Is Random really Random?

[benthetenor]

Did you read my post? What's the point of caring about the odds if you can't adjust your strategy to use them to your advantage? All I wish to know is likely v. unlikely. If a play is likely not going to work, then I'll avoid it. If you want to take the time to discover the odds of all of the relevant plays, you will be able to find the statistically "best" play. Given that, you've just spent 5 to 10 minutes thinking about all of the possible plays at one given instant when you could have arrived at the same conclusion in 15 seconds by simply knowing which play is likely to succeed and which is likely to fail and going with that.

Huh?  You can adjust your strategy to use the odds to your advantage.  It's just that knowing the exact odds of your opponent having certain cards down to the last % doesn't change much about your strategy compared to knowing approximate odds.  For example, knowing that your opponent having a turn 1 Oath happens 35% of the time vs. 15% of the time vs. 3% of the time will make a big difference.  18%, 17%, or 16%- not so much.

I don't know if you mean that people will have an intuitive sense of this probability that is very accurate already, but I think this is probably not the case in general.  Of course, math can't account for opponent's mulligan process, etc., and that's why no one says that you should only use math. 

I'm sorry if I've been unclear. My overall point is that I feel it's unnecessary to compute probabilities outside of normal playtesting. Obviously the difference between 35% and 15% is significant. In my last post I used 5% as a statistically insignificant number. In actuality, with normal playtesting, you should be able to get down to about 10%. In any case, there is no need to know that Oath will get a first turn Oath exactly 35% of the time, and computing that number is a waste of your time. Besides, knowing that it happens only one time in three should not change the way you play your game. You may need to adjust your deck to be able to handle that, but mulliganing into FoW is not a smart play, as often times that will leave you with nothing else. Other times, you will mulligan into FoW and they won't have it. Regardless, this once in three probability will very much reveal itself in normal playtesting. Similarly, 15% (Once in about seven games) will also reveal itself, and reveal itself as being much different than 35%.

In Vintage, if a deck gets the nuts, there is very little that can be done to stop it. Now, if you were to tell me that Oath got a first turn Oath 75% of the time, then I would start to think about mulliganing into FoW, or even better, playing Oath. Even if you said that it got the nuts more than half the time, I'd consider it a potent threat. The odds of losing a match simply to that deck's nuts draws would be > 25%, which is a little scary. (Incidentally, if you look at the 75% first-turn Oath, this leads to the chance of losing the match 56% of the time, which is why it's so scary) One time in three? That's acceptable, because the odds of getting blown out in a match is only about one in nine, or 11%. I'd rather take the 11% auto-loss rate over mulliganing myself into a weaker hand with FoW. This is not to say that mulliganing will always result in a weaker hand, but given an already good hand, I'd rather not take the risk.

[Elric]

benthetenor- I was going to address the point about playtest samples in my last post as well but decided that it deserved its own post.  I agree that playtesting will give you valuable information but playtesting has a very high margin of error.  That is, random chance says that there will be a lot of variation in even what you'd consider a large number of playtest games. 

For this reason, seeing an Oath deck get the nuts draw in 12 out of 30 games doesn't tell you that the deck gets the nuts draw 40% of the time with any degree of confidence- it doesn't even tell you that the deck gets the nuts draw 35-45% of the time with any degree of confidence.  To have a high degree of confidence you need to expand that range to at least 25%-55% and as you said, this range of numbers makes a big difference. 

In the beginning portion of the game everything is most open to mathematical analysis (because you don't have to worry about player choices with the exception of mulligans) and this is where math can be the most useful. 

I started a thread "Margin of Error and Joe Random/Joe Awesome" last year (referring to win percentages, another key component of testing): you can see that thread here
 
When someone says that their Oath deck gets the nuts draw 60% of the time they might be telling the truth about their past games, but that doesn't mean that they should expect to get it 60% of the time in the future, especially if their current 60% was 6 out of 10 games.  If you have an idea of the probabilities based on the underlying math, you're more likely to be able to tell statistical outliers when you see them.

To choose another example, you should always alternate who plays first in testing games.  This is because you know that you are expected to win the coin flip 50% of the time in future matches so you don’t want a string of good or bad luck on coin flips to influence your testing results.  Alternating going first reduces the variance—that is, the margin of error, of your results.  Yet, I have never seen anyone give this reasoning when advocating how to test decks (in fact, I bet a lot of people either flip every time or use the loser plays first rule—those were what I usually did until I thought about this in detail). 

[benthetenor]

I would certainly hope that normal playtesting involves more than 30 games on any one specific matchup...

[Gabethebabe]

It changes the order maybe 30% of the cards in the deck.

That's funny because a single "perfect" crush shuffle changes the position of every card in the deck, so could you explain what you mean by that?

Changing the position of every card in the deck is not equal to randomization. Say you have 10 cards left in your library.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Typically you would grab cards 6-10 and crush them into cards 1-5 achieving:

1, 6, 2, 7, 3, 8, 4, 9, 5, 10.

Typically you would then grab 8, 4, 9, 5 and 10 and crush them into 1, 6, 2, 7 and 3, thus achieving:

1, 8, 6, 4, 2, 9, 7, 5, 3, 10.

Your way is shuffling is flawed. When you have cards:
1-2-3-4-5-6-7-8-9-10 and grab 6-10 to "crush" into 1-5 you should make sure (that is what I do when shuffling in real life) that card no 1 does not stay at first place and card no 10 does not stay at its last place. Crush like this:

6-7-8-9-10
----1-2-3-4-5
leading to (more or less):

6-1-7-2-8-3-9-4-10-5

BTW all the numbers Machinus posted were correct. I took the liberty of checking them :p

[Khahan]

I'm not going to get into the math and statistics because...well I can't.
However, there are a few things worth nothing that I believe some people are overlooking.
The first is this:
If you have a 60 card deck, there are !60N (is that the right way to write the formula...60x59x58x57...x2x1) permutations for the final order of the cards in the deck.

Random means you simply do not know which permutation is the final one. 

That also means that Mox, Mox, Mox, Mox, Mox, Lotus, Land, Land, Land, Land, Land, Land, Land, Land, Land, Land, Land, Land, Orchard x4, Oath x4, Fow x4 etc is a validly 'random' ordering for the deck. You could conceivably shuffle into it. As long as you took no special means to achieve or avoid that result and as long as you have no knowledge of that final result, the deck is random (I'll get more into this later).

Next, somebody early made a comment along the lines of , If Dark Ritual is draw #1, then the chances of it being draw #2 are equal to that of a 1 in 59 chance.

Not quite. Technically, the chances of it being draw #2 are equal to the total number of ways a 60 card deck can be stacked w/ Dark Ritual, Dark Ritual divided by the total number of ways it can be stacked.  Who knows, maybe this is actually equal to the same percentage. But the way you arrive there is important.

There is no reset. You don't suddenly  have a newly randomized 59 card deck. You have a 60 card deck in a set order  and you are looking at the 2nd card from the top (the original top).

As to the question of the thread: Is random truly random?  The answer is No.

There are 2 parts I put out there to describe random (and I'm going to assume there are no objections to those 2)
A) You have no knowledge of the specific final order
B) You took no part in affecting the final order (either purposefully or not)

As for A, an honest player can always answer that part affirmatively. 
As for B, even the most honest player cannot answer affirmatively.

When you build your deck, you start with a set order, everything is clumped somehow. Maybe all lands, then all blue spells, then all black spells. Maybe you sorted to make sure everything is there so you have 4x Mana Drain, 4x chalice etc.  (Perhaps its as messy as 'oh, force of will should go in this deck. And oh, I need 3 Duress. Oh yeah and a Yawgmoth's will. And that means I definitely need 4 Underground sea...so you gather the cards as you think of them) But you've put an initial order to the cards before shuffling.

AFter game 1, you've laid X number of cards out. Lands here, permanents there, graveyard there.  You know all that information. You pick it up, s huffle it into your library and shuffle up. However, once again, you started with some knowledge of the cards (if not their exact order, their basic location....1 trop island, 1 forbidden orchard and 1 Oath were on the board and I set that stuff on top of my library before shuffling).

Until you get to the cut. Here is where randomness may truly be achieved. I have given a set of cards to my opponent. He has no idea about initial order, or approximate place. He either continues to shuffle, or simply cuts. Either way, he creatures a brand new order to the cards with no knowledge of each cards' starting point or final resting place.

I've just got 2 more statements that are seemingly contradictory:
1) There are a number of people discussing this who know the math much, much better than me (obviously since they can discuss the formula involved and understand what they are talking about..heck the fact that even know some of these formulas shows they know more than me about the math)

2) I can honestly say those people are misapplying the math in many cases. I can handle the theory and application pretty well, even if I have no idea what the ! or N in a given formula represents.   

[Elric]

If you have a 60 card deck, there are !60N (is that the right way to write the formula...60x59x58x57...x2x1) permutations for the final order of the cards in the deck.

Random means you simply do not know which permutation is the final one.  ...

Next, somebody early made a comment along the lines of, If Dark Ritual is draw #1, then the chances of it being draw #2 are equal to that of a 1 in 59 chance.

Not quite. Technically, the chances of it being draw #2 are equal to the total number of ways a 60 card deck can be stacked w/ Dark Ritual, Dark Ritual divided by the total number of ways it can be stacked.  Who knows, maybe this is actually equal to the same percentage. But the way you arrive there is important.

This phrase isn't quite correct.  What you're looking for is the total number of ways a 60-card deck can be stacked w/ Dark Ritual, Dark Ritual divided by the total number of ways it can be stacked with Dark Ritual as the top card.  You are "conditioning" on the first Dark Ritual since it's already known.

The way that I use random (and the way that you're using it as well, even if you don't state it explicitly) is that random means that each permutation of the deck is equally likely- not just that you don't know exactly which permutation will come up. 

When you are using this definition of random, you don’t have to worry about all of the possibilities each time.  You can state, “conditional on having already drawn a Dark Ritual as the first card, it’s just as likely that the one remaining Ritual is in any of the next 59 positions.  So the chance for it to be the second card is 1/59.�

[Harkius]

Hello Everyone.

I have been trolling here for a while, and I thought that a few clarifications may be necessary.

1. The probability of your drawing a second Dark Ritual off the top of your deck, provided that it is appropriately randomized, is not 1/59, unless you are running 2 Dark Ritual in your deck. This is because statistics follows a few solid principles.

First, the odds of any event, P(x), are equal to the favorable occurrences (drawing a Dark Ritual) divided by the total occurrences (59 undrawn cards, supposing you run a 60 card deck. So, your probability (P(Dark Ritual) = X/59, where X = number of Dark Rituals in your build, -1). Second, the fact that you drew one Dark Ritual already does not matter. Sounds strange, doesn't it? However, it is a fundamental law in statistics that, unless otherwise stated, statistical trials have no bearing on one another. In fact, that is an important point.

2. Unless otherwise stated, statistical trials have no bearing on one another. There. That's better. For example, let's consider three different tests.

Assume:
-Sufficiently randomized deck
-4 Dark Ritual, 60 Card deck
-No replacement between draws

Test 1:
What is the probability that you will draw a Dark Ritual as your first card?
Answer:
4/60 = 0.06667

Test 2:
What is the probability that you will draw a Dark Ritual in your first seven cards?
Answer:
4/60+4/59+4/58... Provided that you have not drawn a Dark Ritual by the seventh card. Otherwise, substitute 4-Y (where Y is the number of Dark Ritual cards that you have drawn. This is the proper way to calculate this probability. For a four-of card, your odds are approximately 7% if I recall correctly.

Test 3:
What is the probability that you will draw Dark Ritual as your second card if you drew one as your first card?
Answer:
3/59 = 0.05085
Yes. Look at that. Your odds are nearly as good as for the first card. Surprised? That is because this is not the question that you think that it is!

Test 4:
What is the probability that you will draw 2 Dark Ritual cards as your first two cards?
Answer:
4/60*3/59 = 0.00339. This is the test that you have all been looking for.

The reason that Test 3 and Test 4 have different answers, even though they are functionally asking the same question? The method for asking the question is different. The former question is asking what the odds of drawing a Dark Ritual as your second card is, and it is a straightforward question. The second question is asking what the probability is of drawing two Dark Ritual cards, and, as such, is not a set of independent tests. Without the assumption of independence, the variations and the probabilities begin to change.

This is one of the variations that Smmemmen noticed, and one of the reasons that it seems intuitively correct but mathematically flawed. Congratulations, Steven. You noticed that the questions are equally important as the answers. Now, you need to check your question about the Land Grant again. Even without thinning your deck, it could make a difference to shuffle first. It depends on whether your card draws are independent or dependent events. In this case, I would claim that the Land Grant (provided no land is removed), will not alter anything, as the card draws are, functionally, independent events in the realm of Magic.

The times that events are independent or dependent are what concerns whether playing the Land Grant first makes a difference. Off hand, I can't think of a case in MtG where a draw becomes a dependent event, but the places where they mightoccur is within the rules text of specific cards, and my brain is a bit foggy at the moment to figure out a specific card that may make this applicable. In any case, it is doubtful as to whether you could respond to the first card draw, thus negating the ability to alter anything with your Land Grant trick.

Again, hello all. I look forward to joining the community.

Harkius

The three most important tools for posting on forums: Preview, Spell Check, and your Brain. Use them all!

[Jacob Orlove]

Test 2:
What is the probability that you will draw a Dark Ritual in your first seven cards?
Answer:
4/60+4/59+4/58... Provided that you have not drawn a Dark Ritual by the seventh card. Otherwise, substitute 4-Y (where Y is the number of Dark Ritual cards that you have drawn. This is the proper way to calculate this probability. For a four-of card, your odds are approximately 7% if I recall correctly.

The probability of drawing exactly one Dark Ritual isn't quite that easy to calculate--a much better method is to simply calculate the odds of not drawing any rituals in your opening hand (about 60%) and see that you have about 40% odds of drawing at least one ritual.

The precise odds of drawing exactly one of a given four-of, though, is about 33%. For more on the specific math involved, check out this thread.

[Harkius]

You're right. Brain fart, I guess. P(X) = 1- P(!X), after all, and the odds of not drawing a Dark Ritual are much higher, and the odds are much easier to calculate.

To correct myself, the odds of drawing a Dark Ritual card in your hand are ~40%, assuming no replacement, independent draws, etc. The odds of drawing two are correspondingly lower, as you have to calculate that probability based on already having drawn the first.

Test 1:
What is the probability of drawing Dark Ritual as your first card?
Answer:
Approximately 7% (6.66666%)

Test 2:
What is the probability that you will draw a (not only one, but at least one) Dark Ritual in your first seven cards?
Answer:
Approximately 40% (39.951%)

Test 3:
What is the probability that you will draw a Dark Ritual as your second card (assuming that you drew one as your first card)?
Answer:
Approximately 5% (5.085%)
As I had said, not really that much lower.

For those of you who will catch the differences in my numbers, remember that the first post was comprised of raw numbers, and that these represent percentages. The numbers are the same. (With the exception of the 7% brain fart. Thanks Jacob!)

Harkius