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Brainy mathematical folk, can you tell me if I'm calculating this correctly?
I want a single specific card X in my opening hand. I have a standard size deck of 33. I draw my five cards, then mulligan all five away if card X is a noshow. What is the percentage chance I will have cardX in my hand at the start of turn 1?
For a deck with 1 copy of card X, it's (1/33)+(1/32)+...+(1/24) = ~35% And a deck with 2 copies of card X is (2/33)+(2/32)+...+(2/24) = ~70%
And, if still unsuccessful, the very next card you draw would give (if 1 copy in deck) 1/28 = ~3.5% (if 2 copies in deck) 2/28 = ~7%
Am I mathing good?
(For anyone interested, I'm wondering what the probabilities are for getting a Leo di Luca playable in your opening 3 actions.)

Mr. D
United States Oneonta New York

You are not going to add those probabilities together. Instead you need to calculate the odds that you won't draw the card.
1 copy in deck: (32/33)*(31/32)*(30/31)*(29/30)*(28/29) = ~85% that you don't draw the card. Mulligan = ~85% that you don't draw it again.
So 85% x 85% = ~72% that you won't get the card in either the first draw or mulligan.
2 copies in deck: (31/33)*....(27/29) = ~72% to that you don't see it in your first draw. Mulligan = another ~72% chance you don't see it.
So ~51% chance to NOT draw the card.
In summary: With one copy of the desired card, you will be able to start the game with it approximately 28% of the time. With 2 copies of the desired card, you will start the game with it approximately 49% of the time.

Rob Tarr
United States Floyds Knobs Indiana

Not quite. I find it's easier to calculate the chance of something NOT happening, and then subtracting it from 1 to find the chance that it WILL happen. The math is simpler.
The chance of it NOT being in ANY of the first five cards is: (32/33)×(31/32)×(30/31)×(29/30)×(28/29) = 85%
So the chance of it ACTUALLY being in those five cards is 100%85%, or 15%.
With two copies of the card, the numerators change: (31/33)×(30/32)×(29/31)×(28/30)×(27/29) = 72%
And the chance of at least one of the two copies being in your hand is 100%72%, or 28%.
Your math for the very next you draw draw is spot on, though.
Edited to add  Well, totally ed, AND I forgot to consider the mulligan. Must be election day or something.

Rob Tarr
United States Floyds Knobs Indiana

Tubarush wrote: Mulligan = ~85% that you don't draw it again. Keep in mind that the mulligan has you set the first five cards aside. You don't shuffle them back in until after you draw your new starting hand. That would change the second calculation for each case, since you're now starting with only 28 cards.
That ups the chances of starting with the card to 30% with one copy, and 52% with two copies.


Thank you very much! And those numbers feel (not a very mathy term, I know) much better than my 35%/70%; I felt pretty sure I was messing up. Glad I asked.
Pay attention in school kids!


remember that weakness cards cannot be part of your first 5 cards, so redo the math with a 31 denominator


Nushura wrote: remember that weakness cards cannot be part of your first 5 cards, so redo the math with a 31 denominator
Oh, right. Of course. Somehow, this mistake stings the most, ha.
Okay, let's try it.
2 copies in the deck:
1(29/31)*(28/30)*(27/29)*(26/28)*(25/27)*(24/26)*(23/25)*(22/24)*(21/23)*(20/22) = 54.9% before Turn 1.
(With subsequent draws of 1(26/28)= 7.1%, 1(25/27)= 7.4%, etc.)
1 copy in the deck:
1(30/31)*(29/30)*(28/29)*(27/28)*(26/27)*(25/26)*(24/25)*(23/24)*(22/23)*(21/22) = 32.3% before Turn 1.
(With subsequent draws of 1(27/28)= 3.6%, 1(26/27)= 3.7%, etc.)
Nice to know. Thanks again for the lessons.

Mr. D
United States Oneonta New York

tarrkid wrote: Keep in mind that the mulligan has you set the first five cards aside. You don't shuffle them back in until after you draw your new starting hand.
Did not know this. That's what I get for paying attention to a game I don't have yet. Sorry.

Rob Tarr
United States Floyds Knobs Indiana

Tubarush wrote: tarrkid wrote: Keep in mind that the mulligan has you set the first five cards aside. You don't shuffle them back in until after you draw your new starting hand.
Did not know this. That's what I get for paying attention to a game I don't have yet. Sorry. No worries. I had just read through the LearntoPlay PDF last night, and somehow remembered that little factoid.

Darby Harris
United Kingdom Manchester

Another, perhaps easier, way to do the calculation:
For 1 card you just need the probability that the card is in the first 10 of your deck, given how the mulligan works in this game, so we can just use 10/31 = 32.3%
It's a bit more complicated when you have 2 of the card, but the formula would be:
n*(2L  n  1)/(L^2  L)
where n is the number of cards being drawn and L is the size of the deck. Here n = 10 and L = 31, giving 54.8%
This is obtained from writing the expressions from previous posts as a product of a sequence:
1  ∏(Lk1)/(Lk+1), where the product is over k=1 to n.
Then just identify cancelling terms and simplify.

mathew rynich
United States Connecticut

That's interesting. I guess I never notice how generous this game was with it's starting hand. An approximately 55 percent chance to get a must have card in your opening hand is really nice. If all you want is for example a weapon and you include redundant cards of the same type it makes it very likely that you'll have what you need at game start. By this math just including 2x .45 Automatic and 2x Machete means you have an 80 percent chance to have a weapon in your opening hand. Roland has his signature gun which ups his percentage to 88 percent. Pretty cool.
Daisy has a very good chance of grabbing a desired Tome in her opening hand considering she can include two Research Librarians in addition to the two copies of say Old Book Of Lore. Between those two cards you have an 80 percent chance of getting Old Book Of Lore in your opening hand. Throw 2x Medical Texts or 2x Encyclopedia in the deck and now it's a 93% chance that she'll be able to use her ability on turn one. Have all three tomes 2x in your deck plus 2x Research Librarians and you have a 97 percent chance to have a tome opening hand... though now that's eight cards, but the tomes are all good so I guess that's not unreasonable.

Scott Hill
United Kingdom Cambridge Cambridgeshire

Anemone1777 wrote: Another, perhaps easier, way to do the calculation:
For 1 card you just need the probability that the card is in the first 10 of your deck, given how the mulligan works in this game, so we can just use 10/31 = 32.3%
It's a bit more complicated when you have 2 of the card, but the formula would be:
n*(2L  n  1)/(L^2  L)
where n is the number of cards being drawn and L is the size of the deck. Here n = 10 and L = 31, giving 54.8%
This is obtained from writing the expressions from previous posts as a product of a sequence:
1  ∏(Lk1)/(Lk+1), where the product is over k=1 to n.
Then just identify cancelling terms and simplify. So much easier...


