Pokémon TCG: Sword and Shield—Brilliant Stars

Unrandomness on PTCGO?

I actually have noticed this non-randomness of ptcgo as well. My deck has 3 single copy cards and 7/10 of the games i play i start with at least 2/3 single copy cards from my deck
 
There seems to be a pack opening bias that occurs.....7 tornadus in 54 packs of EP or 0 shaymin out of 40 UL.......(6 roserade)
 
Big deal! I opened 80+ Triumphant booster packs and did not pull a single Yanmega, Magnezone or Mew Prime.

This doesn't mean anything, of course, but that's how conspiracy theories get started.
 
I doubt that that heads tails flip will be incorrect.

It is possible to design a bad deck shuffle algorithm. This used to be a big problem in days gone by, but the WWW should have eliminated the bad algorithms from all but the last few adherents to the school of not-invented-here.

The pack opening should absolutely not be random! Google the Coupon Collector Problem to understand why. Unfortunately what I've seen suggests that however pack randomisation has been implemented it is not good enough to avoid justified complaints.

====

I think cabd is making an error in the calculation of how unlikely it is to prize three off of a group of four in pokemon. My own estimate is anything between 1:60 and 1:200. With those odds Pooka's report of 5 games out of 200 shifts from incredibly improbably to just unlucky.
 
Last edited:
I doubt that that heads tails flip will be incorrect.

It is possible to design a bad deck shuffle algorithm. This used to be a big problem in days gone by, but the WWW should have eliminated the bad algorithms from all but the last few adherents to the school of not-invented-here.

The pack opening should absolutely not be random! Google the Coupon Collector Problem to understand why. Unfortunately what I've seen suggests that however pack randomisation has been implemented it is not good enough to avoid justified complaints.

====

I think cadb is making an error in the calculation of how unlikely it is to prize three off of a group of four in pokemon. My own estimate is anything between 1:60 and 1:200. With those odds Pooka's report of 5 games out of 200 shifts from incredibly improbably to just unlucky.

Probably, it was quick tired-calculations without an actual calculator. Also, why is it that everyone typos my name as cadb?
 
I just made 4 mulligans in a row – but that's not the strange thing.

The program always*put back an RDL piece as the last card, which was most surprisingly the first one it drew afterwards (even if the rest of the new hand was different).

I would like to quote myself because I believe this is a common pattern: when you mulligan, the last card you shuffle into your deck is always the same as the first one you draw then.

Anyone noticed this as well?
 
I've prized 3 copies of a card on about five occasions now. Now, having this happen once is extremely improbable, but five times over the course of probably 200 games is a bit absurd. Still, I could be just that unlucky; it's hard to tell.
Think of it like Poker: what are the odds of drawing three of a kind. How many times would you hope to have three of a kind in 200 games of 5 card draw? The Poker analogy works as long as your deck has many cards played as four off copies.
 
Actually, I played only three copies of the card; that's what made it so strange.
that does change things :( The odds go back up towards the highly improbable :( I'd need to see the skeleton of the list to get a better idea of just how improbable
 
Any deck I've built so far that plays a 4-2-4 magnezone line manages to prize at least two zones per game.
Posted with Mobile style...
 
I would like to quote myself because I believe this is a common pattern: when you mulligan, the last card you shuffle into your deck is always the same as the first one you draw then.

Anyone noticed this as well?

I've noticed this too. Happened to me in my last game where I drew Twins 3 times as my first card after 3 mulligans. This is a bug we should report probably.
 
^That is a verified glitch. At least I know I'm not crazy! There are some unrandom things currently in the beta version of PTCGO...
 
Actually, I played only three copies of the card; that's what made it so strange.

Another relevant fact would be whether the card is a basic Pokemon. Because the game doesn't start until you get one, there's a slightly lower chance of having a basic prized than other types of cards.

The core of any computer randomization routine is a seed value-this value determines from what point the algorithm begins generating values. Typically this is based on fractions of a seconds since startup (or some other fixed point in the computer's operation). Generally this is based in some measure on the user's actions (for example, the last time they rebooted the system or when they launched the application in question), since people tend not be able to control their actions that precisely.

On top of these values, there's the shuffling algorithm-in a deck with n cards, there are n! possible arrangements of the cards (some of which would look identical due to multiple copies of cards in the deck). In a perfect world, there would be an equal probability of all n! arrangements, but as you may have noticed from your MP3 player, such algorithms are rarely (if ever) this evenly distributed.

(End of computer science lesson-for now, at least)
 
Yes, I believe the shuffling algorithm is flawed to some degree with the game, mainly because of the mulligan glitch. Once you mulligan, your deck is no longer random because the last card you shuffle in goes to the top of your deck always. Since the first card you draw will not be a Basic Pokemon, it also messes up your odds of having another mulligan; this is not good.
 
Assuming 12 basics in our deck, the probability of NOT drawing a basic in 7 cards is: (48/60)*(47/59)*(46/58)*(45/57)*(44/56)*(43/55)*(42/54)=0.19064 or 19.1%

With the glitch about mulligans, any subsequent attempt to draw a new hand, instead of the same 19.1% will be:0.238 or 23.8%

That may not seem like a bunch, but it's very much so a statistical error.

(I think my math is right, correct me if i'm wrong)
 
You could have also written (48!/41!) / (60!/53!) = 0.19064 = :)

For the 6 card hand you'd write (47!/41!) / (59!/53!) = .238

Yay for math
 
Yeah, I know, but i wanted to write it the long way to illustrate it for people who don't know probability and factorial math.
 
Back
Top