Cantrip Math regarding 3-vs-4 of's.

[tescrin]

The below was derped. Disregard it's statistical inaccuracy until further notice.


I was tinkering with a list that had the slightest blue-splash and was arguing with myself whether or not to include Brainstorm. Certainly the deck made sense as it was without it and that dropping any 4-ofs for it seemed dumb; why reduce consistenc...OHHHHHH


So I mathed it out:
Chance of 4-of in top 8 (you usually don't BS T1) -> ~44%
Chance of a 3-of in top 8 -> ~35%
Chance of brainstorm + 3-of in top 11 -> 24.3%
Chance of Brainstorm and 3-of in hand -> 8.5%

and finally, the actual comparable number!
Chance of Brainstorm in hand and card you need in the top 3 (after drawing a card): 35% + 24.3% - 8.5% = 50.8%!


In other words; it mathematically makes more sense to include BS instead of a one of the 4-of cards you have in every 4-of case that may hinge on it. This also works for Ponder; albeit without all of the nice things BS does for you.



I realize most wouldn't have started at the point I was at; but consider you're deck looks something like:
4 x
4 x
4 x
4 x
4 x
4 x
3 x
1 x
1 x
1 x

it starts to make sense that even though you're trying to build in consistency with 4-ofs, you could eat taxing yourself 1-mana to allow BS into the deck and *increase* the chances of finding any given 4-of you need by making it a 3-of and swapping the card. To me that was a neat revelation and worthy of sharing.


EDIT: Someone may want to math out the probability you'll find a 4-of sideboard (say, you usually have 4 grave-hate cards for dredge/elves/reanimator) in a mulligan vs. keeping a hand with a cantrip and searching for it. I would not be surprised if (unless you need it T1) you're better off keeping the current hand.

[MrShine]

This seems pretty neat... but I guess you also have to compare the chances of BS+a 4-of? Because there must be sometimes when it is better to have BS AND 4-of whatever card you absolutely need to find.

Anyone else with a decent math background out there? My knowledge is rudimentary at best and it would be nice to hear a couple more evaluations.

[Barook]

This seems pretty neat... but I guess you also have to compare the chances of BS+a 4-of? Because there must be sometimes when it is better to have BS AND 4-of whatever card you absolutely need to find.

Anyone else with a decent math background out there? My knowledge is rudimentary at best and it would be nice to hear a couple more evaluations.

Edit: Math was done assuming 4 copies of a cantrip.
That would be Top 8 44.5% + 30.7% in the Top 11, with 14.5% starting chance of both in hand, resulting in or 60.7% total.

If you want to disregard Ponder's shuffle + draw since I can't be assed to do the math for that and if run it alongside 4x Brainstorm, you look at a 45.5% chance to find your stuff. Chance of both at least 1 cantrip in hand + key card is 24.4%, or 65.6% chance total to find your stuff (plus whatever the Ponder shuffle draw brings to the table). This number assumes 4x Ponder + 4x Brainstorm.

If you actually want to mess around with numbers, Magic Workstation has an analysis tool under "Statistics --> Deep Analysis --> Probability Analysis". The OR button is greyed out for whatever reason (premium feature?). That's where I got my numbers from (exact numbers still need to be calculated out with the values given). Since exact numbers slightly differ, I'm also interested who's math is wrong, Tescrin's or MWS'.

[rufus]

Are you sure about that math?

I think the chance with a 3-1 split is much closer to 0.36 and the chance of getting a 4 of in the top 8 is around 0.44.

[wonderPreaux]

I can do a bunch of math and maybe figure out why OP's numbers differ from MWS, but i'll need some time. Consider this a reserved post if that's alright.

[Barook]

Are you sure about that math?

I think the chance with a 3-1 split is much closer to 0.36 and the chance of getting a 4 of in the top 8 is around 0.44.

Depends on what you want to simulate. I didn't notice the talk was about a single cantrip instead of a full playset of 4 copies.

If you would run 1 Brainstorm and 3 copies of the card, you look at 8.4% to have both in the opening hand and 11.7% to have the single cantrip in hand. 35.4% would be your start chance at 8 cards.

So 35.4% + 11.7% - 8.4% = 38.7%

Two cantrips would be 43.8%.

Three cantrips would be 46,7%. (<-- at this point, you're more likely to draw it than the 44.5% of a 4-of)

Four cantrips would be 48.9%.

So, long story short, if you cut 3 cards or above for cantrips, you increase your consistency to draw the card you want.

Edit: Goddamn it, I used the value for 7 start cards instead of the assumed 8 for the multiple cantrip cases. New, rough values are the following (feel free to do the exact math, I'm too tired right now).

2 cantrips: ~ 42,4%

3 cantrips: ~ 43.4%

4 cantrips: ~ 45,4%

[wonderPreaux]

Depends on what you want to simulate. I didn't notice the talk was about a single cantrip instead of a full playset of 4 copies.

If you would run 1 Brainstorm and 3 copies of the card, you look at 8.4% to have both in the opening hand and 11.7% to have the single cantrip in hand. 35.4% would be your start chance at 8 cards.

So 35.4% + 11.7% - 8.4% = 38.7%

Two cantrips would be 43.8%.

Three cantrips would be 46,7%. (<-- at this point, you're more likely to draw it than the 44.5% of a 4-of)

Four cantrips would be 48.9%.

So, long story short, if you cut 3 cards or above for cantrips, you increase your consistency to draw the card you want.

Oh, is this assuming you shoehorn in 4 Brainstorm, or is it just a 1-to-1 swap of Brainstorm and card of note?

[Barook]

Oh, is this assuming you shoehorn in 4 Brainstorm, or is it just a 1-to-1 swap of Brainstorm and card of note?

This is assuming running 3 copies + a 1/2/3/4 cantrips total in your 60 card deck.

Calculating the exact percentages by typing them from MWS is kinda bothersome. Would be nice if somebody could set up an excel sheet for that. I find that topic rather intriguing.

[wonderPreaux]

This is assuming running 3 copies + a 1/2/3/4 cantrips total in your 60 card deck.

Calculating the exact percentages by typing them from MWS is kinda bothersome. Would be nice if somebody could set up an excel sheet for that. I find that topic rather intriguing.

Alright, I'll edit my first post when I'm done, I'll post cases for 3 + 1-4 Brainstorms. Somewhere along the way I might catch why you both got different numbers.

[Barook]

Alright, I'll edit my first post when I'm done, I'll post cases for 3 + 1-4 Brainstorms. Somewhere along the way I might catch why you both got different numbers.

I kinda fucked up cases 2-4 cantrips, so you recalculating them would be nice. Still doesn't change the difference in percentages from tescrin before.

Now a comparisson between 3 copies vs 2 copies would be nice. The MWS probability analysis should put the break even point around 6-7 cantrips, but that needs exact calculation.

[wonderPreaux]

https://docs.google.com/spreadsheets...it?usp=sharing

This took forever because I don't know how to program, but the worthwhile stuff is in the "Resultant % Change" column. Note that having that 4th copy of the card is actually better than the other options, assuming I did this all correctly (there is a nonzero, but, imo, low chance of error unless I theorized something wrong). If you disagree with my mathematics, I can explain/correct on request, im kinda too lazy to do a big write-up atm, cheers.

Edit: In case you're wondering, this is why OP's conclusion is incorrect. If you're looking for a certain "Card A" and you have 4 copies of it, you have the following ways to open with it:
Open copy 1
Open copy 2
Open copy 3
Open copy 4
Open a combination of 2+ copies.

Seems logical, yes? However, if you took out a copy of "Card A" for Brainstorm, suddenly, all the cases where you got "Card A" because you opened the exact 4th copy of it only all got replaced by this ~16% chance of hitting copy 1-3 off Brainstorm. You can't replace sure things with gambles and expect your odds to go up, that makes no sense. When it come to just finding a copy of a certain card, the best odds come from playing the most of it, by far. Moreover, every cantrip is less effective in this scenario because subsequent cantrips only help if you managed to not only fail to find "Card A" (you wouldn't need to cantrip otherwise) but also failed to find your other cantrips (which is harder to do the more you add). I estimate you'd need 10 Brainstorms for the simulated odds to equal just running 4 copies in your deck and at that point things get ridiculous.

[tescrin]

Are you sure about that math?

I think the chance with a 3-1 split is much closer to 0.36 and the chance of getting a 4 of in the top 8 is around 0.44.

I was considering four 3-1 splits (3 of the card, 1 cantrip) instead of four 4-ofs. As in, it's easier to find any of the cards you want by sticking cantrips in their 4th slot places than it is to find the right card with a bunch of 4-ofs.

It intuitively makes sense that in general running cantrips increases your options and thusly increases consistency when finding cards; but it's less intuitive that a 4-of cantrip will help you find a 3-of more than having a 4-of in your deck alone. I hope that makes sense.

EDIT:
I attempted to make that quite clear at the end of my OP given that I show a list with a load of 4-ofs and re-explain the logic. I also tried to make it clear by saying that I was in a splash-blue list without BS and knew I should be running it but couldn't let go of my 4-ofs. However, it turns out it's better to find any given 4-of by swapping to the brainstorm method (of four 3-1 splits) because when you are looking for a given 4-of in your opener you'll have a better chance of finding it after cantripping than just starting with it.

This line of thinking fails if you need to consider a combination of cards, because at that point it's moot that you should usually have 4 of your combo piece.


Examples on my end are things like:

4 SFM
4 Foundry
4 Liliana
*stuff*
23 Lands

became

3 SFM
3 Foundry
3 Liliana
4 Brainstorm
*stuff*
22 Lands

because if I need a SFM, Foundry, or Lili for a particular match; I have a better chance of finding it with a Brainstorm in those slots than having a bunch of 4-ofs while incurring all of the benefits of running Brainstorm; which includes not looking like a complete noob and having more 1-drops in a deck.


The dichotomy came from the necessity of running 4-ofs to have the most consistent finding of certain cards and my thinking of brainstorm as basically replacing a 4-of in the deck due to my unwillingness to make my general strategy worse. However the math shows that this line of thinking is very flawed, because the Brainstorms will get me to my other cards more reliably (even with less of each of them) than not having it (because you play to your outs/best strategy, and thus are always looking for a specific card.)

It should be noted that *this line of thinking is flawed with 1-drops as well, as you probably want them T1; so if you are looking to T1 DRS, it makes no sense to cantrip to a 3-of DRS since the benefit is missing.

[rufus]

...
Seems logical, yes? However, if you took out a copy of "Card A" for Brainstorm, suddenly, all the cases where you got "Card A" because you opened the exact 4th copy of it only all got replaced by this ~16% chance of hitting copy 1-3 off Brainstorm. ...

As far as I was doing things last night, those numbers look the same as the ones I got.

[tescrin]

[url]Stuff about how I'm incorrect.

Disregard the below until further notice:

I disagree. Math says what I said in my post above:

If I take out an SFM, a Liliana, a Confidant, and a Vindicate (non 1-drops) and I need any given of those, I'm more likely to find the one I'm looking for with a 4-of Brainstorm mixed in with four 3-of, than find the appropriate one for the task naturally.

You're under the assumption that holding an SFM, Confidant, or Vindicate is as good as holding a Liliana in this case; but it's not. The four 3-1 splits mean I'm more likely to find the correct card for a given job *assuming I don't need a card from each subset I 3-1'd* using cantrips than having four 4-ofs with no cantrips.

Examples:
-I'm up against burn, I could have 4 Brainstorm and 3 SFM in the deck, or I could have 4 SFM. The first is better; I'm more likely to find SFM with that config than with 4 SFM, 4 DC, 4 Lily, 4 Vindi; because the other cards would not do what I need them to do.

-I'm up against S&T and have Revoker in hand. I need a Lily. Would I rather have 4 Brainstorms and 3 Liliana or would I rather have 4 Liliana? Obviously the former, because the odds are better. The only card here that works is Liliana. SFM, Bob, and Vindi are failures.

-I'm up D&T, I saw T1 they have a Revoker; I need to gain CA. It makes sense that Bob is better here than SFM and Lili, thusly the chance of having a Bob in hand vs. a Bob in the top 11 is better with 4 BS + Bob than 4 Bob alone.


The above examples are irrefutable because your logic implies that all cards taken out are created equal; where if you instead take out cards that are somewhat situationally dependent for your cantrips, you can maximize the chances of having one during the appropriate situation.

EDIT: Further, your conclusion assumes that I made a gamble, but my odds are only the successes; there's no gamble, only statistics. I'm more likely to get a 3-of with 4 Brainstorms in the deck, than to get a 4-of; by T2. If you're going to Mull to hate or try to Brainstorm to it; that's a whole different strategy. The chance, however, of having an appropriate opening for the appropriate opponent (as shown) is higher by using the 4+4x3 strategy than having a simply quadlazer build. [For the fun using 1337 Speak, we can call it Ataxe]

[rufus]

[uncareful use of the quote feature]

For the sake of clarity, please be careful when you use the quote feature in combination with editing. Readers don't always share the same context that you have.

...
So I mathed it out:
Chance of 4-of in top 8 (you usually don't BS T1) -> ~44%
Chance of a 3-of in top 8 -> ~35%
Chance of brainstorm + 3-of in top 11 -> 24.3%
Chance of Brainstorm and 3-of in hand -> 8.5%

and finally, the actual comparable number!
Chance of Brainstorm in hand and card you need in the top 3 (after drawing a card): 35% + 24.3% - 8.5% = 50.8%!
...

I don't think this is an appropriate calculation.
The chance to get at least one of the '3 of' in the top 8 is ~.354
The chance to get none of the '3 of' and at least 1 (of 4) brainstorms in the top 8, and at least 1 of the '3 of' in cards 9-11, is ~0.050
That gives a total probability of ~0.404

Edit:
To compare, the chance to hit a 4 of in the top 8 is ~0.445

[wonderPreaux]

For the sake of clarity, please be careful when you use the quote feature in combination with editing. Readers don't always share the same context that you have.



I don't think this is an appropriate calculation.
The chance to get at least one of the '3 of' in the top 8 is ~.354
The chance to get none of the '3 of' and at least 1 (of 4) brainstorms in the top 8, and at least 1 of the '3 of' in cards 9-11, is ~0.050
That gives a total probability of ~0.404

Edit:
To compare, the chance to hit a 4 of in the top 8 is ~0.445

Yeah, you can't drop from 4 in deck to 3 in deck (which was about a 9% drop in the odds of having it on turn 2) and then expect brainstorm to get you there (the first one makes the greatest % difference and only added ~1.5%)

Brainstorm only adds % in the scenarios where you:
> Didn't get card A already
> Did get Brainstorm
> Find card A in the top 3 (this is what i mean by a gamble, if the copy of brainstorm was actually just the 4th copy of card A, you have card A 100% of the time instead of 16ish % of the time, playing the brainstorm IS the gamble i refer to)

All those conditions have to be true for Brainstorm to have actually made a difference. Tescrin, I believe you may not have correctly factored out all the overlapping occurrences bound by those 3 conditions, hence the difference in numbers. the whole reason i had to route my calcs through ~2 dozen benchmarks (aside from a lack of programming finesse) is to make sure there werent overlapping combinatorics that got counted twice.

[tescrin]

For the sake of clarity, please be careful when you use the quote feature in combination with editing. Readers don't always share the same context that you have.

I almost changed it to something more clear, but I figured the first line of my response of basically "No I'm not" was clear enough I wouldn't cause a flame war by misinterpretation. Your point is fair enough; I just think the context of the line you immediately read in my post is so clear that it's not worriesome.

I'm still thinking about it. It seems I may have had overlap; though I'm not fully convinced of it yet.


EDIT: played with the deckulator a bit more and realized the same error; my middle calculation was "find both in the top 11" which double counts hands that include both "Liliana" and "Brainstorm" as well as hands that have neither but draw into them both as cards 9,10, and/or 11.

Apologies for the sloppy math. It was kinda back-of-the-napkin stuff that seemed intriguing enough to share. I'm still not convinced I'm fully wrong despite looking at it because I'm thinking that the "not knowing I'm looking for this card" bit is relevant; but I'm doubtful enough I'll put a disclaimer in the OP about the bad stats and try to not steer people in the wrong direction.

[wonderPreaux]

ok, fine - let's use these sites and do it your way
http://stattrek.com/online-calculato...geometric.aspx
http://deckulator.appspot.com/

Odds of opening 1+ of a 4-of in your top 8: 44.48%


Now, let's say we drop to 3-of our sought-after card and add 4 Brainstorms.

We have the odds of finding our 3-of in the top 8: 35.42%


Now for Brainstorm, we need the odds that we find it in our top 8, and don't find our wanted card: 44.5%

Multiplied by the chance we find our sought card in the top 3 of the remaining 52 cards: 16.63%

44.5% * 16.63% = 7.4%

35.42% + 7.4% = 42.82%

42.42% < 44.48%

Even your way, the numbers disagree

[tescrin]

It's somewhat unfortunate I made you do all that work while hastily trying to edit my fallacious post. In my endeavor to figure out this issue; I decided to compare with multiple cantrips and similar. Even in our cases we never counted whether having 2 brainstorms made a difference and such; and it seems the net chance is ~2%.

I'm a bit curious where the breaking point is now given multiple cantrips, and whether or not Ponder will be enough to break parity as well.
With that 2% it brings us to about 42.4% from previous calcs, meaning we're getting close to success already.

[wonderPreaux]

It's somewhat unfortunate I made you do all that work while hastily trying to edit my fallacious post. In my endeavor to figure out this issue; I decided to compare with SDT to see if that would be good enough to make the numbers work (because it's still interesting for deckbuilding even if it takes three mana.)

I'm a bit curious where the breaking point is now. It's not 10 cantrips (because this is the assumption that multiple cantrips don't dig deeper) and I notice that none of us cared if you had multiple cantrips in hand in this case. It looks like you gain a net 2% from having 2 brainstorms in hand (you'll need a shuffle too, but we consider that pretty easy to get.)

With that 2% it brings us to about 42.4% from previous calcs, meaning we're getting close to success already. We may consider swapping to ponders here as I imagine if we're this close to breaking parity it must be almost right on with the extra draw. The free shuffle let's us stack the effect without consideration of issues as well.

If you want to add in multiple cantrips, and Ponders no less, it becomes a lot more complex, because you WOULD reasonably play a Ponder t1, for instance and if you want to factor in fetches it becomes even more of a nuisance. However, since your website thing disagrees with my spreadsheet thing, if I continue looking at this at all, it would probably be to find out what is messing up my calcs (though they came to the fundamentally correct conclusion). Still, if you're at the point where you'd play over 4 cantrips to offset dropping 1 of a card, i cant help but wonder why you wouldnt just play 4 of the original thing in the first place (assuming its ok to draw multiples, for instance I can TOTALLY see why you wouldnt run, say, 4 Chrome Mox or something)

edit: as an aside, though, if you want to work together on some sort of complex problem like that, we would likely achieve more than if we just went back and forth with dueling statistics