Pokémon TCG: Sword and Shield—Brilliant Stars

Pokemon Probability

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Dark Ninja

New Member
I just wanted to know if anyone here knows the formula for figuring out the probability of drawing a certain card.
 
It is basic math. Say I ran 4 Celio's Network. I have one in my opening hand and I play it to search my deck. I see that I have three left in my deck, thus I have none in my prizes. There are now 45 cards in my deck: 6 in prizes+7 in hand+1 drawn+ I searched= 15 removed, so 60-15=45. If there are three celio's left and there are 45 cards in my deck then I have a 1/15 or 6.6666....% chance of drawing it.
 
Could I just mention that a very common mistake is not considering the following:

Code: 
Prizes are layed AFTER a basic pokémon is chosen.

What this means is that if you play a single dragonite ex delta in your deck, you can't just say "6 prizes out of 60 = one tenth of the deck therefore 1/10 of the time draggy gets in the prizes". It would however be correct to say "whatever the starting hand, it has a basic pokémon that is laid (the rest of the hand doesn't matter and if you don't you would mulligan so it doesn't change odds) which leaves 59 cards of which 6 are laid. This means draggy has a 6/59 chance of being prized or 10.1%"

Personally, I use a computer program I wrote which does starting hands, topdecks, prizes, master ball and such consistency and saturation across turns. Magic Umbreon's Repeat Draw Equation Routine or M.U.R.D.E.R. for short. I murder decks :)
 
The probability is considerably increased if that card is actually in your deck. If not, then its 0.
 
Dom said:
The probability is considerably increased if that card is actually in your deck. If not, then its 0.
Click to expand...

Uhh....spam?

Anyways, I just do the math in my head for each scenario, almost every turn. If you really want to be able to do that quickly, practice probability! lol, it took a while to be able to do it quickly enough to do it every turn W/O it being considered a stall.
 
I think Dark Ninja is interested in perparation odds. In-game odds are completly different to the odds you consider making a deck. The former are develloped the more you play with the deck.
 
I mean like drawing a certain card in your opening hand. I realize for one card its # of copies of the card in deck/ # of cards in deck. I think it's something like 4/60 for the first draw, 4/59 for the second and so on, but after I do that I don't know what to do.
 
Oh I see, he means like

What is the probability of drawing a basic in my hand of 7 if I have 12 basics in my deck?
Out of that what is the possibility of only drawing 1 basic in my hand of 7?
Out of that what is the probability of that one basic being an absol if i have 3 absol in my deck?

is that what you mean?
 
Absoltrainer said:
Oh I see, he means like

What is the probability of drawing a basic in my hand of 7 if I have 12 basics in my deck?
Out of that what is the possibility of only drawing 1 basic in my hand of 7?
Out of that what is the probability of that one basic being an absol if i have 3 absol in my deck?

is that what you mean?
Click to expand...

exactly.
 
The term you're looking for is hypergeometric distribution.

To solve these, you need to know the standard combination without repetition formula:

[sub]n[/sub]C[sub]r[/sub] = n! / (r! (n - r)!)

Where n is the total size of your set, and r is the number of selections you're making.

Now let's define the following variables:

x = number of a particular card (or type of card) you have in the deck
d = number of cards in the deck (always 60, in Pokemon's case)
z = number of cards you are drawing (7, in the case of your initial hand)
s = number of successes (how many of that particular card you're wanting to get)

And since we can define these questions as a binomial probability (either we got the card we wanted or we didn't), we can simplify the formula for this hypergeometric distribution to:

[sub]x[/sub]C[sub]s[/sub] * [sub](d - x)[/sub]C[sub](z - s)[/sub] / [sub]d[/sub]C[sub]z[/sub]

But that gives you the probability of drawing exactly s. If you're wanting to know the probability of getting at least 1 of something, it's easier to let s = 0 (calculate the probability of not getting the desired cards) and then subtract that probability from 1.

So let's answer the three questions posed already:

What is the probability of drawing a basic in my hand of 7 if I have 12 basics in my deck?

[sub]12[/sub]C[sub]0[/sub] * [sub]48[/sub]C[sub]7[/sub] / [sub]60[/sub]C[sub]7[/sub]
= 1 * 73629072 / 386206920
= 0.1906466927

1 - 0.1906466927 = 0.8093533072

So an 81% chance of drawing at least 1 basic in your initial hand with 12 in the deck.

Out of that what is the possibility of only drawing 1 basic in my hand of 7?

[sub]12[/sub]C[sub]1[/sub] * [sub]48[/sub]C[sub]6[/sub] / [sub]60[/sub]C[sub]7[/sub]
= 12 * 12271512 / 386206920
= 0.3812933854

38% chance of a one-basic start

Out of that what is the probability of that one basic being an absol if i have 3 absol in my deck?

Well, if 3 out of 12 Basics in your deck are Absol, that's 25%. So 25% of your one-basic starts will be Absol.

0.3812933854 * 0.25 = 0.0953233464

Roughly 9.5% overall chance of starting with Absol only.
 
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That was mathy:lol:. I guestimate when I try to see my chances of getting a card. I play Claydol though so that increase my odds:)
 
Remember you're more likely to start with holos than non holos. Factor that into your math.
 
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