RISK probabilities
Shortened version of a more complete analysis located at this web address.
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 conveniences.
In most probability formulas I'll be dealing with, the "numerator" is
defined as a series of terms, and the "denominator" 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.
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:
In a "1 on 1" attack, we obtain:
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
Since the "attacker" is using two dice, we must use the
(N/6)'2 probabilities of throwing "N or less" in the sum.
In a "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
For the "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:
In a "1 on 2" attack, we obtain:
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 [/216] = 161/216 = 996/1296
A summary of the probabilities we've computed thus far, in
descending 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 probability, 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.
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.
The M(N,3) row is the mid-range frequency, neither H or L.
It is derived per column from: 216/2 - H - L.
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
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 the H(N,2)(N-1)'2 [/6'4]
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".
The attacker loses the second dice by:
11(11) + 9(20) + 7(27) + 5(32) + 3(35) + 1(36) [/6'4] = 791/1296
The defense 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).
It's interesting to note that 791 and 505 are the numerators
for the probabilities for BOTH the first and second battles.
But you can't derive the 295, 210, 210, 581 values from them.
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".
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:
The 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
The 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
These values correspond exactly with the computer derived values:
4109 = 2275 + 1834 and 3667 = 2890 + 777
What is the probability the "Attacker" will lose both dice in a "2 on 2"
or "3 on 2" attack?
Without the computer, I'm unable to answer that question!
I'm looking for the formulas that derives 581/1296 for the "2 on 2",
and 2275/7776 for the "3 on 2" attacks, similar to the "3 on 1"
attack formula: P(A loses) = (1/6) ((1+8+27+64+125+216)/216).
I'm unable to determine the formulas! The computer doesn't help
because it only computes the probabilities, not the formulas.
I'm sure the formula is some form of Conditional Probability or
Dependent Probability using formulas like the following:
P(A and B) = P(A) P(B|A)
I'm trying to obtain P(A and B), meaning the probability the
attacker loses the first battle (A) and the second battle (B).
I know the formula to P(A), but not for P(B|A). Most texts that
I've read assume you already know P(A) and P(A and B), and
want to find P(B|A). But I don't know either P(B|A) or P(A and B),
at least not in probability formula notation.
I find it hard to believe that this problem, WITHOUT A COMPUTER,
can't be solved, other than by enumerating all combinations of the
2 or 3 dice thrown by the attacker, and the 2 dice thrown by the
defender. Such tables would have 216 entries for the "2 on 2", and
7776 entries for the "3 on 2". Having to scan such tables by hand
would work, but there must be a mathematical way to do it instead.
August 2014. I tried something to see what effect it would have. I
tried "4 on 2" and "5 on 2" scenarios, still with just two battles. The
odds for the attacker to win BOTH battles went up with 4 or 5 dice.
In other words, the odds the top two dice are high-valued increased.
I haven't yet figured out how to use that knowledge. The M(N,3) row
in the table for three dice represents the 2nd battle dice in a "3 on 2".
That may be what will solve this puzzle for the "3 on 2".