Odds in the game of Risk 3

[karluk]

The "conquer the world" board game, Risk, decides the result of an attack on one territory from another by tossing dice. The attacker gets to toss as many as three dice, while the defender is limited to no more than two. To partly compensate for this advantage, the defender wins all ties.

So, in a typical attack, the attacker will roll three dice and pick out the biggest two numbers showing. The defender rolls only two dice and keeps both numbers. The attacker and defender then compare numbers, high vs. high and low vs. low. There are two armies at stake. If the attacker's high number is bigger than the defender's high number, the attacker wins one army. Otherwise the defender wins one army. Similarly, the comparison of attacker's low number vs. defender's low number also results in a win or loss of one army, depending on how low numbers compare.

The problem for your consideration is to calculate the odds of the attacker winning in a three vs. two dice attack in the game of Risk. This means the attacker's long term odds of success from repeated three vs. two attacks.

When I was in college, I took the trouble to calculate these odds by hand. I haven't worked on the problem since then. Risk is well-enough known that I'm sure it's easily possible to find the odds on the internet, but obviously you are supposed to figure it out yourself.

Unless I missed a possible simplification, the calculations required are difficult to perform by hand, but easily done by computer. I will be interested in seeing if the odds I caculated in college are confirmed or disproven by the members of this forum.

 

[karluk]

If there is enough interest in this game, it would be possible to expand the problem and ask considerably more sophisticated questions about the odds. For example, what are the odds of the attacker conquering a territory when starting out with n armies against a defender with m armies? It seems to me that this question has more practical impact on the game of Risk than a simple calculation of the attacker's long-term odds. Given a real-life situation where I have limited attacking resources, how should I rationally decide whether I should attack immediately continue to build up armies in preparation for a later attack?

 

[kwbMitel]

By my usual Brute force Methdology:
[hide]
Attacker will win 37.17% of the time
Defender will win 29.26% of the time
They will draw 33.58% of the time
[/hide]

**********************************************
What's most important is that you realise ... There is no spoon.

 

[CajunCenturion]

In Risk there are no draws. The defender wins ties. For a good understanding of the game, I invite you to watch this video.

http://www.youtube.com/watch?v=wRUMO0c9mZs

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[karluk]

@kwbMitel
[hide]Your answer looks right. The calculations I did many years ago gave the attacker 2797/5184 = 53.95% (approx.) odds. At the time I must have broken it down separately by win two, lose two, and split 1-1, but I no longer remember those numbers. Your percentages translate into the attacker winning on average 1.0792 armies out of the two armies being contested at every toss of the dice. 1.0792/2 = 53.96%, which is accurate to the fourth significant figure of my calculation.[/hide]

 

[Olaf Doschke]

[hide]
Simply going through all combinations of 5 dices, 6^5=7776 combinations, counting the armies lost on both sides, attacker and defender:

(see h t t p://codepad.org/q2V0q6u9)

Probability per single army to survive is 8391/15552. This is about 0.54, so slightly better odds for the attacker.[/hide]

Bye, Olaf.

 

[karluk]

@OlafDoschke

[hide]Yes, your results exactly match my calculation. 8391/15552 = 2797/5184 after dividing by the common factor of 3. Well done![/hide]

 

[kwbMitel]

@CajunCenturion Re- no Draws

There are 3 potential outcomes to each roll.

1-Attacker Wins Both
2-Defender Wins both
3-Attacker and Defender win 1 each

For the third scenario above, I call that a draw, what would you call it?

@Karluk - Thanks for understanding my presentation method. I have recalculated by simply splitting the draws and the value comes to [hide]0.539544753[/hide]

@Karluk - Re: Real Life Scenario

The decision to attack does not solely depend on relative sizes of armies. If it were that simple, the game would hold no interest for me. Other considerations (in no particular order)
- Can attacking nation be attacked from another direction after this turn? (How many armies need to be left behind as a rear guard)
- How many armies need to be placed on defending nation to be relatively safe on the next turn? (or turns if more than 2 players)
- What is the likelihood of the opposing player getting a set for reinforcements and how many reinforcements would that be?
- Do you have to attack to be able to get your reinforcement card?
- What are the chances that you will get reinforcements your next turn?
- How many continents do you control compared to your opposition. (another reinforcement question)
- Would taking a nation give you control of a continent? Can you hold it?
- Would taking a nation break control of a continent for the opposition?
- Would taking a nation gain you an extra reinforcement the next turn? (total / 3 = reinforcements)
- Would taking a nation reduce the opposition reinforcements?

What calculating the odds has gained me.
Previously, I have typically looked to have 2:1 or 3:1 advantage in armies before attacking depending on a number of factors. 1 Factor that I never considered before now was the actual size of the armies. Once they get to sufficient size (say 50 or more) you can actually attack with a 1:1 ratio and still have some confidence that about 20% will remain afterwards. There are few games where this actually happens but next time it does, I will definitely be considering something I hadn't before.


**********************************************
What's most important is that you realise ... There is no spoon.

 

[strongm]

Red Dwarf: series 4, episode 6, entitled "Meltdown." Behold, The King of Risk, Arnold Rimmer;

RIMMER: So there we were at 2:30 in the morning; I was beginning to wish
I had never come to cadet training school. To the south lay water --
there was no way we could cross that. To the east and west two armies
squeezed us in a pincer. The only way was north; I had to go for it
and pray the Gods were smiling on me. I picked up the dice and threw
two sixes. Caldecott couldn't believe it. My go again; another two
sixes!
LISTER: Rimmer, what's wrong with you? Don't you realize that no one is
even slightly interested in anything you're saying? You've got this
major psychological defect which blinds you to the fact that you're
boring people to death! How come you can't sense that?
RIMMER: Anyway I picked up the dice again... Unbelievable! Another two
sixes!
LISTER: Rimmer!
RIMMER: What?
LISTER: No one wants to know some stupid story about how you beat your
Cadet School Training Officer at Risk.
RIMMER: Then -- disaster! I threw a two and a three; Caldecott picked up
the dice and threw snake eyes -- I was still in it.
LISTER: Cat, can you talk to him?.

CAT is sitting with big pieces of cotton wool plugged in to his ears. As
LISTER talks to him he takes one of the pieces.

CAT: What?
RIMMER: Anyway, to cut a long story short I threw a five and a four which
beat his three and a two, another double six followed by a double four
and a double five. After he'd thrown a three and a two I threw a six
and a three.
CAT: Man, this guy could bore for his country!
LISTER: What I want to know, is how the smeg can you remember what dice
you threw at a game you played when you were seventeen?
RIMMER: I jotted it down in my Risk campaign book. I always used to do
that so I could replay my moments of glory over a glass of brandy in
the sleeping quarters. I ask you, what better way is there to spend a
Saturday night?
CAT: Ya got me.
RIMMER: So a six and a three and he came back with a three and a two.
LISTER: Rimmer, can't you tell the story is not gripping me? I'm in a
state of non-grippedness, I am completely smegging ungripped. Shut the
smeg up.
RIMMER: Don't you want to hear the Risk story?
LISTER: That's what I've been saying for the last fifteen minutes.
RIMMER: But I thought that was because I hadn't got to the really
interesting bit...
LISTER: What really interesting bit?
RIMMER: Ah well, that was about two hours later, after he'd thrown a
three and a two and I'd thrown a four and a one. I picked up the
dice...
LISTER: Hang on Rimmer, hang on... the really interesting bit is exactly
the same as the dull bit.
RIMMER: You don't know what I did with the dice though, do you? For all
you know, I could have jammed them up his nostrils, head butted him on
the nose and they could have blasted out of his ears. That would've
been quite interesting.
LISTER: OK, Rimmer. What did you do with the dice?.
RIMMER: I threw a five and a two.
LISTER: And that's the really interesting bit?
RIMMER: Well it was interesting to me, it got me into Irkutsk.

 

[CajunCenturion]

==> For the third scenario above, I call that a draw, what would you call it?
It's really a terminology issue.

In Risk, you don't ever win an army. Whether you're attacking or defending (assuming two dice for defender), you either lose two armies, lose 1 army, or lose zero armies. You never win an army.

As I said, it's really a terminology issue.


--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
Wise men speak because they have something to say, fools because they have to say something. - Plato

 

[karluk]

Once they get to sufficient size (say 50 or more) you can actually attack with a 1:1 ratio and still have some confidence that about 20% will remain afterwards.

I would be very cautious about allowing a knowledge of the odds to make you more aggressive in your playing strategy. The numbers show that the attacker has an edge, but that it's a rather modest edge. As far as I can see, a 53.95%-46.05% edge would translate into an expectation that the attacker would lose about 8% fewer armies over the course of a long war of attrition. With 50 armies involved, this would mean that the attacker would expect to win, but with only four armies remaining. That's uncomfortably close to not winning at all, so the attacker should probably have a bigger edge, unless he's in bad shape strategically and needs to take chances.

What knowing the odds CAN accomplish is to allow players to make a quick estimate of the likely situation at the end of the battle. With ordinary luck, the attacker can expect to conquer a territory with losses of about 92% of the defender's losses. That's the excess that probably will be left over to occupy the conquered territory. It allows the player to make an informed judgment as to whether he's likely to be able to survive the inevitable counter-attack with whatever he expects to have left.

 

[karluk]

In addition, the players need to know something about the standard deviation of the results. What sort of numerical edge does one need in order to have, say, a 95% level of confidence that an attack will succeed with at least 10 armies left over? I would like to ask a follow-up question along these lines, but I'm not ready to do so right now. I haven't done any work yet on this type of question, so I would have no way to evaluate any answers I might receive.

 

[karluk]

On second thought, I retract my remarks in the post dated 6 Dec 11 15:50. I am interested in ways to translate the attacker's edge into an estimate of the likely outcome of a battle, but I suspect that the correct calculations are a little different than what I did in my earlier post.

 

[kwbMitel]

Karluk - well, regarding my 20% remaining when dealing with large numbers. It's entirely possible that my newer trials have issues but they appear correct as far as I can check. I never bothered to corrolate my latest results with the earlier calculated odds. Basically, I've simply created a dice roll generator and made random rolls. The 20% mark seems to hold true from every power of 10 from 100 - 100,000.

I can't explain why this might be. More investigation on my methods and data are warranted.

**********************************************
What's most important is that you realise ... There is no spoon.

 

[kwbMitel]

So things are looking even more interesting.

I've checked my data and it all looks good. The 20% figure I posted earlier was based on a different stat. Basically it takes typically 80% of your armies in rolls to reduce the opposition to zero with a starting point were the armies are equal.

The remaining armies are actually 40%!!! (approx)

So 50 vs 50 will take 40 rolls and you will have 20 armies left.

Here is my data from 1 such run[tt]

Attacker won 50 times in 39 rolls (64%)
Defender Won 28 times in 39 rolls (36%)

Attack Defend
50 50 A1 A2 A3 A-High A-Low D1 D2 D-High D-Low
48 50 4 1 1 4 1 6 4 6 4
46 50 2 5 1 5 2 6 5 6 5
46 48 5 3 4 5 4 2 1 2 1
45 47 3 3 1 3 3 4 1 4 1
44 46 2 1 3 3 2 6 1 6 1
44 44 1 6 5 6 5 4 3 4 3
44 42 6 4 3 6 4 1 1 1 1
44 40 2 4 1 4 2 3 1 3 1
44 38 6 1 5 6 5 2 4 4 2
44 36 1 5 3 5 3 1 2 2 1
44 34 5 2 4 5 4 2 1 2 1
43 33 5 2 3 5 3 6 2 6 2
42 32 1 3 2 3 2 1 3 3 1
41 31 6 2 3 6 3 5 3 5 3
40 30 3 2 6 6 3 3 3 3 3
39 29 6 4 3 6 4 5 6 6 5
37 29 1 3 3 3 3 4 6 6 4
37 27 4 4 6 6 4 1 5 5 1
37 25 4 3 4 4 4 2 2 2 2
36 24 1 2 4 4 2 5 2 5 2
35 23 3 3 2 3 3 2 6 6 2
35 21 2 6 3 6 3 3 1 3 1
34 20 1 6 3 6 3 6 3 6 3
33 19 1 3 2 3 2 1 6 6 1
32 18 5 5 3 5 5 5 4 5 4
31 17 4 2 3 4 3 4 3 4 3
30 16 3 2 5 5 3 5 4 5 4
29 15 3 3 3 3 3 4 3 4 3
28 14 1 5 1 5 1 4 5 5 4
27 13 1 3 2 3 2 3 1 3 1
26 12 1 5 1 5 1 1 1 1 1
26 10 3 1 5 5 3 3 2 3 2
25 9 4 4 2 4 4 3 4 4 3
24 8 4 3 2 4 3 6 2 6 2
24 6 1 5 6 6 5 4 3 4 3
23 5 2 3 1 3 2 1 5 5 1
23 3 4 3 6 6 4 4 1 4 1
23 1 5 1 4 5 4 1 4 4 1
22 0 3 2 6 6 3 3 4 4 3[/tt]

What I am noticing from the distribution of numbers:
[tt]
Attacker Attacker Defender
rolls Uses Rolls/Uses
qty qty qty
#1 23 3 18
#2 19 8 11
#3 31 24 15
#4 17 16 15
#5 15 15 9
#6 12 12 10[/tt]

I've run the scenario 100 times
Average number of rolls = 41.5
Average Remaining Armies (Attacker) = 18.1

This computes to a win percentage of 60%+

I believe the data before the calculations. Anyone have any insights?

**********************************************
What's most important is that you realise ... There is no spoon.

 

[kwbMitel]

I ran 100 more Trials gathering more data

Again I achieved greater than 60% win percentage.

I think the difference is in the uneven distribution usage of numbers for the attacker which gives him a greater percentage of high roles relatively speaking.

Average Roles this time was 40.68
Average Armies remaining was 19.82

Attacker rolled qty
1 1988
2 1953
3 1980
4 2021
5 1961
6 2001

Attacker Used qty
1 308
2 868
3 1279
4 1664
5 1837
6 1980
Total = 7936

Defender Roled/used qty
1 1358
2 1381
3 1300
4 1240
5 1370
6 1287
Total = 7936

**********************************************
What's most important is that you realise ... There is no spoon.

 

[kwbMitel]

Just for grins I ran 10,000 trials

I forgot to put a low limit on the attacker so the attacker sometimes ended up with negative values.

Even with these in place the averages came out higher than calculated

29 times out of 10000 the attacker failed
Most Roles was 56 (-11) Armies (Whoops)
Least Roles was 30 with 40 armies remaining

Average roles was 41
Average Armies remaining was 19.2

Attacker averaged 60.98%

Still stumped as to why.

**********************************************
What's most important is that you realise ... There is no spoon.

 

[kwbMitel]

rolls not roles

**********************************************
What's most important is that you realise ... There is no spoon.

 

[karluk]

Allow me correct my remarks from yesterday. If both players start with 50 armies and battle until one side runs out of armies, the most likely outcome is that the defender will lose all 50 armies. Because of the attacker's edge, those 50 losses by the defender should be approximately 53.95% of the total losses. That leads to the simple algebraic expression for the total number of armies lost, x:

.5395 * x = 50

So x=92.68 (approximately), and the attacker should be prepared to lose 42 or 43 armies in the battle and be left with seven or eight. That's a bigger margin than I suggested yesterday, but still not particularly comfortable, in my opinion.

 

[CajunCenturion]

@kwbMitel - Based on the 3 vs 2 battle, after every battle, two armies will leave the board. On average, for each roll, the attacker will lose 0.921 armies and the defender will lost 1.079 armies.
0.921 = (2 * 0.2926) + 0.3358
1.079 = 2 - 0.921

==> So 50 vs 50 will take 40 rolls and you will have 20 armies left.
With the defender losing armies at an average rate of 1.079 armies per roll, you'd expect the defender to lose all in (50 / 1.079) rolls, or 46.3392 rolls. After 46.3392 rolls, the attacker would expect to have (on average) 50 - (0.921 * 46.3392) = 7.3216 armies.


--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886
Wise men speak because they have something to say, fools because they have to say something. - Plato