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

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?

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.

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

From my website: Opening Hand Basic Probability: http://www.southernmarylandpokemon.com/assets/documents/openinghandprobability.pdf More numbers can be found on pojo.com http://www.pojo.com/Features/X-Act/index.shtml