Probability and Magic

[Gandalf_The_White_1]

For example, if the Oath player got turn one Oath games one and two, and I have a hand that does not have FOW, I am going to assume that Oath does not have turn one Oath in game three and so I will be more inclined to keep that hand.  LIkewise, if I played against Combo, and he had really shitty draws games one and two, then I am more likely to be conservative about my plays assuming he is likely to get a better hand in game three.  Those are gross examples, but you get the idea. 

These examples don't seem to make any sense to me.  What happened in previous games has nothing to do with what happens in later games.  The probability of a 1st turn oath game 3 is the same as it was game 1 and 2.  Now, if your opponent gets first turn oath all 3 games that is obviously unlikely, but that doesn't mean that getting 1st turn oath in the first 2 games somehow reduces the probability of 1st turn oath the third game.  Likewise, having bad hands in the first two games also has no bearing on your probable hand quality game 3, either.

[Revvik]

Let's say the probability of turn one Oath of Druids is 30%, or .3.
You know he has already had it game one, so the current probability is this:

1     2     3
---------------
.3

if he gets it game two as well, probability will start to look like this:

1     2     3
---------------
.3    .3

The odds of him having first turn Oath both games then can be calculated as .3 X .3.
Add on a turn one Oath on the THIRD game and you get this:

1     2     3
---------------
.3    .3    .3

the odds of him having a first turn Oath of Druids all three games then become .027, or 2.7%.  Which is very unlikely.  Steve's logic is correct then, because while the odds of him having a first turn Oath of Druids for that game is still 30%, this would be his third time in the match with a first turn Oath of Druids, putting it at a 2.7% chance of happening.  A Force of Will is probably unnecessary.

[Machinus]

Actually, it's a fallacy to think it's any less likely to draw the nuts game 3. They are independent probabilities and do not affect each other. Steve should know better.

[The Atog Lord]

For example, if the Oath player got turn one Oath games one and two, and I have a hand that does not have FOW, I am going to assume that Oath does not have turn one Oath in game three and so I will be more inclined to keep that hand.

Wrong, wrong, wrong. Each hand is an independent event and what sort of draw Oath had in games 1 and 2 has no bearing whatsoever on game 3. Steve, I'm surprised that you'd say otherwise.

[Moxlotus]

Let's say the probability of turn one Oath of Druids is 30%, or .3.
You know he has already had it game one, so the current probability is this:

1     2     3
---------------
.3

if he gets it game two as well, probability will start to look like this:

1     2     3
---------------
.3    .3

The odds of him having first turn Oath both games then can be calculated as .3 X .3.
Add on a turn one Oath on the THIRD game and you get this:

1     2     3
---------------
.3    .3    .3

the odds of him having a first turn Oath of Druids all three games then become .027, or 2.7%.  Which is very unlikely.  Steve's logic is correct then, because while the odds of him having a first turn Oath of Druids for that game is still 30%, this would be his third time in the match with a first turn Oath of Druids, putting it at a 2.7% chance of happening.  A Force of Will is probably unnecessary.

This is only when trying to predict the 3 matches before they occur.  I could flip a coin a 100 times.  I flip heads the first 99 times.  The odds of me flipping heads on the 100th time is 50%.  It is basic statistics.

[nataz]

For example, if the Oath player got turn one Oath games one and two, and I have a hand that does not have FOW, I am going to assume that Oath does not have turn one Oath in game three and so I will be more inclined to keep that hand.

Wrong, wrong, wrong. Each hand is an independent event and what sort of draw Oath had in games 1 and 2 has no bearing whatsoever on game 3. Steve, I'm surprised that you'd say otherwise.

Okay, to be fair, I've only got 2 semesters of relevant prob/stats under my belt, but I think steve is using this as an example of conditional probability, not simple independent events. Since (correct me if I am wrong) steve tends to view magic in match %, you can move away from independent events as long as you have new information.

[The Atog Lord]

You have to view each of the games as independent events. Whether viewed before or after the fact, there is no way that the first two games can affect how cards will be drawn in the third.

[Revvik]

true, but I was looking at the match as a whole.  if you're in game three and he has had a first turn oath of druids games one and two, you can actually use that to determine if you need a Force of Will or not:

Your mind can think; "OK, he's had an Oath on the first turn game one and two.  There's no way he'll have it game three, the odds of having it first turn all three games in a match are only 2.7%."
I don't think that's a wrong way to look at the situation, but then to be perfectly fair you don't see me putting too many tournament wins under my belt, now do you 

[The Atog Lord]

I don't think that's a wrong way to look at the situation

It is wrong. You just can't factor those other games in at all in deciding how his draw game 3 will be.

[Machinus]

true, but I was looking at the match as a whole.  if you're in game three and he has had a first turn oath of druids games one and two, you can actually use that to determine if you need a Force of Will or not:

Your mind can think; "OK, he's had an Oath on the first turn game one and two.  There's no way he'll have it game three, the odds of having it first turn all three games in a match are only 2.7%."
I don't think that's a wrong way to look at the situation, but then to be perfectly fair you don't see me putting too many tournament wins under my belt, now do you 

This is why we have a term for logical fallacy, because it's not obvious why your mind makes a mistake. However, in, this scenario is very clearly nothing you can determine about game three from games one and two. It is, in fact, quite wrong to think about the three games as a single entity. This is an arbitrary construction that your mind makes, but it is quite incorrect to adjust your statistics because the third game happens to be played after the first two.

The chance of having first turn oath is exactly the same every game, and is most certainly absolutely unrelated to the happenings of previous games, whether they be immediately prior in the match, in a different match, in a different tournament, played by someone else, in a different deck, in a different format, in a different game...

Hopefully I have made it somewhat clearer how little sense it makes to think that events prior to the game have some effect on what the first seven cards will be.

[Smmenen]

For example, if the Oath player got turn one Oath games one and two, and I have a hand that does not have FOW, I am going to assume that Oath does not have turn one Oath in game three and so I will be more inclined to keep that hand.

Wrong, wrong, wrong. Each hand is an independent event and what sort of draw Oath had in games 1 and 2 has no bearing whatsoever on game 3. Steve, I'm surprised that you'd say otherwise.

It doesn't influence what actually happens in game three, but I expect my matches to mirror my testing experience.  If in testing I find that if a player gets the nuts two games in a row and they don't get it in the third game, the i will presume that it will likely play out that way in matches.  It's not a statistical analysis, but an analysis that I expect my matches to mirror my testing. 

[Elric]

It doesn't influence what actually happens in game three, but I expect my matches to mirror my testing experience.  If in testing I find that if a player gets the nuts two games in a row and they don't get it in the third game, the i will presume that it will likely play out that way in matches.  It's not a statistical analysis, but an analysis that I expect my matches to mirror my testing. 

That's just incorrect.  In your testing the reason why a person who gets the nuts draw 2 games in a row doesn't usually get it in the third game is that the nuts draw happens with a low probability each time!  That doesn't change statistical independence at all.  If you know from your testing that you shouldn't bother to guard against the nuts draw then that is true in every game of the match, regardless of the outcome of previous games.

Time considerations aside, you should play to win each game as if it were the only game.

[Smmenen]

Fair enough.  I always seem to miss that logical fallacy and I never am quite sure why.  I had the same problem with meandeck tendrils and placement of cards in my deck.  In the near future, I'm going to start a thread on probability that goes far beyond what we have talked about in the past.  Specifically, I'm going to argue that a normal distribution may be in the incorrect frame for magic analysis. 

I guess another example might be that my experience with mana drain decks causes me to make base assumptions about what kind of counterspell contingent my opponent is holding. 

I tend to assume that my opponent has a 40% chance of having a FOW or a Mana Drain in their opening hand.  I augment that analysis slightly based upon the number of moxen they drop and other tells.  For example, if my opponent goes: Mox, Land, Pithing Needle, Goblin Welder, then I can pretty much guess that they probably can’t use a FOW, or if they do have one, they don’t have a blue spell.  I may still assume they have a Drain.  The question only becomes critical if I am going to make a key play – like Ritual, Ritual, Mox, Bargain – where a FOW there might just cause me to lose.  I also try to figure out which plays signal which cards an opponent has in hand.  For example, turn one Welder seems to signal a play for turn one Thirst.  If my opponent has very few cards in hand, I’m much less scared of their Thirst. 

When I play against good opponents, and when I’m going to make risky game-breaking plays, I tend to assume that my opponent has a FOW or a Mana Drain.  If my opponent has FOWed once in the early game, then I assume they do not have another until a certain point in the game.  For example, if we get to turn four and they play Brainstorm, then I will think it possible and even likely they have found another.  If they have played Thirst and Brainstorm by turn three despite a turn one FOW, then I will almost definitely assume another FOW.   These situations are very context specific, but I can make them with greater confidence because of my own experience with Drain decks and their basic operation. 

[Smmenen]

Let's say the probability of turn one Oath of Druids is 30%, or .3.
You know he has already had it game one, so the current probability is this:

1     2     3
---------------
.3

if he gets it game two as well, probability will start to look like this:

1     2     3
---------------
.3    .3

The odds of him having first turn Oath both games then can be calculated as .3 X .3.
Add on a turn one Oath on the THIRD game and you get this:

1     2     3
---------------
.3    .3    .3

the odds of him having a first turn Oath of Druids all three games then become .027, or 2.7%.  Which is very unlikely.  Steve's logic is correct then, because while the odds of him having a first turn Oath of Druids for that game is still 30%, this would be his third time in the match with a first turn Oath of Druids, putting it at a 2.7% chance of happening.  A Force of Will is probably unnecessary.

Has this post been refuted?

[nataz]

I tend to assume that my opponent has a 40% chance of having a FOW or a Mana Drain in their opening hand.  I augment that analysis slightly based upon the number of moxen they drop and other tells.  For example, if my opponent goes: Mox, Land, Pithing Needle, Goblin Welder, then I can pretty much guess that they probably can’t use a FOW, or if they do have one, they don’t have a blue spell.  I may still assume they have a Drain

basicly you are talking about conditional prob, whereas most people think of magic as independant events.

I'm not sure if the confusion is coming from people missing the idea that you are lumping events as matches, or if you don't understand independant events. 

//

Steve is *not* asking what the probability is that the oath player has the nut draw (say, 30%)
Steve is asking what the probability is that the oath player has the nut draw 3 times in a row (.3 X .3 X .3)
The individual games ARE independant events, however with in the scope of a match, they are not.

fred and sally both have a deck of playing cards in front of them.

Deck shuffle, fred draws ace of spades. Prob for independant event, 1/52
Deck shuffles again w/ spade back in, sally draws ace of spades. Prob for independant event 1/52, prob that BOTH sally AND fred draw Ace of spades,  (1/52) x (1/52)

and really, most things people should be thinking about when playing any card game are covered in a college level prob and stat course, which for me was a while ago, and I could be wrong

I'm going to have to go check my book.

huggs and kisses
-carter

[Smmenen]

For example, if the Oath player got turn one Oath games one and two, and I have a hand that does not have FOW, I am going to assume that Oath does not have turn one Oath in game three and so I will be more inclined to keep that hand.

Wrong, wrong, wrong. Each hand is an independent event and what sort of draw Oath had in games 1 and 2 has no bearing whatsoever on game 3. Steve, I'm surprised that you'd say otherwise.

Okay, to be fair, I've only got 2 semesters of relevant prob/stats under my belt, but I think steve is using this as an example of conditional probability, not simple independent events. Since (correct me if I am wrong) steve tends to view magic in match %, you can move away from independent events as long as you have new information.

This is correct. I view magic events as match percentages.  I tend to look at the match as a whole, since that is the relevant unit.  This topic deserves its own thread.  Last Jan I wrote a very long piece on probability that I was considering submitting as an article.  Next week I'll pull it up and post in in this forum and we can talk about it.

The question that was debated on my boards was this:  if I have already seen a Dark Ritual, would shuffling my deck make it more likley that I would see another Dark Ritual sooner with Spoils than otherwise.  Logically, the answer is no.  A random deck is a random deck.  But it spurred me to closely analysis probability and magic and I came up with some thoughts worth sharing, I think.  I am very busy at the moment, but I'll share them next week.

[Komatteru]

Let's say the probability of turn one Oath of Druids is 30%, or .3.
You know he has already had it game one, so the current probability is this:

1     2     3
---------------
.3

if he gets it game two as well, probability will start to look like this:

1     2     3
---------------
.3    .3

The odds of him having first turn Oath both games then can be calculated as .3 X .3.
Add on a turn one Oath on the THIRD game and you get this:

1     2     3
---------------
.3    .3    .3

the odds of him having a first turn Oath of Druids all three games then become .027, or 2.7%.  Which is very unlikely.  Steve's logic is correct then, because while the odds of him having a first turn Oath of Druids for that game is still 30%, this would be his third time in the match with a first turn Oath of Druids, putting it at a 2.7% chance of happening.  A Force of Will is probably unnecessary.

Has this post been refuted?

The math is correct in vacuum, but that doesn't help you analyze the situation.  Let's say you have an event with two outcomes A and B, with corresponding probabilities P(A) and P(B).  Suppose the two events are independent -- that is, whether one happens does not explicitly depend on the former (coil flips, for instance).  Then if you run two trials of the event, the probability that A happens in the first and B happens in the second is:
P(A&B) = P(A)*P(B)

However, there is also a probabilty of B given A.  That is, the probabilty of A given that B has occured.  This is denoted P(B|A).  It is computed
P(B|A) = P(A&B)/P(A)

Now, if the events are independent, note that
P(B|A) = P(A)*P(B)/P(A) = P(B)      (assuming P(A) != 0, that is, it can actually happen)

Thus, for independent events, it does not matter what happens previously, as the probability of an event given that some other things happened before it is the same as the probability of the event occuring on its own.

In the scope of the problem, this means you cannot consider what happened in previous games when analyzing what might happen in the next one.  You can determine the odds that your opponent gets "the nutz" two games in a row, but your analysis of whether he gets it in game 2 cannot depend on what happened in game 1.  Games are indeed independent events.  Well, technically, there are ways for them not to be, but they make the probabilistic analysis nearly impossible (you'd be considering things such as the arrangement of the cards in the last game -- did your opponent have 10 land in play?, and how your opponent is shuffling -- things that are nigh impossible to model).

Then this:

if I have already seen a Dark Ritual, would shuffling my deck make it more likley that I would see another Dark Ritual sooner with Spoils than otherwise.  Logically, the answer is no.  A random deck is a random deck.

The mathematics supports this conclusion if you look at the deck as a completely random object. Let's say you have X cards remaining the deck and 3 dark rituals left.  Then the probability that you see a Dark Ritual in top Y cards is:

P = 1 - binomial(X-3, Y)/binomial(X,Y)                                                                         (1)

               

However, you can look at it differently if you start from the beginning of the game.  The odds that there are D Dark Rituals in the top Z cards of your library is

P = binomial(4, D)*binomial(56,Z-D)/binomial(60,Z)                                                       (2)

               

The odds that there are 2-4 Rituals in the top Z cards of your library are then

PD = Sum(binomial(4,D)*binomial(56,Z-D)/binomial(60,Z), D=2...4)                                (3)

Say you've used Y cards already, and you care about the top Z-Y cards of your library.  Also suppose that you've found 1 Dark Ritual.  Now we make a logical assumption.  The assumption is that there is another ritual in the Top Z-Y cards of my library.  We can make this assumption because we are not solving a problem a vaccuum.  We are free to choose Z to be whatever we want when we start the problem.  Thus, when we start, we should have chosen Z large enough to maintain its relevance.  For instance, if we draw on average 4 cards before we play Spoils, then we should choose Z to be at least 4+19 = 23.  This gives us the window within we want to find a second Dark Ritual.  If there is no second Ritual, then we don't care--we lose.  Thus, we can not bother with the analysis for that.  We can do a different analysis to figure out the probablity that we lose to Spoils and proceed with this assumption that we have a second Ritual in the top Z-Y cards. Now, let's calculate the odds that we find it in the top L (where L < Z-Y < Z, corresponding to life total).

PL = 1 - (1/(Z-Y))*((Z-Y-1)/(Z-Y))^min(L-1,Z-Y-1)                                                       (4)

Since we don't care how many Dark Rituals we find in the Top L cards (we only care about not losing the game), this gives us overall odds of:

PTotal = PD*PL                                                                                                        (5)

As a numerical example, suppose you've seen 12 cards and have 1 Ritual.  Using the former approach, the odds that we find it (assuming 20 life) without losing are
P = 1 - binomial(48-3,19)/binomial(48,19) = 78.87%

Using the second set of constraints, we have to set some guidelines.  Let's assume that we'll see 5 cards before Spoils plus our opening hand (as before), meaning we care about what the top 19+5+7 = 31 cards of the library look like (so Z = 31, Y = 12, L = 19) .  By (3),
PD = sum(binomial(4,D)*binomial(56,31-D)/binomial(60,31), D=2...4) = 0.719

By (4), we have
PL = 1 - (1/(31-12))*((31-12-1)/(31-12))^min(19-1,31-12-1)  = 1 - (1/19)*(18/19)^18)) =  98.11%

This this gives us the odds that we don't lose as
P[not lose] = 0.719*0.981 = 70.54%

So why use the second model?  Easy, decks are not completely random.  The second model takes into account the fact that decks tend to be a bit stacked, and cards can often clump together.  You can expand it out to get a better answer, but it's not a fun problem to do.  You could never do that in your head.  The way to do this problem in a more practical setting is to calcuate a percentage difference and use it to bound the error for using the first (simple) model.  If you shuffle the deck first, you can go back to the first model because you can no longer account for clumping.

[forests failed you]

JD I really wish you would have done my math for me at Gencon and just told me that I could have drawn into top eight instead of playing it out!

[Machinus]

You are not allowed to use "conditional probabilty" to predict anything about future games of magic. You can't look at it "as a match" and somehow gain better insight than looking at it as three games. There is no match; the match is meaningless. Just because you are playing three games of magic in a row does not mean anything and it does not change the probabilities. Every single game is identical, no matter what happens in the others.

The probability of opening with FoW is 40% every single game. If you draw one games one and two, the probability for game three is still 40%. It is completely wrong to try to adjust the probabilities for game 3 based on prior events. Do you record every single time you have ever drawn first turn FoW and use that to calculate what it will be in the future? There is no difference between games played right before and games played months before - they both have no significance.

[nataz]

@ mach

keerect, I was wrong.

However, I think the part of steves resp that I quoted would be a good example of conditional probability.

[Elric]

You are not allowed to use "conditional probabilty" to predict anything about future games of magic. You can't look at it "as a match" and somehow gain better insight than looking at it as three games. There is no match; the match is meaningless. Just because you are playing three games of magic in a row does not mean anything and it does not change the probabilities. Every single game is identical, no matter what happens in the others.

I'd word this last sentence as "No game changes based on what happens in the other games."

In an actual 3-game match, your 3 games won't be quite identical because one game will be unsideboarded (50% chance to go first and 50% chance to go second), one game is sideboarded going first and another game is sideboarded going second.  The 3 games are definitely independent, though.

In a given matchup your chance to win the matchup is the chance that you win 2 out of the 3 games.  It’s this that you want to be as high as possible, not the average number of games that you win out of those 3.  This has some minor implications on how to design a deck, but it would only really matter in extreme cases (basically, when your chance to win a sideboarded game going first and chance to win a sideboarded game going second are very different).

[cssamerican]

This is only when trying to predict the 3 matches before they occur.  I could flip a coin a 100 times.  I flip heads the first 99 times.  The odds of me flipping heads on the 100th time is 50%.  It is basic statistics.

I totally agree with this on principle. However, in a real life situation if you call anything other than heads you are an idiot, it's obvious the coin isn't legit

JD I really wish you would have done my math for me at Gencon and just told me that I could have drawn into top eight instead of playing it out!

For all the people who's math isn't up to JD's level you can use the HYPGEOMDIST function in Excel to figure odds on drawing cards out of a fixed population.

So why use the second model?  Easy, decks are not completely random.  The second model takes into account the fact that decks tend to be a bit stacked, and cards can often clump together.  You can expand it out to get a better answer, but it's not a fun problem to do.  You could never do that in your head.  The way to do this problem in a more practical setting is to calcuate a percentage difference and use it to bound the error for using the first (simple) model.  If you shuffle the deck first, you can go back to the first model because you can no longer account for clumping.

I think people tend to forget this point. If someone stacks their deck before a match-up then shuffles it ten times some of the effects of that stacking will remain. When I play a live match I find myself weaving my graveyard into my lands when I pick up my cards in order to prevent land clumping in the following game. It is almost impossible to really randomize a deck in a short amount of time, so with cards like Spoils of the Vault I think the actual playing odds are slightly better than the pure statistical odds. However, the bump is small enough that I don't think it is something people should be worried about.

[The Atog Lord]

if I have already seen a Dark Ritual, would shuffling my deck make it more likley that I would see another Dark Ritual sooner with Spoils than otherwise.  Logically, the answer is no.  A random deck is a random deck.

Yes. A random deck is a random deck. It really doesn't matter if you shuffle a random deck again or not. To think otherwise is exactly the same as thinking that having a little plush monkey on the table next to you will help you get better draws. In other words, wrong.

So, basically, it doesn't matter what your 'testing results' regarding a matchup are, related to Oath's getting a nuts draw game 3. There is some mathematical percentage. This whole matter of conditional probability goes right out the window just as soon as one of those unknown instances becomes a known instance -- in other words, as soon as you know that the opponent did, in fact, get the nuts game 2.

[Smmenen]

Yes, Rich I know.

If I flip a coin a million times and it comes up heads, the probability of the next flip is still 50/50. 

But as human beings, we create conceptual categories that link together unconnected things.  I made a logical fallacy and I admit it. 

[venus to mars]

I'm a bit late in the conversation, but this hasn't been brought up yet. If an event has a 40% probability of occuring, this means that given an incredibly large number of trials, the event will occur in 40% of the trials. 3 is not incredibly large, nor is it even close. We as humans have come to expect such probabilities to be spread out evently, which of course is a highly flawed perception.

[Godder]

The probability of opening with FoW is 40% every single game. If you draw one games one and two, the probability for game three is still 40%.

That ignores the mulligan, though. If a player is willing to mulligan to 6 to get FoW, the probability is about 67% that they will get one. It also doesn't take into account the chance of a hand containing FoW as the only Blue card, which may only be 1 or 2%, depending on the amount of Blue in the list, but is still a not totally insignificant possibility.

[Komatteru]

The odds of finding Force of Will if you mulligan are higher than that.  It's

P(7 cards) + P(6 cards) = 0.399 + 0.3514 = 0.75

This is the odds of seeing a Force of Will in a hand of 6 or 7.  However, once you draw your opener, if you don't have Force of Will, the odds that you get in your new hand of 6 are 0.3514 once again.  It is important to keep things in perspective. The 0.75 number is only useful if you want to think of what the overall odds that you might stop a first turn threat -- a statistical tool that has no use in the real world.

The Hypergeometric distribution (from Excel) is given by this formula, so if you want it and don't want to pull up Excel, use this:
D - number of defective units
N - number of non-defective units
S - sample size
x - number of defective units you wish to pull out

P(x) = binomial(D,x)*binomial(N,S-x)/binomial(D+N,S) = D!/x!/(D-x!) * N!/(S-x)!/(N-S-x)! / ((D+N)!/S!(D+N-S)!)

In a more readable form
                        D! N! (D+N-S)! S!
P(x) = ----------------------------------------------
              x! (D-x)! (S-x)! (N-S-x)! (D+N)!

Or,
D = copies of desired card to be drawn
N = other cards in the deck
S = size hand to be drawn
x = number of copies of desired card to be drawn

Note that if you want to do the odds that you do have a certain card in a sample, it is easiest to do 1 - P(0), as that covers everything.

[Elric]

The odds of finding Force of Will if you mulligan are higher than that.  It's

P(7 cards) + P(6 cards) = 0.399 + 0.3514 = 0.75

This isn't quite right- you can't simply add the probabilities.  You only mulligan to 6 when you don't see Force in the opening 7.

So it should be: P(7 cards) + (1-P(7 cards)) * P(6 cards)= 61.1%

[Komatteru]

If you choose to mulligan, it doesn't matter whether or not FoW was in your opening hand -- since it doesn't affect the overall probabilty of drawing it in a new hand.  Your second probability gives the odds that you will have to mulligan once into FoW.

[venus to mars]

The odds of finding Force of Will if you mulligan are higher than that.  It's

P(7 cards) + P(6 cards) = 0.399 + 0.3514 = 0.75

This isn't quite right- you can't simply add the probabilities.  You only mulligan to 6 when you don't see Force in the opening 7.

So it should be: P(7 cards) + (1-P(7 cards)) * P(6 cards)= 61.1%

If you can't add probabilities why are you adding P(7) to anything? Under that system 3 P(7)s would yield a probability over 1.

Alright:

P(7 cards) = .399
P(6 cards) = .3514


Probability you don't have a force in the opening grip: 1-P(7)
Probability you don't have a force in the first mull: 1-P(6)

Proability that you don't have a force in either the first seven or six:  (1-P(7)) * (1-P(6))

Probability that you have a force in either or both is the composite of this (previous + this = 1.00):
1 - (1-P(7)) * (1-P(6)).