Blocking Probabilities

Hey Wizard & others, I've got a good one for yall!

Series 1: 31,32,33,34
Series 2: 21,22,23,24
Series 3: 11,12,13,14
Series 4: 1,2,3,4

If we randomly are picking 40 numbers from 1 - 80, the probabilities of each of the above series being completed would be equal.

The question I have is, what are the probabilities of Series 2, 3 and 4 being "hit" if only one Series can be counted, and Series 1 takes precedence over Series 2 which takes precedence over Series 3 etc.

So, for instance, if the 40 numbers picked results in Series 1, 3 and 4 being "hit", only Series 1 would be counted in that instance.

Essentially, I'm trying to figure out what the probabilities would be for each Series based on how often they are blocked by the Series that supersede them.


Thanks a lot!

Of the C(80,40) sets of 40 numbers that can be drawn out of the 80:

C(76,36) of them have all four of the Series 1 numbers
P(Series 1) = C(76,36) / C(80,40) = 703 / 12,166 (about 1 / 17.306)

C(76,36) of them have all four of the Series 2 numbers, but this includes the C(72,32) sets that include the Series 1 numbers as well
P(Series 2) = (C(76,36) - C(72,32) / C(80,40) = 122,407 / 2,220,295 (about 1 / 18.139)

C(76,36) of them have all four of the Series 3 numbers, but this includes the C(72,32) sets that include the Series 1 numbers as well and the C(72,32) sets that include the Series 2 numbers - but this is counting the C(68,28) sets that include the Series 1, 2, and 3 numbers twice
P(Series 3) = (C(76,36) - 2 x C(72,32) + C(68,28)) / C(80,40) = 381,215,341 / 7,251,483,470 (about 1 / 19.022)

C(76,36) of them have all four of the Series 4 numbers, but this includes the C(72,32) sets that include the Series 1 numbers, the C(72,32) sets that include the Series 2 numbers, and the C(72,32) sets that include the Series 3 numbers; this is counting the C(68,28) sets that include Series 1, 2, and 4, the C(68,28) sets that include Series 1, 3, and 4, and the C(68,28) sets that include Series 2, 3, and 4 twice - but now you have re-added the C(64,24) sets that include all four series one too many times.
P(Series 4) = (C(76,36) - 3 x C(72,32) + 3 x C(68,28) - C(64,24)) / C(80,40) = 12,170,716,342 / 242,924,696,245 (about 1 / 19.960).

Quote: fitzbean

Hey Wizard & others, I've got a good one for yall!

Series 1: 31,32,33,34
Series 2: 21,22,23,24
Series 3: 11,12,13,14
Series 4: 1,2,3,4

If we randomly are picking 40 numbers from 1 - 80, the probabilities of each of the above series being completed would be equal.

The question I have is, what are the probabilities of Series 2, 3 and 4 being "hit" if only 1 Series can be counted, and Series 1 takes precedence over Series 2 which takes precedence over Series 3 etc.

So, if the 40 numbers picked, if Series 1, 3 and 4 are all completed, only Series 1 would be counted in that instance.

Essentially, I'm trying to figure out what the probabilities would be for each Series based on how often they are blocked by Series that supersede them.


Thanks a lot!

I notice the word, "Blocked." It seems that you don't want other series to hit, but you would rather have Series 4 hit as a best possible result, but no hits from the other Series'?

Okay, let me try to figure that out.

SERIES 4 ONLY

Series 4 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

Because any series, by itself, is just that.

For Series 4 to be blocked, we assume that Series 4 has hit...so that's something that we assume happened. The completion of any other series would, "Block," Series 4, and there are three other series' that are otherwise unrelated:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

That is the probability that ANY other individual series hits in addition to Series 4, (assuming Series 4 has hit already) so now what we are going to do is take that and multiply by the three different series' that would, "Block," Series 4:

0.0459128198133245 * 3 = 0.1377384594399735

Okay, so now what we have to do is take the probability of Series 4 hitting and multiply that by the probability of no other series hitting assuming Series 4 has hit:

0.057783988163735 * (1-0.1377384594399735) = 0.0498249106537644778992779389775

SERIES 3 "ONLY"*

I put, "Only," in this title because it's actually Series 3 OR Series 3 & 4, but as I understand it, Series 3 supersedes Series 4, so it doesn't matter what Series 4 does if Series 3 has hit:

Series 3 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

The next part is exactly the same, except now we only have two series that can block Series 3:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

0.0459128198133245 * 2 = 0.091825639626649

0.057783988163735 * (1-0.091825639626649) = 0.052477936490421318599518625985

SERIES 2 "ONLY"***

***Same thing, this would be for Series 2 hitting which, if assumed, means that neither Series 3 or 4 matters anyway.

Series 2 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

The next part is exactly the same, except now we only have one series that can block Series 2:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

0.057783988163735 * (1-0.0459128198133245) = 0.0551309623270781592997593129925


SERIES 1, WHICH BECOMES THE ONLY THING THAT MATTERS AS IT TAKES PRECEDENCE OVER ALL OTHERS: 0.057783988163735
SERIES 2 ONLY, OR SERIES 2/3/4 OR SERIES 2/3 BUT 2 TAKES PRECEDENCE: 0.0551309623270781592997593129925
SERIES 3 ONLY, OR SERIES 3 & 4 BUT 3 TAKES PRECEDENCE: 0.052477936490421318599518625985
SERIES 4 ONLY: 0.0498249106537644778992779389775

There we go. I hope that I understood your question correctly on at least one of the occasions. Please let me know if I misunderstood your question.

Also, may I ask what the purpose of this was?

It looks like ThatDonGuy and I agree on Series 1 and Series 2 only, but then we start to diverge a little bit for Series 3 & 4:

1/0.057783988163735 = 1 in 17.305832147937410849

1/0.0551309623270781592997593129925 = 1 in 18.1386276928607021585344641492118703

1/0.052477936490421318599518625985 = 1 in 19.055627314586345881430247620677475

1/0.0498249106537644778992779389775 = 20.0702818505595427001792218388242551

ThatDonGuy,

Did I do something wrong, is it rounding, or did we interpret what needed to be done differently?

Quote: ThatDonGuy

Of the C(80,40) sets of 40 numbers that can be drawn out of the 80:

C(76,36) of them have all four of the Series 1 numbers
P(Series 1) = C(76,36) / C(80,40) = 703 / 12,166 (about 1 / 17.306)

C(76,36) of them have all four of the Series 2 numbers, but this includes the C(72,32) sets that include the Series 1 numbers as well
P(Series 2) = (C(76,36) - C(72,32) / C(80,40) = 122,407 / 2,220,295 (about 1 / 18.139)

C(76,36) of them have all four of the Series 3 numbers, but this includes the C(72,32) sets that include the Series 1 numbers as well and the C(72,32) sets that include the Series 2 numbers - but this is counting the C(68,28) sets that include the Series 1, 2, and 3 numbers twice
P(Series 3) = (C(76,36) - 2 x C(72,32) + C(68,28)) / C(80,40) = 381,215,341 / 7,251,483,470 (about 1 / 19.022)

C(76,36) of them have all four of the Series 4 numbers, but this includes the C(72,32) sets that include the Series 1 numbers, the C(72,32) sets that include the Series 2 numbers, and the C(72,32) sets that include the Series 3 numbers; this is counting the C(68,28) sets that include Series 1, 2, and 4, the C(68,28) sets that include Series 1, 3, and 4, and the C(68,28) sets that include Series 2, 3, and 4 twice - but now you have re-added the C(64,24) sets that include all four series one too many times.
P(Series 4) = (C(76,36) - 3 x C(72,32) + 3 x C(68,28) - C(64,24)) / C(80,40) = 12,170,716,342 / 242,924,696,245 (about 1 / 19.960).

How are you calculating that of the C(80,40) that can be drawn, C(76,36) of them will have all four of the Series 1 numbers?

In the same lane, how did you know that of the C(76,36) that have all four Series 2 numbers, C(72,32) of them also include the Series 1 numbers?

Is there a pattern you are following with this or a name/math model for this type of calculation?

Quote: Mission146

It looks like ThatDonGuy and I agree on Series 1 and Series 2 only, but then we start to diverge a little bit for Series 3 & 4:

1/0.057783988163735 = 1 in 17.305832147937410849

1/0.0551309623270781592997593129925 = 1 in 18.1386276928607021585344641492118703

1/0.052477936490421318599518625985 = 1 in 19.055627314586345881430247620677475

1/0.0498249106537644778992779389775 = 20.0702818505595427001792218388242551

ThatDonGuy,

Did I do something wrong, is it rounding, or did we interpret what needed to be done differently?

I think I may have figured out my error and it's that I didn't account for redundancy. That is to say that these hitting:

Series 4, Series 2, Series 1:

In that instance, both Series 2 and Series 1 would, "Block," Series 4...but only one of those actually matters. Series 4 is blocked if either of those things happen. The way I did it ignores overlap blocking...so what I should have done was subtract out any overlap blocking rather than doing a simple *3 on each of the other series' hitting.

That's why I agree with ThatDonGuy on Series 1 and Series 2...overlap blocking does not occur in those instances and Series 1 is never, "Blocked," at all.

Quote: HamHands

How are you calculating that of the C(80,40) that can be drawn, C(76,36) of them will have all four of the Series 1 numbers?

In the same lane, how did you know that of the C(76,36) that have all four Series 2 numbers, C(72,32) of them also include the Series 1 numbers?

Is there a pattern you are following with this or a name/math model for this type of calculation?

Color 31-34 red, and the other 76 balls white. Any draw that contains all of the Series 1 numbers will have 4 red balls and 36 white ones. There is only one way to get all four red balls, and C(76,36) ways to draw 36 of the 76 white ones.

For the Series 1 and 2 sets, color 21-24 blue. There are C(76,36) sets of 40 balls that include 4 blue and 36 non-blue ones. However, every set with both all four red and all four blue balls will have 32 white balls out of 72 possible, so there are C(72,32) sets.

I'm going to do this right and account for redundant blocking because it's bothering me:

(Additions in bold)

I notice the word, "Blocked." It seems that you don't want other series to hit, but you would rather have Series 4 hit as a best possible result, but no hits from the other Series'?

Okay, let me try to figure that out.

SERIES 4 ONLY

Series 4 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

Because any series, by itself, is just that.

For Series 4 to be blocked, we assume that Series 4 has hit...so that's something that we assume happened. The completion of any other series would, "Block," Series 4, and there are three other series' that are otherwise unrelated:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

That is the probability that ANY other individual series hits in addition to Series 4, (assuming Series 4 has hit already) so now what we are going to do is take that and multiply by the three different series' that would, "Block," Series 4:

0.0459128198133245 * 3 = 0.1377384594399735

At this point, we must account for, "Overlap," blocking. That is, if Series 3 also hits as well as Series 2 and/or Series 1 (for example) then the fact that Series 3 has hit is irrelevant. Basically, any occasion where two series, hit, or all three other series' hit, renders the highest numbered series (or Series 2 & 3) irrelevant.

nCr(8,8)*nCr(68,28)/nCr(76,36) = 0.0016048221702069

This can happen where Series 3 & Series 1 hit simultaneously, Series 3 & Series 2 hit simultaneously or Series 1 & Series 2 hit simultaneously, so three possibilities of two other series redundancy or overlap:

0.0016048221702069 * 3 = 0.0048144665106207

Finally, all three other series' hitting would also be redundant, and is represented by:

nCr(12,12)*nCr(64,24)/nCr(76,36) = 0.0000403479115345

Now, we subtract out our overlaps:

0.1377384594399735-0.0048144665106207-0.0000403479115345 = 0.1328836450178183

Okay, so now we figure out the probability of Series 4 hitting alone:

0.057783988163735 * (1-0.1328836450178183) = 0.0501054411928714239496271206495

SERIES 3 "ONLY"*

I put, "Only," in this title because it's actually Series 3 OR Series 3 & 4, but as I understand it, Series 3 supersedes Series 4, so it doesn't matter what Series 4 does if Series 3 has hit:

Series 3 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

The next part is exactly the same, except now we only have two series that can block Series 3:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

0.0459128198133245 * 2 = 0.091825639626649

We must also account for overlap blocking here, which occurs only if Series 1 and Series 2 both also hit simultaneously along with Series 3:

nCr(8,8)*nCr(68,28)/nCr(76,36) = 0.0016048221702069

We subtract both hitting out from our probability of either hitting:

0.091825639626649 - 0.0016048221702069 = 0.0902208174564421

0.057783988163735 * (1-0.0902208174564421) = 0.0525706695157094536246746527565

SERIES 2 "ONLY"***

***Same thing, this would be for Series 2 hitting which, if assumed, means that neither Series 3 or 4 matters anyway.

Series 2 is just this: nCr(4,4)*nCr(76,36)/nCr(80,40) = 0.057783988163735

The next part is exactly the same, except now we only have one series that can block Series 2:

nCr(4,4)*nCr(72,32)/nCr(76,36) = 0.0459128198133245

0.057783988163735 * (1-0.0459128198133245) = 0.0551309623270781592997593129925

SERIES 1, WHICH BECOMES THE ONLY THING THAT MATTERS AS IT TAKES PRECEDENCE OVER ALL OTHERS: 0.057783988163735

SERIES 2 ONLY, OR SERIES 2/3/4 OR SERIES 2/3 BUT 2 TAKES PRECEDENCE: 0.0551309623270781592997593129925

SERIES 3 ONLY, OR SERIES 3 & 4 BUT 3 TAKES PRECEDENCE: 0.052477936490421318599518625985<---This was wrong, correct is 0.0525706695157094536246746527565 or 1 in about 19.02201

SERIES 4 ONLY: 0.0498249106537644778992779389775<---This was wrong, correct is 0.0501054411928714239496271206495 or 1 in about 19.95791

There we go. I hope that I understood your question correctly on at least one of the occasions. Please let me know if I misunderstood your question.

Also, may I ask what the purpose of this was?

Basically, my mistake was that I looked at additional series' hitting along with Series 4 and/or Series 3, when relevant, as an, "Or," question.

The way I originally did it asked this question, "What is the probability that Series 1 OR Series 2 OR Series 3 hits in addition to Series 4?" Or, for the Series 3 question I asked, "What is the probability that Series 1 OR Series 2 hits in addition to Series 3?"

The problem is that any combination of Series' hitting in addition to Series 4 creates redundancy, which means that I should have asked my question like this, "What is the probability that any series hits in addition to Series 4, or that multiple series hit in addition to Series 4?"

To solve that, I need to come up with the following probabilities:

Series 1 & 4 Only
Series 2 & 4 Only
Series 3 & 4 Only
Series 1, 2 & 4
Series 1, 3 & 4
Series 2, 3 & 4

Similarly for the AND question when we look at Series 3 hitting, but the only possible AND that would matter is Series 1 & 2.

Big props to ThatDonGuy who, unlike me, accounted for the possibility of redundancy on the first try. I also owe him a thank you because getting a different answer from him (and correctly assuming that I was the one who did something wrong) led me to discovering my mistake and coming up with the correct answer using my preferred long method.

Also, I'm still curious as to what this was for!

Quote: ThatDonGuy

Color 31-34 red, and the other 76 balls white. Any draw that contains all of the Series 1 numbers will have 4 red balls and 36 white ones. There is only one way to get all four red balls, and C(76,36) ways to draw 36 of the 76 white ones.

For the Series 1 and 2 sets, color 21-24 blue. There are C(76,36) sets of 40 balls that include 4 blue and 36 non-blue ones. However, every set with both all four red and all four blue balls will have 32 white balls out of 72 possible, so there are C(72,32) sets.

Interesting visual that makes this very clear.

When solving an this issue, do you have to calculate it all out by hand, visualizing the blocks or is there a formula of sorts to calculate this? Like what if the same problem came up but there were 20 series? Or 40 series? Would it be necessary to calculate it all by hand?

Quote: HamHands

When solving an this issue, do you have to calculate it all out by hand, visualizing the blocks or is there a formula of sorts to calculate this? Like what if the same problem came up but there were 20 series? Or 40 series? Would it be necessary to calculate it all by hand?

There may be a formula, but I can't think of it off the top of my head. Calculating it "by hand" isn't that hard, except maybe for the combination values themselves.

Quote: ThatDonGuy

There may be a formula, but I can't think of it off the top of my head. Calculating it "by hand" isn't that hard, except maybe for the combination values themselves.

Would you have any suggestions on what to research to understand this more/find a formula?

Quote: Mission146

Also, I'm still curious as to what this was for!

Me and Ham are actually working on a project involving Bingo patterns and probabilities of certain patterns taking precedence over others, but we wanted to present the question as simply as possible, which is how we ended up with just series of numbers and picking a certain amount of numbers against them.

On that note, it is entirely possible that the series would share numbers --

I.E. If the Series were;

Series 1: 31,32,33,34
Series 2: 21,22,33,34
Series 3: 21,22,13,14
Series 4: 1,2,3,4

So in this example, Series 2 shares two numbers with Series 1 and Series 3 shares two numbers with Series 2. We believe dealing with overlapping series is as simple as changing the values in the Combin()s, are we right in that thinking?

Thanks in advance, this is very helpful!

I've not thought about it much but if you were only considering 16 numbers (or fewer) out of 75/80/90 then the other numbers are irrelevant. So if you picked 1 2 3 4 11 12 13 14 21 22 23 24 31 32 33 34 you would not be interested in the other numbers. Also you don't care which order 1 2 3 or 4 come out, so can consider them as AAAA BBBB CCCC DDDD (logically 16 balls). This should make the maths easier. Obviously if you have overlaps, as above, then it's AABB CCBB CCDD EEEE, which has 12 balls (which you can easily do by brute force).