Expected Value of BETA Packs - Steem Monsters š š¾ - POTIONS CALCULATION INCLUDED!
As you might have noticed on stemmonsters.com website, there are only 116,694 packs left, which, according to the banner, is equal to:
As I usually do, I wanted to make some calculation about the expected value of opening beta packs, and how using 100% legendary + gold potions is worth compared to the regular opening.
Price for BETA packs is 2$ per pack, however if you use the 100 packs offer you get 10 free and with the 500 packs offer 75 free. Given that the second one is rather expensive and that packs are also available on Steem-Engine (currently 1.73$), I'll calculate prices based on a cost per pack of 1.8$, which is roughly what you get with the 110 packs for 200$.
I'm also using the @yabapmatt statement of:
Each card has a 20% chance of being rare, 4% chance of being epic, and 0.8% chance of being legendary, as well as a 2% chance of being gold.
IMPORTANT NOTE: price on market vary rapidly, this is calculated as 13th of September, 2019.
Number of Legendaries: 15
Total minimum price of Legendaries: 146,45 USD
Average price per Legendary: 146,45/15 = 9,76 USD
Probability of opening a Legendary per pack: 5x0.8% = 4%
Added value on Legendaries, per booster: 4%x9,76 USD = 0,39 USD
Total minimum price of Gold Legendaries: 2527 USD
Average price per Gold Legendary: 2527/15 = 168,5 USD
Probability of opening a Gold Legendary per pack: 5x0.8%x2% = 0,08%
Added Value on Gold Legendaries, per booster: 0.08%x168,5 = 0,13 USD
TOTAL VALUE OF OPENING A LEGENDARY, GOLD OR NOT, PER BOOSTER = 0,52 USD
Number of Epics: 17
Total minimum price of Epics: 19,95 USD
Average price per Epic: 19,95/17 = 1,17 USD
Probability of opening a Epic per pack: 5x4% = 20%
Added value on Epic, per booster= 20%x1,17 USD = 0,24 USD
Total minimum price of Gold Epic: 296,01 USD
Average price per Gold Epic: 296,01 /17 = 17,41 USD
Probability of opening a Gold Epic per pack: 5x4%x2% =0,4%
Added Value on Gold Epic, per booster: 0.4% x 17,41 = 0,07 USD
TOTAL VALUE OF OPENING AN EPIC, GOLD OR NOT, PER BOOSTER = 0,31 USD
Number of Rares: 23
Total minimum price of Rares: 17,85 USD
Average price per Rare: 17,85/23 = 0,77 USD
Probability of opening a Rare per pack: 5x20% = 100%
Added value on Rare, per booster= 100%x0.95 USD = 0,77 USD
Total minimum price of Gold Rare: 114,04 USD
Average price per Gold Rare: 114,04/23 = 4,96 USD
Probability of opening a Gold Rare per pack: 5x20%x2% = 2%
Added Value on Gold Rare, per booster: 2% x 4,96 = 0,09 USD
TOTAL VALUE OF OPENING AN EPIC, GOLD OR NOT, PER BOOSTER = 0,86 USD
Number of Commons: 24
Total minimum price of Commons: 2,39 USD
Average price per Common: 2,39/24 = 0,10 USD
Probability of opening a Common per pack: 4 per pack
Added value on Common, per booster= 4x0,10 USD = 0,40 USD
Total minimum price of Gold Common: 38,971 USD
Average price per Gold Common: 38,971/24 = 1,63 USD
Probability of opening a Gold Common per pack: 4 x 2% = 8%
Added Value on Gold Common, per booster: 8% x 1,63 = 0,13 USD
TOTAL VALUE OF OPENING A COMMON, GOLD OR NOT, PER BOOSTER = 0,53 USD
BOOSTER VALUE, EXPECTED VALUE = 2,47 USD
DEC cost of 100% Gold Potion = 25 000 DEC
DEC cost of 100% Legendary Potion = 20 000 DEC
DEC cost in Steem = 288 Steem (at 0.00642 price)
DEC cost in USD = 45$
Recalculation of Gold values (4% probability instead of 2%)
Recalculation of Legendary value (1,6% probability instead of 0,8%)
Probability of opening a Legendary per pack: 5x1,6% = 8%
Added value on Legendaries, per booster: 8%x9,76 USD = 0,78 USD
Probability of opening a Gold Legendary per pack: 5x1,6%x4% = 0,32%
Added Value on Gold Legendaries, per booster: 0,32%x168,5 = 0,53 USD
The added value for the other type of Gold cards (Epic/Rare/Common) is doubled.
Added Value on Gold Epic, per booster: 0.8% x 17,41 = 0,14 USD
Added Value on Gold Rare, per booster: 4% x 4,96 = 0,18 USD
Added Value on Gold Common, per booster: 16% x 1,63 = 0,26 USD
BOOSTER VALUE, EXPECTED VALUE = 3,30 USD
Please let me know if I made any mistake in calculation or just review this. I appreciate your comments too.
What do you guys think? Do you open @steemmonsters pack?
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Brought to you by @tts. If you find it useful please consider upvoting this reply.
So, three problems, two that matter for the end math and one that doesn't.
When you do probability of opening a card per pack you're doing conditional probabilities wrong. This would be more obvious if it weren't for the 1 rare guarantee, because without that you wouldn't get 1 rare per pack even though it's 20% likely for any particular card. To get conditional probability you account for the probability that it won't happen and multiply those probabilities together. So the probability of at least one rare showing up in a pack without the guarantee would be (1 - (1-.2)^5): that is, the probability that the first card isn't a rare (80%) times the probability that the second card isn't a rare (same) and so on for all five gets you to probability that no rare has occurred, which is 32.768%, then subtract that from 1 and you get 77.232% probability of getting at least one rare in a pack.
For the end EVs it doesn't matter that you're doing 5 cards at a time, though. It just makes your "probability of opening X per pack" lines inaccurate.
The one that does matter for the end math goes back to that one rare guarantee, because that's an extra condition imposed on the generating algorithm above the 20/40/0.8 numbers, and it makes a difference. @cryptoeater happily calculated this out from real data early on in Beta, and the real percentages per card adjusted for that are:
common 0.689825
rare 0.243135
epic 0.0392
legendary 0.00784
g common 0.014078
g rare 0.004961
g epic 0.0008
g legendary 0.00016
The third problem is that rather than just doing probabilities on commons you're assuming four per pack and of course there are many packs that don't have four commons in them. That may not be a large error.
When I run your average price numbers through my spreadsheet I get a pack ev with no potions of $2.335 and with both potions of $3.369, so despite the errors you're not very far off.
Thanks so much for your accurate reply @tcpolymath.
I got your point on the conditional probability and you are absolutely right, that surely matters. I forgot that you could calculate P(E) by 1-P(Opposite Event), a bit of time passed since my math lessons. At least calculations were not far off!
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