Expected Value of BETA Packs - Steem Monsters 🐉 👾 - POTIONS CALCULATION INCLUDED!

in #steemmonsters5 years ago (edited)
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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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