What Are The Odds?

[retnab]

Does anyone have the equation handy that would let you calculate the odds of drawing certain value or higher cards in your hand each turn?  Specifically I'm thinking of Ramos, who needs 11T or better (four cards in 54) to summon his three Spiders.  6 cards base hand size, up to 8 from Arcane Reservoir and Hannah, up to 10 with them and drawing/discarding 2 cards with a soulstone.  Any math inclined players on here know how to do this?

[Tawg]

4/54 + 4/53 + 4/52 + 4/51 + 4/50 + 4/49 + 4/48 + 4/47 = (7.4% + 7.5% + 7.7% + 7.8% + 7.9% + 8.1% + 8.3% + 8.5%) = 63.2%  Raise that about 18% if you use a Soulstone to draw 2, discard 2.

That isn't exact math though, the probability changes slightly when you account for the fact that you may have drawn one already.  But this math should cover the event that the last card you draw is the one you're looking for, or maybe not.  Each card you draw should have a X/Y chance, where X is how many there are in the deck (4 cards you're looking for) divided by Y or remaining cards in deck.

So higher than half the time, if you have a hand of 8, almost guaranteed if you are stoning for it.  But at the same time, if you're fishing for a Tome on a card by spending Soulstones, you would be just as well off using whatever high card you have in hand and spending the Soulstone to add suits.  Then you could potentially save the Soulstone if you don't end up using Ramos in the way you planned (If he gets paralyzed or a better move comes up).

EDIT: You should be able to do the math out like I did for any given set of cards, especially if you're only interested in drawing a single card from a set of cards, like your Ramos question.  You could even account for cards discarded before drawing by reducing the cards in deck value.  When the math would get more complicated is if you are looking for a specific combination of cards, or you want to know your chances to flip a given card in the middle of a turn.  Granted those probably aren't too hard to figure out, I just don't know any equation off hand, and it's been quite some time since my last statistics course.

[Bengt]

That's not how you calculate this, you have to account for the leafs of the probability tree where you drew desired cards as well. The easiest way is to calculate not drawing any of them and reverse it: 1 - (50/54 * 49/53 * 48/52 * 47/51 * 46/50 * 45/49) = ~38%. Goes up to ~48% with 8 cards. ~57% with 10 cards. This is only valid for the first turn though as in the following turns you'll probably discard some and keep some cards so the deck is a bit shaped and you draw less cards.

[retnab]

Cool, thanks for that!