Hate to say it, but this example is wrong, because in your example you’re using one instance within a string of instances and, as such, isn’t a one off scenario.

A more correct way to put your example would be thus: The average person wins £100 per ticket. Somebody who has NEVER bought a ticket sees this and assumes that if they buy 1 single ticket they SHOULD win £100, but in reality, while the average ticket is worth £100 over a long period of buying tickets, only 20% of tickets win on any given week, so the chances they will win anything with that one solitary ticket are only 20%. Therefore, the likely outcome is they win nothing. Correct, the ticket isn’t worthless, but you must assume it’s value is that of the mostly likely outcome before the event

Mate, this is an odd way to look at statistics. What you are saying is:

Probability of 4 common and 1 rare = 61% (let’s use the actual figure)

Probability of 1 legendary = 5%

You are implying the two probabilities are mutually exclusive. The two kinds of packs are but the probabilities aren’t. Each time you open a pack the expected result is 4 common and 1 rare 61% of the times, 1 legendary 5% and other results 34% of the times. And the brilliant part is that this does include pity timer.

Each pack opening is an independent event (excluding the pity timer that is already added to the frequency distribution, math is a beauty), hence it is as realistic to expect to get a pack with 4 commons and 1 rare 61% of the times as it is to expect 1 legendary 5%… any other way to look at it is ignoring the baseline of statistics and go with your intuition that, since you are a human being, is full of cognitive bias that are just going to misguide you.

You are absolutely right, the expected result is the same. Which means that the variance doesn’t really skyrocket, only our perception of it does. It’s still the same expected distribution. Only the dust value changes.

This is exactly the point I’m making. Because each pack opening is an independent event…if the probability of getting 4 commons and 1 rare is 61% then this is the most likely occurrence when opening 1 single golden pack. As such, when trying to work out the dust value before the event you should assume that this is what you will get from that one pack

Ok, I understand your point but before I leave I want to add that while it is a “safe” or “low expectations” way to look at it, it doesn’t allow a sensible realistic discussion about the value of the packs, which was the point of the thread.

My point is that if an event has a positive expected value, it isn’t a mistake going for it, even if the probability for success is very low.

Maybe I have a flaw in my thinking there that I don’t see. Idk, my track record today is abysmal, so I probably better leave the math heavy topics be for today

@Vlad: I meant that the value difference between the best and worst result is really big there. variance is possibly the wrong term for it

If you want to make it even more complicated, I’m in a position where the EV of my golden pack is more than most. This is because I’m currently saving classic packs up, until they add in HoF replacement cards. Still unlikely that I can guarantee anything, but I can probably increase my odds or getting an epic, maybe a legendary depend on how close to the pity timer I can get before cashing in

Whilst I’ve only read your most recent post, it seems to me that you’re misunderstanding what is meant by the mathematical term ‘expected value’. If the probability of success is low, that means the expected value is failure. The fact that the reward for success is very high has to be offset against the probability of success, as long as we believe there is such a thing as probability in the real world.

the EV isn’t ‘failure’ just because the probability for success is low. I think you mean ‘what is the most likely outcome?’.

EV gives you a value that your event will give out on average while the number of tries approaches infinity. so whether the EV is positive or negative is decided by the ratio of probability for success and payout of success.

I’m beginning to see however why the EV might not be the guiding light in a one-time experiment that I believe it to be. Not fully convinced though

Yes was at the library and didn’t have time to make me reply as accurate as I would have liked. To be more accurate, if the probability of success is low, the expected value is the outcome associated with failure, so for example if the probability of getting a legendary is 1/27, the expected value of legendaries we get from the pack is 1/27 and so in practice we expect to get 1 legendary from every 27 packs (I’m not saying this is the actual probability of getting a legendary: I’m just using the first number I thought of for the sake or argument).

Maybe I should read the earlier comment you made which you were referring to, but when you say

it doesn’t seem to make sense. I mean in the example I give, the event of getting a legendary has a positive expected value and the probability of success is very low and so if we accept that there is such a thing as randomness, then logically it is indeed a mistake to go for it.

Also, the way you put it, it makes it sound like the expected value is somehow divorced from the probability for success, when in actual fact they are the same thing in the case of getting a legendary from a pack.

In general, they are intrinsically linked:

So the expected value is the sum of: [(each of the possible outcomes) × (the probability of the outcome occurring)].

I took that from a webpage on expectation and variance.

I’m not sure I understand your example with the event of getting a legendary in this context.

The way I see it, I’m talking about a decision whether or not to buy something, say a golden pack (assuming that’s actually possible). So in my mind the golden pack becomes like a lottery ticket: you buy one and find out what it pays out later.

Then we calculate the dust EV of the pack by multiplying all possible outcomes with their respective probabilities. Say the result will be X dust. Since you bought the golden pack with Y €, you can compare X to the dust EV of the number of regular packs that you could buy with Y € and choose whichever is higher.

That was my initial idea anyway, but as I said, I’m beginning to see flaws in this theory regarding a one-time event, like buying one golden pack.

However, my original point in this thread was a little different than the above.

It was about if it is better to buy the adventure with real money (option A) or buy 15 packs with the same amount of real money (option B). The paid part of the adventure gives you 12 regular packs and a golden one. So the difference between A and B is that A has one golden pack more and 3 regular packs less than B.

So as long as the EV of the golden pack is higher than that of 3 regular packs, A is the better option (dust wise only). So it’s not necessary to exactly know the EV of the golden pack, but it’s only important to know if it’s more or less than 3 regular packs. I calculated a golden pack EV of 5.5 times the EV of a regular pack.

So the golden pack has the EV of 5.5 regular packs, which is more than 3, so when using real money, it’s more dust efficient to buy the adventure instead of 15 packs. And if my calculation is wrong, which might well be because it’s also based on a couple of assumptions and shaky data, I feel the room for error (difference between 3 and 5.5) is big enough to still come to the same conclusion.

I have no problem with this either, as long as the real money cost of the full adventure is the same as that of 15 packs.

The only issue is, as you say, what the EV of a golden pack is. I seem to remember reading in someone’s comment on this thread that the chances of getting rares, epics and legendaries is different for golden packs. I don’t know if this is true, but if it isn’t true and one knows the chances of getting them in a non-golden pack it seems straight forward enough just to replace the dust values of non-golden cards with golden in the calculation.

Having read your last reply it’s clear you know how to do EV calculations now and maybe you knew this all along. It just wasn’t clear when I read the reply of yours to which I initially responded for the reasons I gave and either way, the way you put it didn’t sound right, as I explained.

It seems to me that the ‘probability for success’ is actually what determines the EV and thus whether or not it’s a mistake to ‘go for it’, so a low probability of success is likely to mean it is a mistake to go for it, although taken on its own, this statement of yours probably needs some clarification, unless it’s clear from reading your earlier posts, but I couldn’t be bothered to do that - well not everything anyway.

My example about the expected value of number of legendaries from a given number of packs was not intended to be especially related to your calculations, which I was unaware of at the time. I was just plucking the first thing that came to mind to try to clarify what I was saying. It was probably a bit simplistic to be truthful. A better example would be the EV of number of heads from 2 tosses using a biased coin with probability 2/3 of getting a head, which would be 2 * P(H/H) + 1 * P(H/T) + 1 * (T/H) = 2 * 4/9 + 1 * 2/9 + 1 * 2/9 = 12/9

Thanks. I realise it may not have been obvious from my earlier reply, which was rushed as I was out of time on a library pc, but I really don’t need anyone to explain it to me, not was I seeking anyone’s explanation of it, holding a First Class Honours in Maths, Stats and Operational Research as I do. This said, I must admit I did need to refresh my memory on the subject, as it’s been a long time since I’ve looked at it and the page I looked at was fine for that: all I needed was the definition/formula I quoted from it in my earlier reply, which I then applied to the biased coin example that was conjured up in my mind after a streamer was talking about the likelihood of getting a taunt from using Conurer’s Calling on a Mountain Giant earlier today, so I don’t feel any need to look at the wikipedia page you linked, informative as it may be. Actually I initially considered a fair coin and then modified it to a biased to see if that would shed any more light on the matter or refresh my memory any further.

Anyway, my point here, if I may be so bold and without wanting to be rude or nasty, is that I never asked for an explanation of the subject. My comment was about how something khisana said doesn’t sound right and I don’t think this means that I’m not understanding the subject: I think it means that what he wrote is incorrect, or at best, is poorly expressed.

I actually find it odd how it’s so often the case when I criticise someone’s words that they or someone else seem to think I’m asking them to explain the subject to me as if I’m a student asking a teacher for an explanation, when I don’t recall ever coming across as asking for help, but rather I came across as disagreeing with and criticising what they said, actually putting me in the position of the teacher rather than a poor, bewildered student as they seem to see me.

I’ve never seen any evidence that the golden packs have different odds but tbf there’s nowhere near enough data to know either way.

However I’m pretty sure they “can’t” have because of potentially breaking the pity timers, which would screw up their reported odds. So my plan is to wait until I’ve gone 9 packs without an epic and then the golden pack must contain an epic (or better).

Still even with being able to “rig” the golden pack, your still massive down in value buying it with gold. £££ wise, I always think of it as £1 = 1 pack = 100 dust, which is actually only a minor simplification (sales/average pack dust value). So you’d be close to even, if you wait for the epic pity timer, which shows how you would still be down if you don’t have that luxury/patience and just roll the dice

It’s also rigged so it’s not a flat percentage (it changes exponentially after pack 32). It’s statistically likely that I never have and never will hit pack 39 in the classic set without a legendary. If you get to pack 38 the odds have actually increased to the point where you have a 50% chance of getting the legendary in the next pack as opposed to it still being 5% and then 100% in pack 40. I’ve never seen any evidence of this being true with the epics, so presume they work differently and are just a flat 20% and then 100% for pack 10.

Interesting. By the way, when you talk of being close to breaking even if you hit a golden epic, I don’t follow. Unless I’m mistaken, it’s far from breaking if you buy the adventure with gold (see the edit to my OP for details of my calculations on this and then just add the 400 dust gained from disenchanting the golden epic to the figure for the dust gained from buying the adventure), whereas if you buy it with real money it seems to me that you already break even without hitting a golden epic.

Ah ok thanks for getting back to me. I was actually using £15.72 (well £15 to be exact), by just converting the USD price to GBP. I didn’t realise they had a different GBP price, but I guess £15.72 would be a bit of a weird price, so what not round it up to £17