 attempts yielded silver jackpots. This is 20 overall jackpots. The OP stated an outcome of 2 gold jackpots and 4 silver jackpots in 84 attempts or 6 jackpots overall. 5(84/100) = 4.2 gold jackpots and 15(84/100) = 12.6 silver jackpots were expected in 84 attempts. Overall, the OP would reasonably expect 20(84/100) = 16.8 jackpots in those attempts. I would agree with the OP that they didn't receive as many jackpots as expected, but I believe that if more attempts were made the numbers would tend toward the expected outcomes. This player just had worse than average/expected outcomes. -
Also @Cronus your math was to be 99% certain of a SINGLE jackpot when as I've shown in 84 trials nearly 17 were expected. Once you realize this I'm certain that you'll agree that the OP had worse than average luck.
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Nope, your assumptions are incorrect. You're treating the probability of the event being dependent on a prior result. In this case getting either jackpot the first time does not affect the probability of getting a jackpot the next time, which makes the events independent. You need to multiply the product of the probabilities. If you have a dependent event then you can do exactly what you did which is to take the % and multiply by the total number of events to get the expected outcome (100 events * 15% chance = 15 expected occurrences).
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I'll send you my thoughts on through Line. One sec.
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On average 5 out of 100 would be gold. The chance to get exactly that, isn't as big as one might think. It is the biggest chance though.
Chronus's explained is funny.
If hundred people would do 21 missions. 99 of them would get at least 1 jackpot (but probably more) 1 person would get screwed and got f'in none...
All one average though, because the chance if 100 people would do 21 missions and (only) 1 would get no jackpot would be a very small chance of actually happening.
Chances are so much fun...
And very counter intuitive for anyone in general... -
If the RNG is independent for determining chances (meaning rerolled after every mission) I stand by what I wrote. That is the correct calculation. If the RNG is somehow dependent, then the math changes. If after you complete your first mission the next 20 outcomes after a mission is determined the math will be much closer to 5 gold jackpots per 100 missions. But even this becomes a mix of dependent and independent events since the next 20 outcomes after initial 20 behave independently. To be truly dependent NG would somehow have to predict the future and know in advance how many missions you'd complete. Then they could distribute the chances among that exact number of events.
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I believe it's feasible that post mission crate types and amounts are predetermined X number out. I also believe that NG uses this tactic to "guarantee" the percentages they post in radio calls and in post mission crate outcomes, including jackpot events.
Anyone want to perform a ridiculously large number of missions next jackpot event and gather all of the data as far as which number mission lead to a silver or gold jackpot? From that it might be possible to determine whether or not they predetermine the outcomes and how many outcomes they predetermine before they are regenerated. I fully expect them to use this approach.
State lotteries often do this to split their jackpots. For instance, if a scratch off game is released with 5 1 million dollar jackpots and they intend to print 10 million tickets, they might include 1 jackpot in each set of 2 million tickets. If they front loaded the tickets and put all 5 jackpots in there, they could/would lose money since people would know there are no more jackpots after the first set of 2 million tickets and all 5 jackpots have been claimed. -
Ask my bookies at NG
They know all about the RNG's and 1,2,3's -
Que Scott Stiener Sacrifice...
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