Regarding the article

I have some doubts about the formula in here

-First of all, the formula presented is missing a parameter say “M” to represent the number of groups ( committees we have), where the probability P of myopic represents having say “X” malicious such that

P=X/NM

So there must be something wrong in this formula

-True at the end the shuffling algorithm chooses one at random, this could mean to divide the resulting probability by 1/M, but not to execlude M at all; I don’t think things are that simple.

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-Bernoulli trials ( which the binomial distribution is based on) only have two outcomes ( like throwing a coin), but we here have more than 2 groups; it looks like the number of slots was mixed up with the number of groups or maybe the simple example above it.

-It could be more viewed like throwing a dice only with M outcomes (I mean throwing a dice is a special case or example with M=6)

»»»I think there maybe(probably) a known formula or distribution for such probability that I can’t recall now

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-Another way of viewing it is like u have an N*M bits number where u know have exactly X=P*N*M 1s (or 0s), we want to know the probability that when partitioning the number to groups of N contiguous bits, at least N/2 of the X 1s will be in one of those groups.

Till we are able to derive a general formula ( or someone remind us of known one from the literature), please follow this simple examples with me to realize the error in the equation when we have more than 2 groups; ie. M≥3

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-Let M=3, N=5 (15 attestor divided into 3 groups of 5 each)

-say X=4 ( we have 4 malicious/myopic persons) where 0.25≤(P=4/15)≤0.33

P~=0.2666…

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The 4 malicious (consider them persons, ie identity matters let’s call them A,B,C,D)

Have 3⁴=81 ways to be distributed over the 3 groups (each has 3 choices to one of the groups, and in general these choices are indep as long as N≥X)

-Possible outcomes are:

4,0,0 ==⟩ can happen in 3 ways

(which of the 3 groups they will be in)

3,1,0 ==⟩ can happen in 4*3*2=24 ways:

Pick each of the 4 be the single one and choose its group in 3 different ways, the remaining have 2 choices of the 2 remaining groups; in detail

ABC,D,0/ABC,0,D/0,ABC,D/0,D,ABC/D,ABC,0/

D,0,ABC

Then repeat for

ABD,C,0

BCD,A,0

ACD,B,0

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Till now prob =(3+24)/81=27/81=1/3

That’s larger than initial probability (if we re-divide by3 for the shuffler selection we get 1/9 still not the same as the presented formula).

We can check the rest of options (less than 50% of N) to be sure the enumeration of all cases add up to 81 so we are not wrong.

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-2,2,0 ==⟩3*6=18

-Divide them into 2 groups in 3 ways?

AB,CD/AC,BD/AD,BC

that’s it

-now arrange the 3 partitions (these two& zero)in the 3 groups in 3!=6 ways

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2,1,1==⟩ 6*6 =36*

1st two& their group in 123/2=6ways

The arrangement in the groups in 3!=6 ways

AB,C,D/AB,D,C/C,AB,D/C,D,AB/D,AB,C/D,C,AB

AC,B,D/AC,D,B

AD,B,C/AD,C,B

BC,A,D/BC,D,A

BD,A,C/BD,C,A

CD,A,B/CD,B,A

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Total=3+24+18+36

=81 ✓

1st two& their group in 12

If u find this example hard to follow check the case when X=3 (P=3/15=0.2)

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A,B,C

A(1,2,3),B(1,2,3),C(1,2,3) 27 ways

3,0,0 ⟩ 3 ways

Prob(≥50%)=3/27=1/9

check the rest:

2,1,0 ⟩ 3*3*2=18ways

1,1,1⟩ordering=3!=6ways

## total=3+6+18=27 ✓

If u get it now, check the case where P=1/3, ie X=5 like N, we have the possibilities:

5,0,0 ==⟩ could happen in 3 ways

4,1,0 ==⟩ 30

5 ways to choose the single one* 3!=6 to arrange

3,2,0 ==⟩60

5*4/2=10 to choose 3 out of 5, then * 3!=6 to arrange
3,1,1 ==⟩60
5*4/2=10 to choose 3 out of 5, then * 3!=6 to arrange

2,2,1 ==⟩90

5 to choose the single one, * selecting 2of4 4

*3/2=6, /2 divided by 2 for the order of the 2 elements partitions doesn’t count,*6/2=90

*3!=6; ie 5*6Total no of ways =3⁵=243

Prob(≥50%)= (3+30+60+60)/243 =153/243 =51/81 … ≥ 0.5

If we divide by 3 again for the o/p of the shuffling is only 1 committee

51/243=0.209

Now let’s get back to the formula in the article & substitute with N=5 for both values of P to see the difference in the results:

*When X=3, p=0.2
the summation has only one term
Prob=(0.2)³ *64/10⁴= 0.0512

*(0.8)²*10 =8

»»»» Does NOT equal 1/9

If we tried to divide by 1/3 (maybe this is his point)

1/27~0.037037…

still NOT the same

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*When X=4, P=4/15, 1-P=11/15*

The summation has 2 terms 4,1 and 3,2

Prob= (4/15)⁴(11/15)

The summation has 2 terms 4,1 and 3,2

Prob= (4/15)⁴

*5 + (4/15)³*(11/15)²*(5

*4/2)*

=(4³/15⁵)2]

=(4³/15⁵)

*11*5[4+11=(64

*11*26)/(3*15⁴)

=18,304/151,875

=0.12052

»»»» Does NOT equal 1/3

If we tried to divide by 1/3 (maybe this is his point)

1/9~0.11111…

still NOT the same

*When X=5, P=1/3, 1-P=2/3
The summation has 3 terms 5,0 & 4,1 & 3,2
Prob= (1/3)⁵ + (1/3)⁴*(2/3)

*5 + (1/3)³*(2/3)²*(5

*4/2)*

=(1/3)⁵10]

=(1/3)⁵

*[1+ 2*5 + 4=51/243

Only in this case when X=N, the probability is the same as after selecting a group at random, but this is a special case; and yet the probability is not very small though

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I know I didn’t derive a fixed formula or recall an existing one from the literature yet, but at least this to point out the written one is not accurate?