Hi there,
I would like to know and learn how to calculate the odds of the following cases:
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have 2 copies of the same (any) card in the opening hand. The deck is build of 2 copies of every one of the 15 cards it holds.
Opening hand should be calculated twice: for the one that goes first, and the one that goes second.
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Same deck as before, having 2 copies of the same card after drawing 13 cards. (including the cards that are being drawn for the opening hand.
(if there is anything missing - ask. these issues really bother me)
Thank you!
I’m going to do this pre-mulligan, since post-mulligan depends on your decisions.
The easiest way to compute the odds of having 2 copies of at least one card is to compute the odds of it NOT being the case.
Going first, to NOT have 2 copies of any card, you need to (1) first draw any card (100% chance), (2) then draw any of the remaining 14 cards, of which there now are 28 in a 29 card deck (so 28/29 chance), and (3) finally draw any of the remaining 13 cards, of which there are 26 in a 28 card deck. Total odds to NOT have any duplicated card pre-mulligan is 1 * 28/29 * 26/28 = ~89.7%. So the chance to HAVE two copies of any card pre-mulligan is 100 - 89.7 = ~10.3%.
Going second, the computation is the same but now the fourth card has to be one of the 12 remaining: 24 out of 27. So we get 1 * 28/29 * 26/28 * 24/27 = ~79.7%. And the chance you DO have a duplicate is now ~20.3%.
Again the same trick, but now the formula becomes a lot longer. But it’s still the same pattern.
To not see a single card twice: 1 * 28/29 * 26/28 * 24/27 * 22/26 * 20/25 * 18/24 * 16/23 * 14/22 * 12/21 * 10/20 * 8/19 * 6/18 = ~0.0898%.
Which makes the chance that you DO see two copies of at least one of your cards equal to ~99.9102%.
Hope this helps.
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you can just google a probability calculator. there are such developed already for other card games
using the standard yugioh probability calculator we use at yugioh pro level (note that it can calculate everything for hearthstone except taking mulligan into account)
opening:
1st: 1,38%
2nd: 2,3%
13 card: 17,93%
17 (13+4 since your were vague): 31,26%
Thank you very much .
It did help. It was very clear !
I have one additional question: In your last calc you say there is almost 100% to have a double copy of one of the cards. that is for any card. does it change if I want a specific card? lets remember the context of this math: heathstone. So having 2 copies of some card could mean nothing while having 2 copies of another could mean a win. I have a feeling its a diff chance but I cant do the math to prove it… am I correct?
The chance to have 2 copies of a specific card from your deck after 13 draws is definitely different than the chance of having 2 copies of any card after 13 draws.
First, we need to work out in how many orders you can draw 13 cards to include those 2. So the first could be the first card drawn, or the second, … or the last. And the second could be any except the one the first was. That means that there are 13 x 12 = 156 possible ways to draw 13 cards and have those specific two included. (If we only look at “the card” vs “other card” and don’t care about what other card that is). Except that the two cards are identical so we counted double (since “card found on draws 7 and 12” or “on 12 and 7” are the same). So there’s actually 78 options.
For each of those 78 ways, the chance that it happens is computed effectively the same. E.g. to get those two as 12 and 13, the chance is 28/30 (since the first draw has 28 “good” cards out of 30) * 27/29 * … * 18/20 (for the 11th card) * 2/19 * 1/18 (for the final two cards that are the ones we want). This can be rewritten as:
(28 * 27 * 26 * 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18 * 2 * 1) / (30 * 29 * 28 * 27 * 26 * 25 * 24 * 23 * 22 * 21 * 20 * 19 * 18). For each other combination, the order of the terms in the multiplication changes, but the terms are the same; since order is irrelevant in multiplication that means that the above formula is the chance for each of the 78 possible ways.
Using rules of simplification, we can remove the terms that are on both sides of the division, for a much simpler actual calculation: (2 * 1) / (30 * 29).
And since there are 78 ways, each with a (2 / 870) chance of happening, the chance that one of them happens is 78 * (2 / 870) = exactly 20%.
If you put two copies of a card in your deck and you need to have both in your hands after drawing 13 cards, then that will happen in one out of five of your games.
Obviously, smart mulligan and running card draw in the rest of your deck will significantly improve your chance. And adding survivability cards will help you live longer then 13 card draws, which will also increase the chance to get those specific two.
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Dude!
That was a great explanation!
Thank you!
This basically means I can calculate the chances to get any card and any combo at any given turn right ? just need to tweek the calculation…
Thanks!
Basically, yes. But with some caveats. Mulligan changes the odds (which is why I focused on seeing certain cards after X draws, not turns). Card draw is a thing. There are some decks / cards that affect your deck. On the positive side, cards such as Tracking or Sightless Watcher. For your opponent, cards such as Bad Luck Albatross or Gnomeferatu.
But as long as you look at number of draws and assume no cards being shuffled into, destroyed from, or changed in your deck, you can do the math for any combination of cards and turn number.