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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