I also have some questions about "maths".

[DragoFireheart]

I'm running a rock deck. I have 4 Thoughtseizes in it.

I want to know the % of me drawing at least 1 Thoughtseize in my initial 7 card opening hand.

Any help is appreciated.

[Sanguine Voyeur]

I do believe, although I'm not positive, that the chance of it not happening is 88.7%. So the chance of it happening should be 22.3%.

But I could be way off.

[porcupinetreeman]

I'm running a rock deck. I have 4 Thoughtseizes in it.

I want to know the % of me drawing at least 1 Thoughtseize in my initial 7 card opening hand.

Any help is appreciated.

4/60 * 7 = ~ 47%

[Dilettante]

4/60 * 7 = ~ 47%

Different equation. You remove a card with each draw till you reach a success...

4/60+4/59+4/58+4/57+4/56+4/55+4/54=49.2%

[on1y0ne]

Different equation. You remove a card with each draw till you reach a success...

4/60+4/59+4/58+4/57+4/56+4/55+4/54=49.2%

This equasion assumes you are not drawing a Thoughtseize with any of your draws. If draw number 1 or 2 or 3, etc. was a thoughtseize, then the odds change...

[MathMan]

The way to do this one is to figure out the probability of NOT getting at least one...

For the following calculations, aCb is the binomial coefficient defined by b!/a!(b-a)!

We get 7C53 / 7C60 (number of hands without a Thoughtseize divided by number of total possible hands). With a little Excel help, that comes out to just over 60%; so the probabililty of at least one Thoughtseize (or any 4 of) in your opening hand of 7 is about 40%.

This goes down to about 35% if you mulligan to 6 cards.

Jim

[Finn]

about 40%.This is accurate. Not having the same mathematical expertise, I tested this in practical terms a very long time ago. It came up almost exactly that after hundreds of shuffles and draws. It is actually a pretty valuable thing to know, and I am wondering why the question isn't more common.

[Nightmare]

This is accurate. Not having the same mathematical expertise, I tested this in practical terms a very long time ago. It came up almost exactly that after hundreds of shuffles and draws. It is actually a pretty valuable thing to know, and I am wondering why the question isn't more common.Because the answer is a known quantity. The math on this specific question has been done so many times that it's considered a given.

Your opponent will have the Force approximately 40% of the time.

[Finn]

Well, yeah. It is one of those rules of thumb. But all new-ish players have a habit of asking the same questions. So you would expect it to come up every so often. It's sorta like the entire Burn thread. Oh well.

The much more important question is why the British pluralize "maths" and we do not. Anyone got the equation for that handy?

[DragoFireheart]

Well, yeah. It is one of those rules of thumb. But all new-ish players have a habit of asking the same questions. So you would expect it to come up every so often. It's sorta like the entire Burn thread. Oh well.

The much more important question is why the British pluralize "maths" and we do not. Anyone got the equation for that handy?

Would you feel better knowing I am not British?

Also, what makes you think a player is new by asking a % based question?

[Finn]

You did use quotes around the word so I assumed it was a specific reference to Mr. Heffer's usage of the word. Maybe he isn't a fish-and-chips guy either. Dunno. But yeah, I feel better.
Also, what makes you think a player is new by asking a % based question?Maybe I am guessing this for the same reasons Adam thinks it is common knowledge. It's possible that I didn't mean to imply that at all. Perhaps I took a wild guess. Most likely though, I picked up some contextual clues from the past 481 posts.

[DragoFireheart]

Maybe I am guessing this for the same reasons Adam thinks it is common knowledge. It's possible that I didn't mean to imply that at all. Perhaps I took a wild guess. Most likely though, I picked up some contextual clues from the past 481 posts.

It's not that I don't understand the concepts behind statistics: I simply never had a statistics course. I imagine I could figure it out on my own, but I'd rather ask someone that knows more.

Also, are you talking about the burn thread? :laugh:

[Bovinious]

The much more important question is why the British pluralize "maths" and we do not. Anyone got the equation for that handy?

I think the title of the thread is a reference to my thread in here from a few days ago in which I ask for help with "maths". That wasnt me being a Brit or anything, just internets speex.

[Kuma]

According to the Hypergeometric Distribution, the odds of drawing in your opening hand at least one of a card you have four of in a 60 card deck is ~ 39.95%. This increases to 44.48% on the draw.

Taking mulligans makes things more interesting. If you mull to six, the odds of getting at least one Thoughtsieze in your six is 35.14%, 39.95% on the draw. (As if you didn't mulligan on the play.) So, if you're willing to mull to six to see a Thoughtsieze, the odds it will be in either your seven or your six (Assuming you won't mulligan if it is in your seven) is 61.05% [ P(In your seven) + P(Not in your seven, but is in your 6) ].

The odds of seeing a card somewhere along the way if you're willing to mull to one, i.e. to get Leyline vs. Ichorid, is 86.15%

[Pinder]

We get 7C53 / 7C60 (number of hands without a Thoughtseize divided by number of total possible hands).

Wait...isn't it 7C56, because you're making combinations of 7 cards from the other fifty-six cards that aren't Thoughseize? Or are you running the seven Thoughtseize secret tech in the main?

Either way though, it won't change the math a ton. 40% sounds right.

[Brehn]

7C56 is correct. 7C56/7C60 = 0,6005 -> 39,95 % chance of drawing Thoughtseize.

(It would change math a ton. If 7C53 was correct, you'd have 60,09% chance... that's the 7 Thoughtseize-tech.)

[Pinder]

I'm at work right now, but I do think it would be interesting to see the math on a 3-of, 2-of, or singleton as well. I can do it when I get home, I suppose, but I'm lazy and impatient, so someone should do it now. It's the same math that Brehn posted, but with 57, 58, and 59 in place of the '56' in 7C56.

I wonder if we should get a probabilities thread stickied somewhere with a comprehensive list? In Community? Format Discussion?

[Brehn]

3-of: 31,54%
2-of: 22,15%
1-of: 11,67%

[Sanguine Voyeur]

I wonder if we should get a probabilities thread stickied somewhere with a comprehensive list? In Community? Format Discussion?I'd rather have instructions so I can compile statistics of my own. For example, suppose I wanted to know what the chance of having a Moon effect resolve on turn one, on the draw, against a deck with four Forces and four Dazes.

That sounds complicated and if I want the answer, I should work for it, but I can't.

[entity]

I do believe, although I'm not positive, that the chance of it not happening is 88.7%. So the chance of it happening should be 22.3%.

But I could be way off.

:laugh: Now THAT is the real calculation power...

[Maveric78f]

http://img89.imageshack.us/img89/5286/combinatoirecartesby6.jpg

[Finn]

the chance of it not happening is 88.7%. So the chance of it happening should be 22.3%.Lol. Don't hurt yourself working this one out.

[Sanguine Voyeur]

Lol. Don't hurt yourself working this one out.Missed that error. When I typed it, I was in a rush.

Not only that but the underlying math is wrong. I did [56/60]*[56/59]*[56/58]*[56/57]*[56/56]*[56/55]*[56/54]. I now see that I should have done [56/60]*[57/59]*[56/58]*[55/57]*[54/56]*[53/55]*[52/54]. I think.

I really would like to learn probability stuff, I'm clearly clueless.

[Shtriga]

I was always terrible at probabilities in particular

[Nihil Credo]

I really would like to learn probability stuff, I'm clearly clueless.
Recommendation: An Introduction to Probability Theory and Its Applications by William Feller. One of the best, if not the best.

[thefreakaccident]

Well, firts you will set up a confidence interval for the chance of success/failure (per draw per possible thoughtseize), then you will do a standard P-test, set up the statistic and solve for a 95% confidence level, of coarse this is all assuming that the deck isn't stacked or manipulated by some means.

[Pale Moon FTW]

The formula isn't that advanced:
Chance of drawing exactly k cards of which are m copies of it in the deck, from a deck of N cards when drawing a hand of n cards.
Prob = (m!/k!(m-k)! * (N-m)!/(n-k)!((n-m)!-(n-k!))) / N!/n!(N-n)!

To calculate the chance of drawing at least one copy of the card, solve for k=0 (chance of not drawing any) and subtract from 1. (0! is defined as 1)
The only problem is that the factorials produce obscurely large numbers so you'll need a calculator that can show numbers in the order or 10^100.

[Maveric78f]

No need to go crazy with 10^100 numbers...
http://img141.imageshack.us/img141/9228/relationcombincr0.jpg

[Pale Moon FTW]

hmm something's flawed in that formula, it gives a chance of about 53% of not drawing a card which there are 4 of in a 60 cards deck, and the chance of not drawing any is in fact 60%.

[Maveric78f]

Right misses +1 in the lower bounds of the products.

Take w=0 and read my formula, you'll understand my mistake.

Edit : corrected

http://img403.imageshack.us/img403/6790/relationcombintf9.jpg