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I have an equation I use that's a bit more limited, but simpler to use :) . All it will tell you is the probability of drawing at least one of a certain card which you have a certain amount of in a certain number of draws,
p = ( ( (t-c)! / (t-c-d)! ) / ( t! / (t-d)! ) )
p = Probability
c = Number of the type of card you're testing for in the deck
d = Number of cards you're drawing
t = Number of cards in the deck
It looks more complicated than it really is, since I can't use fraction bars :P .
So if I had six precognitions in a 30 card deck, and I wanted one in my hand of seven, then the probability would be:
p = 1-((30-6)!/(30-6-7)!) / ((30)!/(30-7)!)
p = 1-.17
p = .83
p = 83%
| p(X=1) | 39,669 |
| p(X=2) | +31,317 |
| p(X=3) | +10,439 |
| p(X=4) | +1,491 |
| p(X=5) | +0,081 |
| p(X=6) | +0,001 |
| p(X>0) | =82,998 |
http://elementscommunity.org/forum/index.php/topic,7893.0.htmlAre you able to do this? I cant... ;). My knowledge in informatics is too limited. I elected mechanical engeneering and maths as advanced courses. But at least I can take the work typing all this numbers on. Too I guess for pillars it can be more intersting to know the chance getting at least 3 or 4.(I know about the QI and the QI is working pretty well, but it could be a nice addition)
I actually think we're probably all using the same formula to get the same answers, but that table puts it nicely too. I was almost envisioning a 3D graph with the x axis being the number of cards wanted included in the deck, the y axis being the number of cards in the deck, and the z axis being the number of that card wanted in a hand...
Actually, I'm calculating for "At least 1," (I should edit that in), and you're right that the "-1" isn't needed.Ohh, sorry, I thought I read "exactly 1" or rather I misinterpreted your sentence:
So if I had six precognitions in a 30 card deck, and I wanted one in my hand of seven, then the probability would be:
The draw chance of any card depends not only on the number of copies of the card and the size of the deck, but also the number of zero-cost cards in the deck.
Because the mulligan is a bit weird in EtG (it leaves the original hand on top of the deck), I don't see an easy way to calculate the correct probabilities. Populating the tables using a simulator may be the only real solution.
The game reshuffles the deck after the mulligan, doesn't it?We don't know if it shuffles again or if the cards are sent to the bottom of the deck. Take the option you like most, the result won't change considerably
[15:39:38] Chapuz: 1) the deck is shuffled. 2) You draw 7 cards. 3) if you didn't get any 0-cost cards, the deck is reshuffled and you draw another 7 cards
[15:39:45] Chapuz: am I right?
[15:39:53] ColorlessGreen: yeah, 4 year old outdated thing in tutorials != current thing in stats forum
[15:39:55] ColorlessGreen: no
[15:40:16] Chapuz: no?
[15:40:19] ColorlessGreen: (3) if you didn't get any 0-cost cards, a new hand is drawn out of the cards you don't currently have in your opening hand and your opening hand is put back on top of the deck
[15:40:30] ColorlessGreen: meaning the top seven cards are guaranteed non-zero-cost
[15:40:41] ColorlessGreen: and the second opening hand has a higher than normal chance of zero cost cards