Probability notes:

The game of RISK is played such that two players are allowed to "attack"
and "defend" in the following ways:

a) the "attacker" may roll 1, 2, or 3 dice.
b) the "defender" may roll 1 or 2 dice.
c) the highest rolled attack dice is compared to
   the highest rolled defense dice, and
   1. the attacker loses if the attack dice <= defense dice,
   2. the defender loses if the attack dice > defense dice.
d) The compared dice are discarded, and if each player
   still has dice left, repeat c) above.

Standard dice have six faces numbered from 1 through 6.  The probability
of throwing a specific number is 1/6.  If we call any specific number N,
then the probability of throwing "N or less" with a single dice is the
sum of the probabilities for each value from 1 to N.  This yields N/6 as
the probability of throwing "N or less" with a single dice.  Likewise,
the probability of throwing "more than N" is 1-(N/6) or (6-N)/6.

When rolling two dice, the probability of rolling "N or less" is given
by the multiplicative law as:  (N/6)(N/6).  For example, the probability
of throwing 5 or less with two dice is 25/36.  Similarly, for three
dice:  (N/6)(N/6)(N/6).  The general form is:  (N/6) to the i-th power,
where "i" is the number of dice thrown.

At this point, I'd like to introduce some notational convieniences.
In most probability formulas I'll be dealing with, the "numerator" is
defined as a series of terms, and the "demoninator" is a single value.
For example, if probability P is defined as n/d (P=n/d), and Q is
defined as 1-P, then we have the following:

     P = n/d
     Q = 1 - P
       = 1 - (n/d)
       = (d/d) - (n/d)
       = (d-n)/d
     Q = d-n [/d]

The last line shows the special notation of [/d] meaning:
"all divided by d".  This makes the numerator easier to read.
You can think of the numerator as being the "frequency" of
occurrence out of all possible outcomes.

The other convention is for "x to the i-th power", which will be given
as:  x'i

For example:  x'0  = 1
              x'1  = x
              x'2  = x squared
              x'3  = x cubed
              x'4  = x to the 4-th power
etc.

Now let us examine some RISK cases.  First, the "1 on 1" attack, where
the "attacker" rolls only one dice and the "defender" only rolls one
dice.  If the "defender" rolls a 6, then the "attacker" loses regardless
of what value he throws.  The probability of this event is:  (1/6)(6/6)
since (1/6) is the probability the "defender" rolls a 6, and (6/6) is
the probability the "attacker" rolls "6 or less".  If the "defender"
rolls a 5, then the "attacker" loses if he rolls a 5 or less.  The
probability of this event is:  (1/6)(5/6) since (1/6) is the probability
the "defender" rolls 5, and (5/6) is the probability the "attacker"
rolls "5 of less".  The sum of all these events (for N=1 to 6) yields
the probability the "attacker" will lose, which is:

   P(A loses) = (1/6) [(1+2+3+4+5+6)/6]
              = 1+2+3+4+5+6 [/6'2]    = 21/36  = 756/1296

In a "2 on 1" attack, we obtain:

   P(A loses) = (1/6) [(1+4+9+16+25+36)/36]
              = 1+4+9+16+25+36 [/6'3] = 91/216 = 546/1296

Therefore, the probability the "defender" loses is:

   P(D loses) = 1 - P(A loses)
              = 216 - 91 [/6'3]      = 125/216 = 750/1296

Since the "attacker" is using two dice, we must use the
(N/6)'2 probabilities of throwing "N or less" in the sum.

Likewise, in the "3 on 1" attack, we obtain:

   P(A loses) = (1/6) [(1+8+27+64+125+216)/216]
              = 1+8+27+64+125+216 [/6'4]       = 441/1296

In a "1 on 2" attack, we can compute the probability the "attacker"
loses by computing the probability the "defender" loses and subtract
that from "unity".  The "defender" loses when the "attacker" rolls N
with his dice, and the "defender" rolls "N-1 or less" with both of his
dice.  The probabilities for the "defender" are:  (N-1/6)'2 for N=1 to
6, which yields:

     P(D loses) = (1/6) [(0+1+4+9+16+25)/36]
                = 1+4+9+16+25 [/6'3] = 55/216  = 330/1296

Therefore, the probability the "attacker" loses is:

     P(A loses) = 1 - P(D loses)
                = 216 - 55 [/6'3]   = 161/216  = 996/1296

A summary of the probabilities we've computed thus far, in
decending order by "attacker" losing, would look like this:

  Attack    Defend    P(A loses)  P(D loses)

     1   on   2       966/1296    330/1296

     1   on   1       756/1296    540/1296

     2   on   1       546/1296    750/1296

     3   on   1       441/1296    855/1296

When rolling two dice, the probablity, H(N,2), that the
"highest" number rolled is "N", and the probability, L(N,2),
that the "lowest" number rolled is "N" is given by:

   N      1       2       3       4       5       6
 H(N,2)   1       3       5       7       9      11   [/36]
 L(N,2)  11       9       7       5       3       1   [/36]

These probabilities are determined as follows:  The probability of
throwing N on one dice AND "N or less" on the other is:  (1/6)(N-1)/6
for N on the first dice and less than N on the second, plus (1/6)(N-1)/6
for N on the second dice and less than N on the first dice, plus
(1/6)(1/6) for N on both dice.  Thus:

   H(N,2) = (1/6)(N-1)/6 + (1/6)(N-1)/6 + (1/6)(1/6) = (2N-1)/36

Since the probability of throwing "more than N" on a dice is (6-N)/6,
the probability of throwing N on one dice AND N or more on the other is:

   L(N,2) = (1/6)(6-N)/6 + (1/6)(6-N)/6 + (1/6)(1/6) = (13-2N)/36

Notice that for any particular N, the sum of H(N) and L(N) is 12/36 or
1/3.  Also, when two dice are thrown, H(N) represents the probabilities
of rolling N as the "highest" dice, and L(N) represents the probabilities
of rolling N as the "lowest" dice.

We could have arrived at these probabilities another way.  The
probability of throwing N as the "highest" dice with two dice must the
probability of throwing "N or less" MINUS the probability of throwing
"N-1 or less" with two dice.  We already know that (N/6)'i gives the
probability of throwing "N or less" with "i" dice, so the probability we
desire is:  (N)'i - (N-1)'i [/6'i].  (Use M=N-1, calculate with M, then
substitute N-1 for M.)

    i       H(N,i) = (M+1)'i - M'i = (N)'i - (N-1)'i [/6'i]

    1         1                      =  1            [/6]
    2         2(N-1) + 1             =  2(N-1) + 1   [/36]
    3         3(N-1)'2 + 3(N-1) + 1  =  3N(N-1) + 1  [/216]
  etc.

Notice that for i=2 we derive (2N-1)/36 as the probability of
throwing N as the "highest" of two dice.

In general, each column in a table of outcomes totals to (i/n)(n'i).
For n=6, two dice sum to 1/3 of 36, 3 dice sum to 1/2 of 216, etc.

With three dice, we get the following table:

   N      1       2       3       4       5       6
 H(N,3)   1       7      19      37      61      91   [/216]
 M(N,3)  16      40      52      52      40      16   [/216]
 L(N,3)  91      61      37      19       7       1   [/216]

Notice that the sum of the H(N,i) probabilities for N from N=1
through N=N is:  N'i [/6'i]  when "i" dice are thrown.
This is a consequence of throwing "N or less" with "i" dice.

In the "3 on 2" case, each column of probabilities under N for H(N,3)
and L(N,3) sum to the same amount: 3 x 6'2 = 108.  Since H(N,3) is the
reverse of L(N,3), we can determine the M(N,3) row in the table without
a computer.  The Attacker table then becomes:

  M(N,3) = 108 - H(N,3) - L(N,3)

Let's return to the game of RISK and examine the probabilities for
the "2 on 2" attack, where the attacker loses the first dice by:

   1(1) + 3(4) + 5(9) + 7(16) + 9(25) + 11(36) [/6'4]  = 791 [/1296]

and the defense loses the first dice by:

   1(0) + 3(1) + 5(4) + 7(9)  + 9(16) + 11(25) [/6'4]  = 505 [/1296]

What you see above are what I call "probability formulas", consisting
of a numerator with a series of terms in which "multiple" means "and",
and "sum" means "or".  These are "(A and B) or (C and D) ..." events.

These probabilities were derived as follows:  The probability the
attacker loses the first dice is the sum of the probabilities of the
defender throwing N as his "highest" dice AND the attacker throws N or
less as his "highest" dice.  The defender throwing N as his "highest" is
given by H(N,2) [/6'2].  For example, if the defender throws 6 as his
"highest" (with 11/36 probability), the attacker can throw 6 or less
(anything) and loses.

The probability the attacker throws N or less as his "highest" dice is
the sum of the "highest" dice probabilities up to and including N.
That's the sum of the H(N,2) probabilities thru N, which we already know
is given by:  N'2 [/6'2].

So the terms of the attacker loses formula are H(N,2)(N'2) [/6'4] for N
from 1 thru 6.  For example, if the defender throws 6 as his "highest"
(with 11/36 probability),the attacker can throw "6 or less"
(anything=36/36) and loses.  That's the "11(36) [/6'4]" term in the
attacker loses formula.

Similarly, we can determine the probability that the defender loses the
first dice.  If the attacker throws N as his "highest" dice, then the
defender must throw N-1 or less as his "highest" dice to lose.  The N-1
or less probabilities are again the sum of H(N) terms thru N-1, and the
attacker's probability is H(N).  Again, the sum of these terms yields
the probability that the defender loses the first dice.

Once the first dice (highest) has been decided, we must determine the
outcome of the second dice.  The second dice is always the "lowest" for
both players.  L(N) gives the probabilities for each N being the "lowest".

(defender wins) or (attacker loses) the second dice by:

   11(11) + 9(20) + 7(27) + 5(32) + 3(35) + 1(36) [/6'4] = 791 [/1296]

(attacker wins) or (defender loses) the second dice by:

   11(0)  + 9(11) + 7(20) + 5(27) + 3(32) + 1(35) [/6'4] = 505 [/1296]

These probabilities were derived in a manner similar to the first dice
losses.  For example, the 5th term in the attacker equation is 3(35)
which is the probability of the defender throwing 5 as his "lowest" dice
(3/36), AND the attacker throws 5 or less as his "lowest" dice, which is
the sum of L(N) from 1 thru 5 (35/36).  The sum is given by:  (12-N)(N)
[/36]  for each value of N.  That is equivaluent to:  36 - (6-N)'2 [/36].

A computer program was written to compute the probabilities for
both the "2 on 2" and "3 on 2" attacks.  The results were:

     Attacker loses:    "2 on 2"    "3 on 2"   Attacker wins:
        neither           295         2890       both (1st+2nd)
        only 1st          210         1834       only 2nd
        only 2nd          210          777       only 1st
        both (1st+2nd)    581         2275       neither
     ----------------------------------------------------------
     Total events:       1296         7776

The computer derived counts for Attacker loses were 791 = 210 + 581
(1st + both), and 505 = 295 + 210  (neither + 2nd) in the "2 on 2"
attack, which corresponds to the probabilities that were computed for
the "attacker" and "defender" losing the first compare (highest).
Likewise, 791 = 210 + 581  (2nd + both), and 505 = 295 + 210
(neither + 1st), which are the probabilities for the "attacker"
and "defender" losing the second compare (lowest).
A summary of the computer derived probabilities yields:

                      Defend +2    Each +1    Attack +2
  Attack    Defend    Attack -2    Each -1    Defend -2

     2   on   2       581/1296    420/1296    295/1296
     [/6'4] = 1296          1001/1296    715/1296

     3   on   2       2275/7776   2611/7776   2890/7776
     [/6'5] = 7776          4886/7776   5501/7776

The two probabilities listed on lines by themselves are the sums of the
two nearest terms above them, and they represent the "attacker" or
"defender" loses "at least one".

For the "3 on 2" scenario, my computer program "rolls" all combinations of
3-dice for the attacker, like an odometer; sorts each result into a separate
descending list, and then does all combinations of 2-dice for the defender,
and sorts each result into another list.  It compares 1st-to-1st in each list
(1st battle), then 2nd-to-2nd (2nd battle), and then takes 0 and adds 1 for
the 1st attacker win, and 2 for the 2nd attacker win.  The result is 0,1,2,3,
an index into a table of four counters that record:  attacker wins another at
this index.  After doing all combinations for the defender, I then go back and
get another attacker list, and run the defender combinations against that, etc.
When I'm all done (all attacker with all defender combinations), I have a
table of four-values for attacker wins:  none, 1st only, 2nd only, or both.
That table contains:  2890, 1834, 777, 2275 which matches the Attacker wins
column shown earlier.  Many sites which discuss Risk Probabilities sum together
the 1834 and 777 to give a 2611 probabality for each winning only one battle.

It's interesting to note that by using the H(N) values for the "highest"
of three dice, along with the "N or less" figures for both i=2 and i=3,
we can compute the probabilities of the "attacker" losing the first
dice, and the "defender" losing the first dice in the "3 on 2" attack as
follows:

(defender wins) or (attacker loses) the first dice by:

D(sA) = 1(1) + 3(8) + 5(27) + 7(64) + 9(125) + 11(216) = 4109
A(sD) = 1(36) + 7(35) + 19(32) + 37(27) + 61(20) + 91(11) = 4109

(attacker wins) or (defender loses) the first dice by:

A(sD) = 1(0) + 7(1) + 19(4) + 37(9) + 61(16) + 91(25)  = 3667
D(sA) = 1(215) + 3(208) + 5(189) + 7(152) + 9(91) + 11(0) = 3667

The notations: (sA) and (sD) mean "partial sum of terms" for A and D.

These values correspond exactly with the computer derived values:

   4109 = 2275 + 1834   and   3667 = 2890 + 777

To see if I could shed more light on this problem, I wrote another
computer program which not only computed the probabilities for the
four cases: Attacker wins both, Attacher wins 2nd only, Attacker wins
1st only, and finally, Attacker wins none; but also output the number
of times each dice value was rolled for each case.  By that I mean, it
showed the number of times 1, 2, 3, 4, 5, 6 was rolled in the event.
I output these as tables, six wide by four deep.  There are tables for
all 3 of the Attacker's dice, and 2 tables for Defender's dice.
Here are the tables:

 Dice number thrown       1    2    3    4    5    6
                                                            Summation
 First battle (1st die)
 Attacker wins both       0    4    55   255  766  1810          2890
 Attacher wins 2nd only   0    40   184  414  616  580           1834
 Attacker wins 1st only   0    3    21   78   210  465           777
 Attacher loses both      36   205  424  585  604  421           2275

 Second battle (2nd die)
 Attacker wins both       0    220  568  846  856  400           2890
 Attacher wins 2nd only   0    220  472  558  424  160           1834
 Attacker wins 1st only   165  270  210  105  27   0             777
 Attacher loses both      411  730  622  363  133  16            2275

 Attacker 3rd die
 Attacker discard both    885  811  622  384  163  25            2890
 Attacker discard 2nd     630  550  376  198  70   10            1834
 Attacker discard 1st     465  210  78   21   3    0             777
 Attacker discard none    1296 625  256  81   16   1             2275
                                                                   ^
 Summation                3888 3888 3888 3888 3888 3888 <- 7776/2  ^
                                                                 7776
 First battle (1st die)
 Defender loses both      200  552  774  800  564  0             2890
 Defender loses 2nd only  0    8    54   192  500  1080          1834
 Defender loses 1st only  15   72   171  264  255  0             777
 Defender wins  both      1    16   81   256  625  1296          2275

 Second battle (2nd die)
 Defender loses both      1420 880  432  142  16   0             2890
 Defender loses 2nd only  780  560  324  138  32   0             1834
 Defender loses 1st only  75   192  243  192  75   0             777
 Defender wins  both      101  312  513  608  525  216           2275
                                                                   ^
 Summation                2592 2592 2592 2592 2592 2592 <- 7776/3  ^
                                                                 7776

Notice that each row of the four events add up to the four probabilities.
Also notice that the columns for the three Attacker dice add up to 7776/2,
and the Defender dice columns add up to 7776/3.  When the Attacker wins with
a 6, the Defender didn't roll a 6, otherwise the Defender would win.  That's
why you see 0 under the 6 for Defender losing both and 1st in the first battle,
and all three losing cases in the second battle.  There must be a clue here.

Don't misinterpret the numbers in the rows of these tables.  For example, in
the First battle, the Defender loses 2nd only, the Defender rolled a 6 and won
the 1st battle, but not the 2nd battle.  Each set of four rows shows Attacker
wins: both, only 2nd, only 1st, or neither.  The Attacker and Defender dice for
BOTH battles populate these four rows in every table, with one table for each
die rolled by each player.  So when the Attacker rolled a 6 in the 1st battle,
and the Defender rolled 5 or less (not a 6), the Attacker won the 1st battle.
This is a complex matrix to understand.

What is the probability the "Attacker" will win both dice in a "2 on 2"
or "3 on 2" attack?  Without the computer, I'm unable to answer that
question!  What laws apply?



I then wrote another computer program designed to show the probabilities
of rolling N as the "highest", "next highest", etc. down to "lowest"
when rolling "i" dice.  I got:

              N       1       2       3      4      5      6

   For i=2   H(N)     1       3       5      7      9     11
   [/36]     L(N)    11       9       7      5      3      1

   For i=3   H(N)     1       7      19     37     61     91
   [/216]    M(N)    16      40      52     52     40     16
             L(N)    91      61      37     19      7      1

   For i=4   H(N)     1      15      65    175    369    671
   [/1296]           21     123     261    363    357    171
                    171     357     363    261    123     21
             L(N)   671     369     175     65     15      1

Notice that each column adds to the same amount, which turns out to be:
i x 6'(i-1) [/6'i]  (expressed as probabilities).  These turn out to be
simply i/6, yielding the series: 2/6, 3/6, 4/6, 5/6.  You can also see
that the sum of H(N) terms from 1 thru N is equal to N'i at all times.
For example, for i=4 at N=3 you have: 3'4 = 1+15+65 = 81.

From what I've observed, it appears reasonable to generalize the
probability the "attacker" (and "defender") lose the first dice compare
("highest") when rolling n-sided dice where each side has a different
number, and all sides are equally probable.  If the "attacker" rolls "a"
dice, and the "defender" rolls "d" dice, and they are n-sided dice
(n>=2), then the probabilies for the "attacker" and "defender" losing
the "highest" are:

   P(A loses) = 1 - P(D loses)  or
   P(A loses) = Sum {(k'd - (k-1)'d) (k'a)}[/n'(a+d)]  for k=1 to n

   P(D loses) = 1 - P(A loses)  or
   P(D loses) = Sum {(k'a - (k-1)'a) ((k-1)'d)}[/n'(a+d)]  for k=1 to n

Each sum is divided by [/n'(a+d)] to make it a probability.

When k=1, the P(A) term equals one, and the P(D) term is zero.  If all
the terms for both P(A) and P(D) are added together, you will notice
that almost all terms cancel.  Only (n'd)(n'a) is left, which is the
same as n'(a+d)[/n'(a+d)], or total of 1.

Using these formulas I was able to obtain the same values of probability
for P(A loses) and P(D loses) for all cases where either A or D throws
only one dice, that is, all "a on 1" or "1 on d" attacks.

We could devise a case with n=2 using coins where a "head" is considered
to be higher in value than a "tail".  In such a case, if the "attacker"
tossed 2 coins, and the "defender" also tossed 2 coins, we'd get:

    P(A) = 1 +  (2'2-1)(2'2) [/2'4] = 13/16
    P(D) = 0 +  (2'2-1)(1)   [/2'4] =  3/16

as the probabilities of losing the first compare.  Determining the
probability of loss for the second compare can be done like the "2 on 2"
dice problem, but a generalization for the second (or subsequent)
compare, especially with "a" or "d" being 3 or more, does not seem easy.
In fact, deriving the probability the "attacker" loses ALL compares,
when there are two or more, does not seem possible!  Any ideas?



In the "3 on 2" attack, each column of probabilities under N for H(N,3)
and L(N,3) sum to the same amount: 3 x 6'2 = 108.  Since H(N,3) is the
reverse of L(N,3), we can determine the M(N,3) row in the table without
the computer.  The Attacker table then becomes:

  M(N,3) = 108 - H(N,3) - L(N,3)

    N        1       2       3      4      5      6
  H(N,3)     1       7      19     37     61     91   [/216]
  M(N,3)    16      40      52     52     40     16   [/216]
  L(N,3)    91      61      37     19      7      1   [/216]

The Defender table is as follows:

   N      1       2       3       4       5       6
 H(N,2)   1       3       5       7       9      11   [/36]
 L(N,2)  11       9       7       5       3       1   [/36]

We already computed 1st dice losses earlier, as follows:

(defender wins) or (attacker loses) the first dice by:

D(sA) = 1(1) + 3(8) + 5(27) + 7(64) + 9(125) + 11(216) = 4109
A(sD) = 1(36) + 7(35) + 19(32) + 37(27) + 61(20) + 91(11) = 4109

(attacker wins) or (defender loses) the first dice by:

A(sD) = 1(0) + 7(1) + 19(4) + 37(9) + 61(16) + 91(25)  = 3667
D(sA) = 1(215) + 3(208) + 5(189) + 7(152) + 9(91) + 11(0) = 3667

The notations: (sA) and (sD) mean "partial sum of terms" for A and D.

We can determine that the defender will lose the 2nd dice
compare when the following conditions are true:

      Defender rolls a 1, and Attacker rolls more than 1,
  or  Defender rolls a 2, and Attacker rolls more than 2,
  or  Defender rolls a 3, and Attacker rolls more than 3,
  or  Defender rolls a 4, and Attacker rolls more than 4,
  or  Defender rolls a 5, and Attacker rolls more than 5,
  or  Defender rolls a 6, and Attacker rolls more than 6. (0)

The Defender terms are L(N,2), and the Attacker terms are the
sum of the M(N,3) terms above the given N.  Therefore:

D(sA) = 11(200) + 9(160) + 7(108) + 5(56) + 3(16) + 1(0) = 4724
A(sD) = 16(0) + 40(11) + 52(20) + 52(27) + 40(32) + 16(35) = 4724

This is the probability the Defender loses the 2nd dice, and
it matches the computer derived value:  4724 = 2890 + 1834.

The Attacker loses the 2nd dice can be simply determined as:

    P(A loses) = 1 - P(D loses) = 7776 - 4724 = 3052 [/7776]

This probability also matches the computer derived value:

    3052 = 2275 + 777

This probability could also be derived from L(N,2) and M(N,3)
for Defender and Attacker starting with the following:

    Defender rolls a 1, and Attacker rolls 1,
 or Defender rolls a 2, and Attacker rolls 2 or less,
 or Defender rolls a 3, and Attacker rolls 3 or less,
 or Defender rolls a 4, and Attacker rolls 4 or less,
 or Defender rolls a 5, and Attacker rolls 5 or less,
 or Defender rolls a 6, and Attacker rolls 6 or less (anything).

Therefore:

D(sA) = 11(16) + 9(56) + 7(108) + 5(160) + 3(200) + 1(216) = 3052
A(sD) = 16(1) + 40(4) + 52(9) + 52(16) + 40(25) + 16(36) = 3052

We now have the following compare probabilities:

    4724 = Defender loses 2nd, Attacker wins 2nd
    3667 = Defender loses 1st, Attacker wins 1st
    3052 = Attacker loses 2nd, Defender wins 2nd
    4109 = Attacker loses 1st, Defender wins 1st

Therefore:

    4724 = Defender loses both  +  Defender loses only 2nd.
    3667 = Defender loses both  +  Defender loses only 1st.
    3052 = Attacker loses both  +  Attacker loses only 2nd.
    4109 = Attacker loses both  +  Attacker loses only 1st.

When the Attacker wins both, the Defender loses both.
When the Attacker wins only the 1st, the Defender wins only the 2nd.
When the Attacker wins only the 2nd, the Defender wins only the 1st.
When the Attacker loses only the 1st, the Defender loses only the 2nd.
When the Attacker loses only the 2nd, the Defender loses only the 1st.
When the Attacker loses both, the Defender wins both.

Unfortunately, these are all dependent equations, and you can't figure
out "Attacker loses both".

What we are looking for is a solution for the following table:

     Attacker loses:    "2 on 2"    "3 on 2"   Attacker wins:
        neither           295         2890       both (1st+2nd)
        only 1st          210         1834       only 2nd
        only 2nd          210          777       only 1st
        both (1st+2nd)    581         2275       neither
     ----------------------------------------------------------
     Total events:       1296         7776

Determine above given what we derived from the probability formulas:

    "2 on 2"    "3 on 2"
      505         4724     Defender loses both + Defender loses 2nd
      505         3667     Defender loses both + Defender loses 1st
      791         3052     Attacker loses both + Attacker loses 2nd
      791         4109     Attacker loses both + Attacker loses 1st
    ---------------------------------------------------------------

SPECIAL NOTE:  I can't explain why, but H(6,3)[6] = 91, and also
36 - H(6,2)[6] = 25, and the product is 2275, Defender wins both battles.
H(N,3)[6] is the probability 91/216 that the Attacker rolls a 6.
H(N,2)[6] is the probability 11/36  that the Defender rolls a 6,
Subtraclting that from 36 yields 25/36, the probability Defender
does NOT roll a 6 with either of his dice. Apparently, 25(91) = 2275
which is the probability the Attacker will LOSE BOTH battles. Why?
If this can be explained, this problem can be solved by four equations
with four unknowns, where the fourth unknown is now known to be 2275.

  Defender loses in "3 on 2"
     W   loses both
     X   loses 2nd only
     Y   loses 1st only
     Z   wins both (1st+2nd)

  W+X = 4724
  W+Y = 3667
  Z+Y = 3052
  Z+X = 4109

  Replace Z by 2275, and solve for X, Y, and finally W.  Result should be:

     W = 2890
     X = 1834
     Y =  777
     Z = 2275

Honestly, I think this is just a coincidence because the same "trick"
doesn't work for the "2 on 2" tables.



Thanks to the web site "Murderous Maths: The Unknown Formula!"
I found this formula:  (S+N-1)! / ((N-1)! S!)
where N is the total number of items to choose from,
and S is the number of items you're allowed to choose.

This did help in solving this Risk problem.  When we throw
3 dice, we get 56 unique combinations of the 6 numbers.
That's derived: (8)! / (5! * 3!) = 40320 / (120 * 6)
because S=3 (selectors) and N=6 (choices per dice).
Likewise, throwing 2 dice gives 21 unique combinations.



I wrote another computer program to examine the unique pairs of dice used in
the compare processes, both for the 2 on 2, and 3 on 2 cases.  What I discovered
is that there are only 21 unique pairs for both cases.  However, some pairs
occur many times, especially for the 3 on 2 case.  That's because the 56 unique
triplets collapse down to just 21 pairs because the 3rd dice is always less than
or equal to the 2nd dice, which is always less than or equal to the 1st dice.
Here are the two tables, first for the 3 attacker dice in the 3 on 2 scenario, and
then the 2 dice for the defender in that case, and both players in the 2 on 2 case.
I also discovered I could construct these pairs manually, without a computer.

   Times   1st    2nd   3 dice rolled - 216 combinations - 21 unique

     1      1      1
     3      2      1
     4      2      2
     3      3      1
     9      3      2
     7      3      3
     3      4      1
     9      4      2
    15      4      3
    10      4      4
     3      5      1
     9      5      2
    15      5      3
    21      5      4
    13      5      5
     3      6      1
     9      6      2
    15      6      3
    21      6      4
    27      6      5
    16      6      6

   Times   1st    2nd   2 dice rolled - 36 combinations - 21 unique

     1      1      1
     2      2      1
     1      2      2
     2      3      1
     2      3      2
     1      3      3
     2      4      1
     2      4      2
     2      4      3
     1      4      4
     2      5      1
     2      5      2
     2      5      3
     2      5      4
     1      5      5
     2      6      1
     2      6      2
     2      6      3
     2      6      4
     2      6      5
     1      6      6

The "Times" column indicates how many times the 1st & 2nd dice occur in the 21
cases.  For example, the 6,5 pair occurs 27 times out of the 216 combinations.
The 3rd dice contributes to these pairs once the 3 dice are sorted to get the
highest and next highest (1st & 2nd).  These tables are the genesis for the
H,M,L tables for both 3-dice and 2-dice.  For each dice number, 1st or 2nd,
sum the "Times" values and you'll get the values in the H,M or H,L tables.
What's missing in the H,M,L tables is the frequency of each pair, 1st and 2nd.
For the attacker to lose both battles, all 21 of the 3-dice table must be matched
with all 21 of the 2-dice table, multiplying their frequencies according to the
win/loss nature of the comparisons of the two 1st's and two 2nd's.  That should
give the four numbers for win-both, win-1st, win-2nd, lose both.  Use the 2-dice
table twice, once with another 2-dice table, and once with the 3-dice table.

           Attacker Loses Tables for both 2x2 and 3x2 cases

 2x2       1   2   1   2   2   1   2   2   2   1   2   2   2   2   1   2   2   2   2   2   1
 3x2       1   3   4   3   9   7   3   9  15  10   3   9  15  21  13   3   9  15  21  27  16
         1,1 2,1 2,2 3,1 3,2 3,3 4,1 4,2 4,3 4,4 5,1 5,2 5,3 5,4 5,5 6,1 6,2 6,3 6,4 6,5 6,6

 1  1,1   LB  L2  LN  L2  LN  LN  L2  LN  LN  LN  L2  LN  LN  LN  LN  L2  LN  LN  LN  LN  LN
 2  2,1   LB  LB  L1  L2  LN  LN  L2  LN  LN  LN  L2  LN  LN  LN  LN  L2  LN  LN  LN  LN  LN
 1  2,2   LB  LB  LB  L2  L2  LN  L2  L2  LN  LN  L2  L2  LN  LN  LN  L2  L2  LN  LN  LN  LN
 2  3,1   LB  LB  L1  LB  L1  L1  L2  LN  LN  LN  L2  LN  LN  LN  LN  L2  LN  LN  LN  LN  LN
 2  3,2   LB  LB  LB  LB  LB  L1  L2  L2  LN  LN  L2  L2  LN  LN  LN  L2  L2  LN  LN  LN  LN
 1  3,3   LB  LB  LB  LB  LB  LB  L2  L2  L2  LN  L2  L2  L2  LN  LN  L2  L2  L2  LN  LN  LN
 2  4,1   LB  LB  L1  LB  L1  L1  LB  L1  L1  L1  L2  LN  LN  LN  LN  L2  LN  LN  LN  LN  LN
 2  4,2   LB  LB  LB  LB  LB  L1  LB  LB  L1  L1  L2  L2  LN  LN  LN  L2  L2  LN  LN  LN  LN
 2  4,3   LB  LB  LB  LB  LB  LB  LB  LB  LB  L1  L2  L2  L2  LN  LN  L2  L2  L2  LN  LN  LN
 1  4,4   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  L2  L2  L2  L2  LN  L2  L2  L2  L2  LN  LN
 2  5,1   LB  LB  L1  LB  L1  L1  LB  L1  L1  L1  LB  L1  L1  L1  L1  L2  LN  LN  LN  LN  LN
 2  5,2   LB  LB  LB  LB  LB  L1  LB  LB  L1  L1  LB  LB  L1  L1  L1  L2  L2  LN  LN  LN  LN
 2  5,3   LB  LB  LB  LB  LB  LB  LB  LB  LB  L1  LB  LB  LB  L1  L1  L2  L2  L2  LN  LN  LN
 2  5,4   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  L1  L2  L2  L2  L2  LN  LN
 1  5,5   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  L2  L2  L2  L2  L2  LN
 2  6,1   LB  LB  L1  LB  L1  L1  LB  L1  L1  L1  LB  L1  L1  L1  L1  LB  L1  L1  L1  L1  L1
 2  6,2   LB  LB  LB  LB  LB  L1  LB  LB  L1  L1  LB  LB  L1  L1  L1  LB  LB  L1  L1  L1  L1
 2  6,3   LB  LB  LB  LB  LB  LB  LB  LB  LB  L1  LB  LB  LB  L1  L1  LB  LB  LB  L1  L1  L1
 2  6,4   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  L1  LB  LB  LB  LB  L1  L1
 2  6,5   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  L1
 1  6,6   LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB  LB

The table above is for both cases, and the top two lines show the frequencies of the attacker pairs,
which follow on the line below from 1,1 through 6,6.  The left side of the table shows the frequencies
for the defender pairs which follow in the neighboring column.  This is a 21 by 21 table, with 441 spots
filled with codes:  LB, L1, L2, LN  meaning Attacker Loses (Both, 1st, 2nd, Neither).  The L2 entries are
highlighted to help you focus on how the table works.  For each line with L2's, scan across the line and
sum the frequencies above each L2.  Then multiply that by the frequency at the start of the line.  There
are 15 rows with L2's, so you should get 15 products.  Sum them and you should get 777 which is the number
of times the Attacker loses ONLY the 2nd battle.  You can do the same thing with L1, LB, and LN.  That's
the simplest method, without using a computer, to solve this problem.  From these earlier formulas:

(attacker wins) or (defender loses) the first dice by:

A(sD) = 1(0) + 7(1) + 19(4) + 37(9) + 61(16) + 91(25)  = 3667
D(sA) = 1(215) + 3(208) + 5(189) + 7(152) + 9(91) + 11(0) = 3667

We can derive the "Attacker wins both" as follows:  3667 - 777 = 2890.  That's for 3x2.

If you use the 2x2 frequencies, L2 yields 210. Thus:  505 - 210 = 295.  That's for 2x2.

Remember, we've already determined the following identities are true:

When the Attacker wins both, the Defender loses both.
When the Attacker wins only the 1st, the Defender wins only the 2nd.
When the Attacker wins only the 2nd, the Defender wins only the 1st.
When the Attacker loses only the 1st, the Defender loses only the 2nd.
When the Attacker loses only the 2nd, the Defender loses only the 1st.
When the Attacker loses both, the Defender wins both.

For the 3x2:   3667 = Defender loses both  +  Defender loses only 1st.

To obtain "Attacker wins both" we start with the above combined value to find "Defender loses both".
We must subtract "Defender loses only 1st" which is equivalent to "Attacker loses only 2nd", to
obtain "Defender loses both", which is equivalent to "Attacker wins both", our final answer.



The 21 unique pairs shown above are determined as follows:  x,x for 1<=x<=6 creates 6 unique combinations
with a frequency of 1.  x,y and y,x create 2 occurrences which sort into x,y with x > y.  There are 15
combinations: 2,1 3,1 3,2 4,1 4,2 4,3 5,1 5,2 5,3 5,4 6,1 6,2 6,3 6,4, 6,5.  The number of pairs is found
using the "Triangular Number" formula:  n(n+1)/2  where n = 5  for the five values of x involved.  When we
combine the x,x and x,y pairs in order, we get the Defender's combinations AND frequencies shown on the
left side of the table.  The Attacker's combinations of two dice are identical for both the 2x2 and 3x2
cases, with the same frequencies for 2x2.  But 3x2 has different frequencies.

In 3x2, three-dice are tossed.  They can be x,x,x or x,x,y or x,y,z.  x,x,x are unique, and there are 6 of
them, and when you drop the last x you get x,x with a frequency of 1 for each.  x,x,y can occur three ways:
x,x,y or x,y,x or y,x,x.  Sort these placing the highest dice first.  If you got x,x,y after the sort, you
drop the y, and the frequency is 3 TIMES (x-1).  If y is first, relabel y,x,x into x,y,y and drop the last y,
The frequency of this x,y is 3 TIMES y.  Finally, there's the x,y,z group where all three dice have different
values, like 2,1,6.  There are 6 possible arrangements:  x,y,z x,z,y y,x,z y,z,x z,x,y z,y,x.  When sorted,
you get one arrangement with x > y > z, like 6,2,1.  When you drop the last dice (the lowest), you get x,y
with a frequency that varies by the discarded dice, which must be 1<=z<y, or y-1 times.  So each x,y has
a frequency of 6 TIMES (y-1).  The minimum y is 2, and maximum is 5.  When you combine these x,y frequencies
with the ones from x,y above, you get the total frequency for these x,y pairs.  Likewise, combine all the x,x
frequencies for the same x, including the unique ones from x,x,x.

If you examine the frequencies for 6,1 6,2 6,3 6,4 6,5 for the 3x2 case, you'll see 6,1 starts at 3 for the
three combinations: 6,1,1 1,6,1 1,1,6.  Then each of the following is 6 more than the one before, ending with
the 27 combinations of 6,5 with 1 through 5.  Remember, 6 can occur in any position, and move to the front.
Basically, you're looking at all permutations of x,5,6 for 1<=x<=5, three for 5,5,6 and six for each of
the rest (x<=4).  Thus, 6(4)+3 = 27.  For 6,6 you get three 6,6,y for 1<=y<=5, plus one for 6,6,6 (16).



Definition:  An n-sided die has "n" faces (n=2x for x>0), and
each face has a different numerical value, N, in the range:  N=1
to n.  Each die is assumed to be balanced so that the probability
of any face occurring when a die is tossed is given by:  1/n.   A
6-sided die is called a "standard die".  The word "dice" will be
used to mean one or more n-sided balanced die.

Theorem-1: The probability, E(N,i), of throwing "N or less" using
"i" n-sided dice for any N in the range: N=0 to n, is:

   E(N,i) = (N/n)'i  or  N'i [/n'i]

Theorem-2: The probability, H(N,i), of throwing N as the "highest"
valued dice using "i" n-sided dice for any N in the range:
N=1 to n, is:

   H(N,i) = E(N,i) - E(N-1,i)

Definition:  An "attacker", A, throws "a" dice.  A "defender", D,
throws "d" dice.  The "highest" dice rolled by each are compared,
and A loses one point if the value of D's "highest" dice is equal
to or greater than the value of A's "highest" dice; otherwise, D
loses one point.  The compared dice are discarded, and if both A
and D still have dice left, we repeat the compare process above.
This is the "game of RISK" in which there is an "a on d" attack.

Problem:  What is the probability that A will lose two points in
a "2 on 2" attack using standard dice in the game of RISK?

Theorem-3: The probability, P(a,d,n), that A will lose the first
compare (first "highest") using n-sided dice when A throws "a"
dice and D throws "d" dice is:
   P(a,d,n) = Sum:  H(k,d) x E(k,a) [/n'(a+d)]  for k=1 to n

Theorem-4: The probability, Q(a,d,n), that D will lose the first
compare (given the same circumstances) is:
   Q(a,d,n) = Sum:  H(k,a) x E(k-1,d) [/n'(a+d)]  for k=1 to n

Note:  P(a,d,n) + Q(a,d,n) = 1



The following tables are H(N) thru L(N) for 2 or more dice which
have n-sides.  The tables show the count of times the dice had
1 thru N as their highest value for each set of dice.  H(N) is
always line 1, and L(N) is always the last line in each table.
Theorem-3,4 can compute the probabilities for other than the
first compare by using the H and E values from the i-th row
in these tables, where "i" represents the i-th compare.
The tables show the H values, and the E values are the sum
of the H values up to the desired column position (k or k-1).
N is 6, and the column index is the number of dice thrown.


   |      1      2      3      4      5      6      46656
---|------------------------------------------
 1 |      1     63    665   3367  11529  31031
 2 |     31    801   4271  11281  17991  12281
 3 |    406   4266  11366  15706  12006   2906
 4 |   2906  12006  15706  11366   4266    406
 5 |  12281  17991  11281   4271    801     31
 6 |  31031  11529   3367    665     63      1



   |      1      2      3      4      5      6       7776
---|------------------------------------------
 1 |      1     31    211    781   2101   4651
 2 |     26    326   1106   2126   2666   1526
 3 |    276   1356   2256   2256   1356    276
 4 |   1526   2666   2126   1106    326     26
 5 |   4651   2101    781    211     31      1



   |      1      2      3      4      5      6       1296
---|------------------------------------------
 1 |      1     15     65    175    369    671
 2 |     21    123    261    363    357    171
 3 |    171    357    363    261    123     21
 4 |    671    369    175     65     15      1



   |      1      2      3      4      5      6        216
---|------------------------------------------
 1 |      1      7     19     37     61     91
 2 |     16     40     52     52     40     16
 3 |     91     61     37     19      7      1



   |      1      2      3      4      5      6         36
---|------------------------------------------
 1 |      1      3      5      7      9     11
 2 |     11      9      7      5      3      1



The following table is for 4-sided dice (1-4).


   |      1      2      3      4     4096
---|----------------------------
 1 |      1     63    665   3367
 2 |     19    429   1739   1909
 3 |    154   1254   1994    694
 4 |    694   1994   1254    154
 5 |   1909   1739    429     19
 6 |   3367    665     63      1



   |      1      2      3      4     1024
---|----------------------------
 1 |      1     31    211    781
 2 |     16    176    456    376
 3 |    106    406    406    106
 4 |    376    456    176    106
 5 |    781    211     31      1



   |      1      2      3      4      256
---|----------------------------
 1 |      1     15     65    175
 2 |     13     67    109     67
 3 |     67    109     67     13
 4 |    175     65     15      1



   |      1      2      3      4       64
---|----------------------------
 1 |      1      7     19     37
 2 |     10     22     22     10
 3 |     37     19      7      1



   |      1      2      3      4       16
---|----------------------------
 1 |      1      3      5      7
 2 |      7      5      3      1



   |      1      2      4
---|--------------
 1 |      1      3
 2 |      3      1



There are several observations that should be made about these
tables.  First, L(N) is always the reverse of H(N), and H(N)
always has values determined by Theorem-2 for "i" dice.  Also,
the tables are "saddle shaped" in that they are symetric from
upper left to lower right, and upper right to lower left.
Each column sums to: i x n'(i-1)  for "i" dice with "n" sides.

If the dice have "n" sides, and there are "i" dice involved,
the left column's values are determined as follows:
Place n-1, i downto 1, 1 to i in columns as shown below, and
multiple by the previous entry (starting with 1), and add the
result to previous table entry to obtain the next table entry.

i=5| n=6 sided dice
---|-------------------------------------
 1 |                               >    1
   >  5 * 5 / 1 *     1 =     25  <
 2 |                               >   26
   >  5 * 4 / 2 *    25 =    250  <
 3 |                               >  276
   >  5 * 3 / 3 *   250 =   1250  <
 4 |                               > 1526
   >  5 * 2 / 4 *  1250 =   3125  <
 5 |                               > 4651
   >  5 * 1 / 5 *  3125 =   3125  <
   --------------------------------> 7776

These same values can be obtained by using the standard formula
for obtaining "j" events within "m" trials for events where "p"
is the probability of success, and "q" is "1-p".  For six sided
dice, use p=1 and q=5 [/6].  The formula is:

      Sum of terms for r=j thru r=m where each term is:

      f(m,r) p'r q'(m-r)

  and f(m,r) is the factorial expression:  m! / ((m-r)! r!)
===============================================================
Summary:

The attacker loses the first 6-sided dice by:

D(sA) = 1(1) + 3(8) + 5(27) + 7(64) + 9(125) + 11(216) = 4109
A(sD) = 1(36) + 7(35) + 19(32) + 37(27) + 61(20) + 91(11) = 4109

The attacker loses the second 6-sided dice by:

D(sA) = 11(16) + 9(56) + 7(108) + 5(160) + 3(200) + 1(216) = 3052
A(sD) = 16(1) + 40(4) + 52(9) + 52(16) + 40(25) + 16(36) = 3052

The defender loses the first 6-sided dice by:

A(sD) = 1(0) + 7(1) + 19(4) + 37(9) + 61(16) + 91(25)  = 3667
D(sA) = 1(215) + 3(208) + 5(189) + 7(152) + 9(91) + 11(0) = 3667

The defender loses the second 6-sided dice by:

D(sA) = 11(200) + 9(160) + 7(108) + 5(56) + 3(16) + 1(0) = 4724
A(sD) = 16(0) + 40(11) + 52(20) + 52(27) + 40(32) + 16(35) = 4724

We now have the following compare probabilities:

    4109 = Attacker loses 1st, Defender wins 1st
    3052 = Attacker loses 2nd, Defender wins 2nd
    3667 = Defender loses 1st, Attacker wins 1st
    4724 = Defender loses 2nd, Attacker wins 2nd

     Attacker loses:    "3 on 2"   Attacker wins:
        neither           2890       both (1st+2nd)
        only 1st          1834       only 2nd
        only 2nd           777       only 1st
        both (1st+2nd)    2275       neither
     ----------------------------------------------------------
     Total events:        7776

   |      1      2      3      4      5      6        216
---|------------------------------------------
 1 |      1      7     19     37     61     91
 2 |     16     40     52     52     40     16
 3 |     91     61     37     19      7      1


   |      1      2      3      4      5      6         36
---|------------------------------------------
 1 |      1      3      5      7      9     11
 2 |     11      9      7      5      3      1



With 4-sides dice (pyramid shape):

The attacker loses the first 4-sided dice by:

D(sA) = 1(1) + 3(8) + 5(27) + 7(64) = 608
A(sD) = 1(16) + 7(15) + 19(12) + 37(7) = 608

The attacker loses the second 4-sided dice by:

D(sA) = 7(10) + 5(32) + 3(54) + 1(64) = 456
A(sD) = 10(1) + 22(4) + 22(9) + 10(16) = 456

The defender loses the first 4-sided dice by:

A(sD) = 1(0) + 7(1) + 19(4) + 37(9) = 416
D(sA) = 1(63) + 3(56) + 5(37) + 7(0) = 416

The defender loses the second 4-sided dice by:

D(sA) = 7(54) + 5(32) + 3(10) + 1(0) = 568
A(sD) = 10(0) + 22(7) + 22(12) + 10(15) = 568

We now have the following compare probabilities:

    608 = Attacker loses 1st, Defender wins 1st
    456 = Attacker loses 2nd, Defender wins 2nd
    416 = Defender loses 1st, Attacker wins 1st
    568 = Defender loses 2nd, Attacker wins 2nd

     Attacker loses:    "3 on 2"   Attacker wins:
        neither            314       both (1st+2nd)
        only 1st           254       only 2nd
        only 2nd           102       only 1st
        both (1st+2nd)     354       neither
     ----------------------------------------------------------
     Total events:        1024

   |      1      2      3      4       64
---|----------------------------
 1 |      1      7     19     37
 2 |     10     22     22     10
 3 |     37     19      7      1


   |      1      2      3      4       16
---|----------------------------
 1 |      1      3      5      7
 2 |      7      5      3      1



Definitions in Probability Theory:

1.  An "experiment" is the process by which an observation
    (or measurement) is obtained.

2.  Each experiment may result in one or more outcomes which are
    called "events";  so an "event" is a collection of outcomes,
    called A, B, etc.

3.  A "simple event" is an event that can't be decomposed;
    it is one and only one outcome of an experiment, called E.

4.  A "sample point" is a simple event.

5.  The set of all sample points for an experiment is called
    the "sample space" called S, consisting on N sample points.

6.  If all sample points are equally probable, then the
    "probability" associated with a sample point is: 1/N

7.  The probability of event A is equal to the sum of the
    probabilities of the sample points that compose A.
    P(A) = n/N   where "n" sample points compose A.

8.  The "complement" of event A is the collection of all
    sample points in S that are not in A; denoted by: ~A.
    Note:  P(A) + P(~A) = 1.

9.  If there are M sample points in one experiment, and
    N sample points in another, then there are MN sample
    points in the combined experiment.  For example,
    there are 6 possible outcomes tossing a single dice,
    and 36 outcomes tossing two dice (6x6).

10. A "permutation" is an ordered arrangement of "r" distinct
    objects.  The number of ways of ordering "n" distinct objects
    taken "r" at a time is given by:  R(n,r) = n!/(n-r)!

11. The number of "combinations" of "n" objects taken "r" at a
    time is given by:  C(n,r) = R(n,r)/r! = n!/r!(n-r)!

12. If A and B are two events in S, then the "intersection"
    of A and B is defined to be the event consisting of
    all sample points that are in both A and B; denoted by AB.

13. The "conditional probability" of A given that B has
    occurred is denoted by:  P(A|B) and is given as:
    P(A|B) = P(AB)/P(B)  and  P(B|A) = P(AB)/P(A)

14. Two events, A and B, are said to be "independent" if
    either  P(A|B) = P(A)
    or      P(B|A) = P(B)
    Otherwise, the events are said to be "dependent".

15. P(AB) = P(A) x P(B|A)
    or    = P(B) x P(A|B)
    If A and B are independent, P(AB) = P(A) x P(B)
    Stated another way, if p1, p2, ... pi are the separate
    probabilities of the occurrence of "i" independent
    events, then the probability, P, that "all" of these
    events will occur in a single experiment is:
    P = p1 x p2 x ... x pi

16. In the case of "i" dependent events, if the probability
    of the occurrence of the first event is p1, and if,
    after this event has occurred, the probability of the
    occurrence of the second event is p2, and if, after the
    first and second events have occurred, the probability
    of the occurrence of the third event is p3, and so on,
    then the probability, P, that all events will occur
    in the specified order is:
    P = p1 x p2 x ... x pi
    Example:  What is the probability that from amoung 23
    randomly selected people at least two have birthdays
    falling on the same day, that is, on the same month
    and day (not necessarily the same year)?
    Solution:  Failure occurs if all 23 people have
    birthdays falling on different days of the year.  We
    shall assume 365 days in a year, and that the probability
    of a person having a birthday on any of these days is the
    same.  Now for all 23 people to have different birthdays, all
    of the following 23 dependent events must occur:  the first
    person has a birthday on any day of the year, which occurs,
    of course, with probability 1;  the second person has a
    birthday on any day of the year EXCEPT the one on which the
    first person has a birthday, for which the probability is
    364/365; the third person has a birthday on any day of the
    year EXCEPT those of the first two, for which the probability
    is 363/365; and so on.  Therefore, the probability, Q, that
    all 23 people have different birthdays is:
        Q = p1 x p2 x ... x p23 = 0.493
    so the desired probability, P , that at least two people
    have birthdays falling on the same day is:  ~Q  or
        P = 1 - Q = 1 - 0.493 = 0.507
    a better than even chance!
    Example:  A, B, and C in order toss a coin.  The first one
    to throw heads wins.  What is A's probability of winning?
    Solution:  If H stands for head, and T stands for tail,
    then the following sequences indicate success for A:
        H                  p1 = 1/2
        TTTH               p2 = 1/16
        TTTTTTH            p3 = 1/128
        ...etc...
    Therefore, P(A) = p1 x p2 x ... p(infinity)  which is
    the geometric series:  a + ar + ar'2 + ar'3 + ... = a/(1-r)
    where r, the ratio between any two consecutive numbers, is
    between -1 and 1.  In this case, we have: a=1/2 and r=1/8,
    so P(A) = (1/2)/(1-(1/8)) = (1/2)/(7/8) = 4/7

17. Two events, A and B, are said to be "mutually exclusive"
    if the event AB contains no sample points; P(AB) = 0.

18. If A and B are two events in S, then the "union" of
    A and B is defined to be the event consisting of all
    sample points in A or B or both; denoted by:  AuB.

19. P(AuB) = P(A) + P(B) - P(AB)
    If A and B are mutually exclusive, P(AB) = 0 and
    P(AuB) = P(A) + P(B)
    Stated another way, if p1, p2, ..., pi are the separate
    probabilities of the occurrence of "i" mutually exclusive
    events, then the probability, P, that some "one" of these
    events will occur in a single experiment is:
    P = p1 + p2 + ... + pi

20. Bayes' Rule.  Consider an experiment that involves the
    selection of a sample from one of "k" populations, call
    them H1, H2,...,Hk.  The sample is observed, but it is
    not known from which population the sample was selected.
    Suppose that the sample results in event A.  Then the
    problem is to determine the population from which the
    sample was selected.  This inference will be based on
    the conditional probabilities, P(Hi|A), i=1,2,...,k.
    To find the probability that the sample was selected
    from population "i" given that event A was observed,
    P(Hi|A),i=1,2,...,k, note that A could have been
    observed if the sample were selected from population
    1 or 2 or any one of the "k" populations, H1, H1,...,Hk.
    The probability that population "i" was selected AND
    that event A occurred is the intersection of the events
    Hi and A, i.e., (AHi).  These events, (AH1), (AH2), ...,
    (AHi), are mutually exclusive and hense:
        P(A) = P(AH1) + P(AH2) + ... + P(AHi)
    Then the probability that the sample came from the "i"
    population is:
          P(Hi|A) = P(AHi) / P(A)
    or    P(Hi|A) = P(Hi) x P(A|Hi) / P(A)
    or    P(Hi|A) = P(Hi) x P(A|Hi) / Sum j=1 to k, P(AHj)
    or    P(Hi|A) = P(Hi) x P(A|Hi) / Sum j=1 to k, [P(Hj) x P(A|Hj)]
    Therefore, finding P(Hi|A) requires knowledge of the
    probabilities, P(Hi) and P(A|Hi), i=1,2,...,k.
    P(Hi) is the probability that Hi is the contributing population,
    and P(A|Hi) is the probability of A occurring in that population.
    The sum of P(Hi),i=1,2...k, must total to unity.

    Example:  Three machines produce similar parts.  Machine E
    produces 40% of the total, machine F, 25%, and machine G,
    35%.  On the average, 10% of machine E's parts are defective;
    likewise, 5% for machine F, and 1% for machine G.  One part
    is selected at random from the total output and is found to
    be defective.  What is the probability that is was produced
    by machine E?  Solution: Let D represent the event, selecting
    a defective part.  We need to determine the probability of
    the part having been selected from machine E given that the
    part was defective, which is P(E|D) = P(E) x P(D|E) / P(D)
    and P(D) is the sum of the following probabilities:
        P(DE) = P(E) x P(D|E) = (0.40)(0.10) = 0.0400
        P(DF) = P(F) x P(D|F) = (0.25)(0.05) = 0.0125
        P(DG) = P(G) x P(D|G) = (0.35)(0.01) = 0.0035
    so          P(D) = P(DE) + P(DF) + P(DG) = 0.0560
    Therefore,  P(E|D) = (0.40)(0.10) / (0.0560) = 0.714
    which shows that on the average machine E produces
    71.4% of the defective parts.   (junk it!)

    Example:  Suppose you are shown two rings that appear to be
    identical.  Upon close examination of one of them you find
    that it is a diamond.  Some time later, you again have an
    opportunity to examine one of the rings (not knowing which
    one) and this too is a diamond.  Still later, one of them
    is lost.  What is the probability that the lost ring was
    NOT a diamond?
    Solution:  We know at least one ring was a diamond, so
    either both were diamonds, or one was a diamond and one
    was not a diamond.  In two previous examinations, we
    examined a diamond.  The probability of this occurring
    under the first assumption (both diamonds) is unity.
    This same probability under the second assumption is
    (1/2)(1/2) = 1/4.  Thus, the probability that both of
    the rings were diamonds is:  P(H2) = (1)/(1+1/4) = 4/5.
    Likewise, the probability that one is not a diamond is:
    P(H1) = (1/4)/(1+1/4) = 1/5.  These sum to unity.
    The probability a diamond was lost under the first
    assumption equals unity, P(A|H2) = (1).  Under the
    second assumption, the probability is: P(A|H1) = 1/2.
    So the probability a diamond was lost because of any
    one of these assumptions is:
        P(A) = P(H2) x P(A|H2) + P(H1) x P(A|H1)
        P(A) = (4/5)(1) + (1/5)(1/2)
        P(A) = 4/5 + 1/10 = 9/10
    So the probability that the lost ring was NOT a diamond
    is only 1/10.  Not very good odds.

21. A "binomial experiment" is one that possesses the following
    properties:
    a. The experiment consists of "n" identical trials.
    b. Each trial results in one of two outcomes, success
       or failure, denoted by S and F respectively.
    c. The probability of success, p, remains the same from
       trial to trial.  The probability of failure is equal
       to (1-p) = q.
    d. The trials are independent.
    e. We are interested in "y", the number of successes
       observed during the "n" trials.
    The probability associated with a particular value of y
    is simply the term involving p to the power y in the
    expansion of (q+p)'y, which is written as:
       P(y) = C(n,y) x p'y x q'(n-y)  for y=0,1,2,...,n

    Example:  A rifleman has a consistent probability of 0.8
    of hitting a target with a single shot.  If he fires four
    shots at the target,
    a) what is the probability of exactly two hits?
       P(2) = C(4,2) x (.8)'2 x (.2)'2
            = 4!/(2!x2!) x (.64) x (.04)
            = 0.1536
    b) what is the probability he will hit the target
       at least two times?
       P(at least two) = P(2) + P(3) + P(4)
                       = 1 - P(0) - P(1)
                       = 1 - 0.0016 - 0.0256
                       = 0.9728
    c) what is the probability of all four hits?
       P(4) = C(4,4) x (.8)'4 x (.2)'0
            = 4!/4!0! x (.8)'4 x (1)
            = (.8)'4 = 0.4096

22. Expected Value.  If "y" is a discrete random variable
    with probability P(y), then the expected value of y is:
       E(y) = Sum over all y, P(y) x y

    Example:  8000 tickets are sold in a lottery at $1.00 each.
    The prize is a $3000 automobile.  If you buy 2 tickets,
    what is your expected gain?
    Solution:  Either you will lose $2.00 (gain -$2.00) or
    win $2998 ($3000 - $2) with probabilities of 7998/8000
    and 2/8000 respectively.
    E(y) = [-2 x (7998/8000)] + [2998 x (2/8000)]
         = -$1.25
    Therefore, if this lottery were repeated infinitely,
    and you always bought two tickets, your average expected
    gain would be:  -$1.25  (a loss).

    Example:  On an average, 1 out of 50 policy holders dies
    during a year.  What should be the yearly premium for
    every $1000 of insurance coverage?
    Solution:  We want to calculate E(y)=0.  If the event
    does not occur during the year, the insurance company
    will gain the premium, or y=C dollars.  If the event does
    occur, the gain (a loss) will be: y=-(1000-C) dollars.
    The probabilities of these two values is 49/50 and 1/50.
    E(y) = 0 = C(49/50) + (C-1000)(1/50)
           0 = C(50/50) - (1000)(1/50)
          20 = C
    So the insurance company must charge $20 to break even.



 PL360 Program to determine Attacker 3x2 win table.

 BEGIN  ARRAY 5 INTEGER CELL = 5(0);
        INTEGER Y1 SYN CELL, Y2 SYN Y1(4),
             Y3 SYN Y2(4), Y4 SYN Y3(4), Y5 SYN Y4(4);
        ARRAY 3 INTEGER ASET;  ARRAY 2 INTEGER DSET;
        BYTE A SYN ASET(3), B SYN A(4), C SYN B(4),
             D SYN DSET(3), E SYN D(4);
        ARRAY 133 BYTE OUTPUT = 133(" ");
        EQUATE SIDES SYN 6;

    FOR R1 := 1 STEP 1 UNTIL SIDES DO
    FOR R2 := 1 STEP 1 UNTIL SIDES DO
    FOR R3 := 1 STEP 1 UNTIL SIDES DO
    BEGIN  STM(R1,R3,ASET);
    X: IF A < B THEN
       BEGIN  A := A XOR B;
              B := B XOR A;
              A := A XOR B;
       END;
       IF B < C THEN
       BEGIN  B := B XOR C;
              C := C XOR B;
              B := B XOR C;
              GOTO X;
       END;
       FOR R6 := 1 STEP 1 UNTIL SIDES DO
       FOR R7 := 1 STEP 1 UNTIL SIDES DO
       BEGIN  STM(R6,R7,DSET);
           IF D < E THEN
           BEGIN  D := D XOR E;
                  E := E XOR D;
                  D := D XOR E;
           END;
           |- COMPUTE PROBABILITIES -|
           R10 := 1 + Y5 =: Y5;
           IF A > D THEN  |- Attacker wins 1st dice -|
           BEGIN  IF B > E THEN R10 := 1 + Y1 =: Y1 |- wins both -|
                  ELSE  R10 := 1 + Y3 =: Y3; |- wins only 1st -|
           END ELSE IF B > E THEN R10 := 1 + Y2 =: Y2  |- 2nd -|
           ELSE R10 := 1 + Y4 =: Y4;   |- Attacker loses both -|
       END;
    END;
    R1 := @OUTPUT(1); R2 := 1;  R3 := 9;  R5-R5;
    FOR R4 := 1 STEP 1 UNTIL 5 DO
    BEGIN  R0 := Y1(R5);  VALTOBCD;
       R1 := @B1(R3);  R5 := @B5(4);
    END;  R0 := @OUTPUT;  WRITE;
 END.

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