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Does this make sense (probabilities)
Drez
Registered User regular
Registered User regular
This is in reference to a game. I keep seeing people (everywhere) post what I think is incorrect information. I'm trying to explain why I think it's wrong. I'm not sure I'm doing the best job here.
This is what I've written:
The long and short of it is this: You get three abilities in the game. Let's call them A, B, and C for the sake of brevity. And lets call a common event that occurs "EVENT."
With A you have a 2/100 chance of EVENT occurring when A is happening.
With B you have a 8/100 chance of EVENT occurring when B is happening.
With C you have a 5/100 chance of EVENT occurring when C is happening.
When you acquire B, you now have a 50% chance of A happening and a 50% chance of B happening.
When you acquire C, you now have a 40% chance of A happening, a 40% chance of B happening, and a 20% chance of C happening.
So, with A only, you have a straight 2% chance of EVENT occurring.
With A and B, you have an overall 5% chance of EVENT occurring. As they are weighted out the same, the overall percent chance is 5%, isn't it?
And as the chance of EVENT occuring within C on its own is 5%, you still have an overall 5% with A, B, and C...right?
Maybe I'm just being ignorant. I forget how to do this formulaically.
This is what I've written:
So when I was trying to get all the forma I needed, I saw a lot of posts suggesting that when you finally get Enemy Search C, it made it more difficult to find Fiends.
Is this really true?
From what I understand:
- Enemy Search A allows a 2% chance to encounter a Fiend in Carina, Delphinus, Eridanus, and Fornax. If you only have Enemy Search A, 100% of enemy searches will be Enemy Search A-type encounters, so you have a 2/100 chance in finding a Fiend with only Enemy Search A.
- Enemy Search B allows an 8% chance to encounter a Fiend in Carina, Delphinus, Eridanus, and Fornax. When you get Enemy Search B, you now have a 50% chance of an enemy search being of the Enemy Search A type and a 50% chance of an enemy search being of the Enemy Search B type.
So you have a 50% chance that you will have a 2% chance of finding a Fiend as well as a 50% chance that you will have an 8% chance of finding a Fiend. Overall, this is a 10/200 chance, or 5%. So, from my understanding, once you obtain Enemy Search B, you now have a 5% chance of finding a Fiend.
- Enemy Search C allows a 5% chance to encounter a Fiend in Carina, Delphinus, Eridanus, and Fornax. When you get Enemy Search C, you now have a 40% chance of an enemy search being of the Enemy Search A type, a 40% chance of an enemy search being of the Enemy Search B type and a 20% chance of an enemy being of the Enemy Search C type.
So you have a 40% chance that you will have a 2% chance of finding a Fiend, a 40% chance that you will have an 8% chance of finding a Fiend, and a 20% chance that you will have a 5% chance of finding a fiend. As the chance to encounter Enemy Search C is weighted at half the chance to encounter either Enemy Search A or B, you now have a 15/300 chance to encounter a fiend, which is 5%.
Isn't that right?
The long and short of it is this: You get three abilities in the game. Let's call them A, B, and C for the sake of brevity. And lets call a common event that occurs "EVENT."
With A you have a 2/100 chance of EVENT occurring when A is happening.
With B you have a 8/100 chance of EVENT occurring when B is happening.
With C you have a 5/100 chance of EVENT occurring when C is happening.
When you acquire B, you now have a 50% chance of A happening and a 50% chance of B happening.
When you acquire C, you now have a 40% chance of A happening, a 40% chance of B happening, and a 20% chance of C happening.
So, with A only, you have a straight 2% chance of EVENT occurring.
With A and B, you have an overall 5% chance of EVENT occurring. As they are weighted out the same, the overall percent chance is 5%, isn't it?
And as the chance of EVENT occuring within C on its own is 5%, you still have an overall 5% with A, B, and C...right?
Maybe I'm just being ignorant. I forget how to do this formulaically.
Switch: SW-7690-2320-9238Steam/PSN/Xbox: Drezdar
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.4*.02 + .4*.08 + .2*.05 = .05 = 5%.
(40 percent chance of type A * 2% chance of success given type A + the same for B and C)
What arguments are people giving for why it's not 5%?
Most FAQs list you as having a 2% chance of getting a Fiend with Enemy Search A, 8% chance of getting a Fiend with Enemy Search B, and 5% chance of getting a Fiend with Enemy Search C. But those are the chances to encounter a fiend within the set of possible encounters once the game decides if you've encountered A, B, and C. So first the game decides out of encounter types {A, B, and C} and then within each encounter type you have a certain chance to see a fiend (2%, 8%, and 5% respectively). But since the C-type is weighted at half of what A and B each are, it doesn't actually effect the overall chance.
Thanks for the confirmation.
I think people are just looking at the FAQs and seeing "oh, 5%, so your chance goes DOWN?!?!" without thinking it through.
The game, by the way, is Shin Megami Tensei: Strange Journey.
The other possibility is that the listed 2%, 8%, and 5% in the descriptions are actually correct, in that they already take into account the splitting of the encounter type lists.
The other other possibility is that there are other encounter types and the game might actually make its selections differently from how we've assumed. For example if you have Enemy Search C, the game might first give you a 40% chance of getting a type-A encounter, and if that fails, give you a 40% chance of getting a type-B encounter, and if that fails, giving you a 20% chance at a type-C encounter, and if that fails, it gives you a "normal" or some other kind of encounter. The correctness of the sort of calculation I posted above assumes the game isn't doing this (and instead assumes that if you have Enemy Search C, you are guaranteed either a type A, B, or C encounter, and type A and B each occur 40% of the time and type C occurs 20% of the time). The description makes it sound like the game isn't doing this, but if it were it wouldn't be the first time an in-game description doesn't match reality.
You are right, in one sense, that Enemy Search C doesn't "help" encounter Fiends, but that is not its entire purpose. Each new Enemy Search ability increases the pool of enemies you can encounter. You have a 100% chance of encountering an enemy, but the Fiend is the rare type.
Also, you can never have only B or only C or only B and C - the Enemy Search abilities are cumulative. By the end of the game, you either have A and B or A, B, and C as C is optional.
So with Enemy Search A, you have a certain set of enemies you can encounter. Lets look at it this way, in crudest terms:
Enemy Search A: {F, 1, 2, 3}
Enemy Search B: {F, 3, 4, 5}
Enemy Search C: {F, 6, 7, 8}
F is the Fiend for that location. Each number represent a different unique enemy.
So with Enemy Search A alone, you have access to four enemies in that location, and the chance of encountering the Fiend in an A-type encounter is 2%. The set is {F, 1, 2, and 3}.
With Enemy Search A and B, your set is {F, 1, 2, 3, 4, and 5}. But some of these overlap (like F and 3) so the overall percentage of getting the Fiend within that set is really 10 out of 200, as I visualize it, because the possibility of getting set A is equivalent to getting set B. So the chance of getting an F when you have A and B is 5%.
And so on.
And this makes more sense than the developers actually making rare enemies rarer when you get the optional Enemy Search.
I think people are thinking that the larger pool of potential enemies to encounter makes is more difficult to encounter a Fiend, but given the actual statistics, that isn't true, because the encounter probability of the Fiend is increased when you get B and kept even when you get C. It was confusing me that so many people felt that acquiring Enemy Search C made their chances of encountering a Fiend worse, when it doesn't seem to be true. It seems as though the encounter rate for a Fiend is low with A, then raised a little with B by making it more common within the B set, and then kept the same when the player gets C.
Does that make sense?
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Let me explain the mechanics a little better.
When you enter a new dungeon level, the game instantiates and randomly places a number of enemy search locations on the map which the player can visibly see and then choose to initiate (or not initiate). If the player only has Enemy Search A, the game makes every encounter Enemy Search A. If the player has Enemy Search A and B, then the game flips a coin (50/50) between A and B for each encounter. If the player has A, B, and C, then the game decides if the encounter is A, B, or C via 40%/40%/20% percent chances. There are no other possibilities - it is always one of the three, one of the two, or just A depending on which Enemy Search abilities the player has.
Once the type of encounter is decided, there is a certain percent chance within the set of a given enemy within the set, and some of these overlap with other sets.
So the bottom line is that with Enemy Search A, B, and C, the player has a 40% chance of having a 2% chance of getting a Fiend, a 40% chance of having an 8% chance of getting a Fiend, and a 20% chance of having a 5% chance of getting a Fiend. I am 100% positive that this is how the game functions. But my interpretation of this, without having actually done the math, is that THIS situation is mathematically/statistically equivalent to having just Enemy Search A and B with a 50% chance of getting a 2% chance of finding a Fiend, and a 50% chance of getting an 8% chance of finding a Fiend.
I am trying to compare the probability of encountering a Fiend in this method if the player has A, A+B, or A+B+C, which are the only three possibilities for a player.
So, based on your formula earlier:
Only A: 1.00 * 0.02 = 0.02 (or 2%)
Only A+B: (0.50 * 0.02) + (0.50 * 0.08) = 0.05 (or 5%)
A+B+C: (0.40 * 0.02) + (0.40 * 0.08) + (0.20 * 0.05) = 0.05 (or 5%)
Right?
Yes. It's perfectly clear to me now. Thanks.
Yes.
Thank you!