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# Odds in the game of Risk3

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## Odds in the game of Risk

(OP)
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.

### RE: Odds in the game of Risk

(OP)
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?

### RE: Odds in the game of Risk

By my usual Brute force Methdology:

#### Hidden:

Attacker will win 37.17% of the time
Defender will win 29.26% of the time
They will draw 33.58% of the time

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

### RE: Odds in the game of Risk

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

--------------
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To get the most from your Tek-Tips experience, please read
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### RE: Odds in the game of Risk

(OP)
@kwbMitel

#### Hidden:

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.

### RE: Odds in the game of Risk

#### Hidden:

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.

Bye, Olaf.

### RE: Odds in the game of Risk

(OP)
@OlafDoschke

#### Hidden:

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

### RE: Odds in the game of Risk

@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

#### Hidden:

0.539544753

@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.

### RE: Odds in the game of Risk

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.

### RE: Odds in the game of Risk

==> 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
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### RE: Odds in the game of Risk

(OP)

#### Quote (kwbMitel):

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.

### RE: Odds in the game of Risk

(OP)
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.

### RE: Odds in the game of Risk

(OP)
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.

### RE: Odds in the game of Risk

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.

### RE: Odds in the game of Risk

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

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

What I am noticing from the distribution of numbers:

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

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.

### RE: Odds in the game of Risk

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.

### RE: Odds in the game of Risk

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.

### RE: Odds in the game of Risk

rolls not roles

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

### RE: Odds in the game of Risk

(OP)
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.

### RE: Odds in the game of Risk

@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.

--------------
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### RE: Odds in the game of Risk

My data suggests that the margin is different from the calculated.

Can you see anything wrong in the one example I posted?

the 100 trial exercise was to prove that the distribution of numbers was more even than the one I posted

The 10,000 trials of 50 against 50 was to get an average that you could be confident in.

I'm sure many in this forum can duplicate my trials. I'd be interested to know if they get similar results. If I've made a mistake, I can't see it.

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

### RE: Odds in the game of Risk

==> I am interested in ways to translate the attacker's edge into an estimate of the likely outcome of a battle,
In a 3 vs 2 battle, provided the attacker beings with the same or greater number of armies, the likely outcome is that the attacker will win.  In every 3 vs 2 roll, the attacker loses 0.921 armies and the defender loses 1.079 armies.

Here are two fun follow up questions.  What is the minimum number of armies that the attacker must have to have a better than 50% of winning, given that the attacker will lose the 3 vs 2 advantage once the attacker army count drops to three and below?  Secondly, what is the minimum number of armies (A) the attacker must have, so that even if the defender has one more army (A + 1), the odds still favor the attacker?

--------------
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To get the most from your Tek-Tips experience, please read
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### RE: Odds in the game of Risk

==> 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.
That's consistent with my post 7 Dec 11 10:05.  7.3216 would qualify as seven or eight.

--------------
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To get the most from your Tek-Tips experience, please read
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### RE: Odds in the game of Risk

(OP)

#### Quote (kwbMitel):

Can you see anything wrong in the one example I posted?
Yes.  It looks to me as if you are not calculating the results of the dice throws correctly.  In your post dated 7 Dec 11 1:12, I see seven lines where you appear to be crediting a 1-1 split outcome when the numbers on the dice actually give the defender a win on both.  The corrected results give the attacker a 43-35 edge, or 55.13%, which is reasonably close to the predicted odds.
The lines that appear to credit incorrect outcomes are:

#### Quote (kwbMitel):

39     29     6     4     3     6     4     5     6     6     5
36     24     1     2     4     4     2     5     2     5     2
34     20     1     6     3     6     3     6     3     6     3
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

### RE: Odds in the game of Risk

==> Can you see anything wrong in the one example I posted?[/i]
You're problems seem to begin in this area:
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

The 41, 31 looks correct, as does the 40, 30, but the 39, 29 doesn't look correct.

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
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### RE: Odds in the game of Risk

Well,

the probabilities already give the mathematical expected values of armies lost on both sides per round of 3 vs 2 dice.

You can calculate the mathematical expected value of lost armies after N rounds quite easy from it's probability as 2*N*p.

Let's say the probibility of the attacker to lose an army per dice comparison is pa and the probability of the defender to lose his army is pd=1-pa.

If defender has D armies and attacker has A armies, the goal is to kill D armies. Then the expected loss after N rounds should match D=2*N*pd. The attacker loses 2*N*pa armies at the same time. In the inverse calculation the average number of rounds needed to beat the defender would be D/(2*pd).

In extreme situations the number of needed rounds as minimum is of course D/2 or as the opposite, with a streak of luck the defender kills thee attacker in A/2 rounds.

And the maximum number of rounds will be (A+D)/2, anyway the dices are rolled, always 2 armies are lost per round (neglecting the cases at the end, when less dice are used).

You would need to compute the probabilities for each number of rounds between min(A,D)/2 and (A+D)/2 rounds (more exactly the rounded values, as you can't play half a round of course). This leads to a certain distribution, you would need to figure out the probabilities for each outcome and then could compute a confidence to win in eg 90% of the cases.

There is no need to go into monte carlo simulations, but of course that helps to support theoretical values.

Bye, Olaf.

### RE: Odds in the game of Risk

==> Can you see anything wrong in the one example I posted?[/i]You're problems seem to begin in this area:
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
The 41, 31 looks  correct, as does the 40, 30, but the 39, 29 doesn't look correct.

Agreed Defender 6-5 should defeat Attacker 6-4 both times

I'll try and see why my formulas thought otherwise.

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

### RE: Odds in the game of Risk

Actually things get very bad at around:
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

These should all have been 2 win for defender and the program split them.

Just goes to show that another pair of eyes pays off sometimes.

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

### RE: Odds in the game of Risk

(OP)

#### Quote (CajunCenturion):

What is the minimum number of armies that the attacker must have to have a better than 50% of winning, given that the attacker will lose the 3 vs 2 advantage once the attacker army count drops to three and below?
In my opinion, the right way to approach this question is to figure out the minimum number of armies the attacker needs in order to retain a better than 50-50 chance that the defender will lose the right to roll two dice before the attacker loses the right to throw three dice.  If the defender starts with 50 armies, he will lose the right to throw two dice when he has lost 49.  It seems to me that the critical number of dice tosses occurs at the point when the defender's expected losses are greater than 48.5.  That's when it's a better than 50-50 chance that the defender will have fewer than two armies remaining.
But the defender, on average, expects to lose 2*2797/5184 armies per dice toss.  So we need

2*2797/5184 * x > 48.5

x > 44.95

Therefore, after 45 dice tosses, the odds favor the defender having either lost all his armies, or having only one left.

Now we need to calculate the attacker's expected losses after 45 dice tosses.  The attacker's expected losses are 2*2387/5184 armies per dice toss.  So, after 45 tosses, the attacker's expected losses are

2*2387/5184 * 45 = 41.44

If the attacker starts with 45 dice, he expects to have about (45-41.44) = 3.56 armies remaining after 45 tosses.  It should be a better than 50-50 shot that he still has four armies left and retains the right to throw three dice.

### RE: Odds in the game of Risk

If what you're looking for is that point where the defender is reduced to rolling one die while you still retain the option to roll all three, assuming each party starts with the same number of armies, then I would work that as follows:
Given that the defender loses 1.079 [ (2 * 0.3717) + 0.3358 ] per roll, and the attacker loses 0.921 [ (2 * 0.2926) + 0.3358 ] per roll, work the following simultaneous equations:
x - 1.079r = 1
x - 0.921r = 4
Solving for r gives up
1 + 1.079r - 0.921r = 4 ==> r = 18.9873, which means that you can expect to achieve that 4:1 advantage after 19 rolls of the dice.  The minimum number of armies that you'd need in order to engage is:
x - (0.921 * 19) = 4  ==> x = 21.499, which means the minimum number of armies you need is 22.

--------------
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### RE: Odds in the game of Risk

(OP)

#### Quote (CajunCenturion):

x - 1.079r = 1
x - 0.921r = 4
Cajun, your simultaneous equations are similar to the way I would approach the same problem, but with two exceptions.  The first is minor - you are using approximate expected losses per dice throw of 1.079 and 0.921 instead of the exact 2797/2592 and 2387/2592.  But your approximations are close and in all likelihood will yield the same result.
But I'm not convinced that your constants of 1 and 4 are correct.  I would use 1.5 and 3.5 for the same reasons I gave when I calculated that 45-attack vs. 50-defend slightly favors the attacker.  I will think about it overnight and see if my logic still looks right in the morning.  If so, I calculate that the attacker can proceed when both players have 16 armies and still have favorable odds of success.

### RE: Odds in the game of Risk

Fixed my issues with my program. 1 missing = sign, sigh. The good news is I learned some cool new tricks in the process so it's not all bad.

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

### RE: Odds in the game of Risk

(OP)
I stand by my earlier statement that a 16 vs. 16 battle favors the attacker.  The numbers, as far as I have been able to calculate them, support this.  After 13 dice throws, the attacker's expected number of armies remaining is 4.03 and the defender's expected number is 1.97.  So 4 armies attack vs 2 armies defend is likely, but if either side falls below this, it is more likely to be the defender, who has an expectation of only 1.97 armies.

After 14 tosses, the expected number of remaining armies is 3.11 attacker and 0.89 defender.  The attacker is more likely to still have four armies remaining than the defender still having two, and if it comes down to 3 armies vs. 1, the attacker will be able to throw two dice against the defender's one, a battle that favors the attacker.

So the attacker's edge in a 16 vs. 16 battle isn't very large, but it seems quite clear.

### RE: Odds in the game of Risk

==>  I would use 1.5 and 3.5
That's certainly an option, and you can make a case for it, but it's a strategy that's dependent on favorable truncation.

Another way to look at is that the attacker is gaining an advantage of 0.158 armies per roll and to meet the stated objective of ending up with a 3 dice vs 1 die roll, you need to obtain an advantage of >= 3 armies.  The number of rolls required for that is 3 / 0.158 = 18.9873 rolls, i.e., 19 rolls which requires 22 armies to engage.

By using 3.5 and 1.5, you're calculating the number of rolls needed to obtain an advantage of >= 2 armies, which only projects to a 3:1 advantage, or a two vs one dice roll.  And yes, that happens in 12.6582 rolls.  Indeed, if each party starts with 16 armies, then after 13 rolls you'd expect the attacker to have 4.027 armies and the defender 1.973.  The attacker has > 4 armies and the defender < 2, but mathematically, you're advantage is only 2.054 armies, not 3.  For you to achieve your stated goal of having a 3 vs 1 dice advantage on the next roll, you need an advantage of 3 armies and you can only have that if you truncate away .973 armies from the defender.  You cannot truncate away any armies before the 13th roll, but you have to after the 13th roll to meet the objective.

It's a valid strategy, but one which assumes that once the attacker's advantage is greater than 2 armies that it's equivalent to an advantage of 3 armies.  That's not an assumption that I would to require as part of my strategy, but again, that's doesn't mean that a strategy that requires that assumption is an invalid strategy.

--------------
Good Luck
To get the most from your Tek-Tips experience, please read
FAQ181-2886: How can I maximize my chances of getting an answer?
Wise men speak because they have something to say, fools because they have to say something.  - Plato

### RE: Odds in the game of Risk

As the Math in this case eludes me, I've resorted to re-writing my simulator.

The simulator now agrees with the 53.9% calculated earlier

I've added the factors for reducing armies below 4(attacker) and 2(defender)

Averaging out a 1000 trials each:

16 Armies comes in well ahead of 50%
15 55%
14 54.8%
13 52%
12 51%
11 Armies appears to be doing OK: 49% on first but over 50% on next 4 (51.6 over 5000)
10 49%,52.6%,48.5%,48.4%,50.1%  = Not enough

Again, my confidence in the simulator is high but I admit I've made mistakes already with this. How does 11 fair against your calculations?

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

### RE: Odds in the game of Risk

(OP)
I don't really think the work Cajun and I have done is sufficient to either confirm or refute your simulations.  I would regard what we've done more as setting an upper bound on when the advantage in an equal armies battle switches over from defender to attacker.  My earlier post presents, I believe, a convincing argument that 16 vs. 16 favors the attacker, but that doesn't eliminate the possibility that 15 vs. 15 (or less) also favors the attacker.

In my view, getting a precise limit on when the advantage switches from defender to attacker is likely to require a detailed enumeration of cases, including an analysis of the probabilities when the attacker is tossing fewer than three dice and/or the defender is tossing only one.  I am interested in this question, but haven't done any real work on it yet.

### RE: Odds in the game of Risk

OK,

I'm seeing something in my results that doesn't look right anyway. I can get results for all attacker remaining armies from 1 to 16. The results for the defender however always seem to be odd numbers. 1,3,5,7,9... (except for 0 of course).

Maybe this is a result of the Attacker needing to stop with 1 army remaining instead of 0. Actually, that makes sense to me. Does that make sense to you?

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

### RE: Odds in the game of Risk

The odd number result for Defender makes complete sense now that I've thought about it. This adds confidence for me in the program.

Here are some results from 10,000 trials of 16 vs 16

Average amount of rolls - 15.16 (high = 20, Low = 10)
Average remaining armies for Attacker - 4.0844 (high 16, Low 1)
Average remaining armies for Defender - 2.1345 (high 15, Low 0)
Average Wins for Attacker - 55.35%
Average Wins for Defender - 44.65%

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

### RE: Odds in the game of Risk

(OP)

#### Quote (kwbMitel):

The results for the defender however always seem to be odd numbers. 1,3,5,7,9... (except for 0 of course).
If I'm understanding you correctly, this looks like an error in your simulations.  It's certainly possible for the defender to be left with an even number of armies following a 16 vs. 16 battle.  In the most extreme case, for example, the defender might roll double sixes every single time, reducing the attacker to a single army while still retaining all 16 defending armies.

### RE: Odds in the game of Risk

Thus is the nature of my ability to rationalize. I will try to locate the error. It looked strange (and is), which is why I brought it up.

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

### RE: Odds in the game of Risk

As it turns out it was a pretty minor thing. In those cases where the attacker was only throwing 1 die, while the defender was throwing 2, I had placed a marker of X for the Attackers second throw. As it turns out, X is considered greater then digits 1 to 6 and was being registered as a loss for the defender. This is now corrected. I'm still trying to wrap my head around why this caused the remainder to always be odd but lucky that it did or I wouldn't have seen the error.

This does not significantly change my stats except for the defender remaining armies when the defender wins.

I also spotted 1 other minor error in that I was counting the rolls incorrectly. The limits are actually 8-18 for rolls not 10-20

So This is the updated data after these corrections:

Average amount of rolls - 13.23 (high = 18, Low = 8)
Average remaining armies for Attacker - 4.0847 (high 16, Low 1)
Average remaining armies for Defender - 2.3815 (high 16, Low 0)
Average Wins for Attacker - 54.77%
Average Wins for Defender - 45.23%

Note: I had to run 20000 to get a result of 16 remaining armies for the defender. Result was 2 in 20000 vs 10 in 20000 for the attacker. This seems about right.

Does anyone have a formula to see how close the numbers above are?

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

### RE: Odds in the game of Risk

Re-running the numbers for lowest Equal Match where the odds favor the attacker, 11 no longer makes the cut.

12 appears to be the lowest.

The other part about starting with 1 less requires a much higher number apparently. Still working on that.

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

### RE: Odds in the game of Risk

#### Quote (CajunCenturion):

Secondly, what is the minimum number of armies (A) the attacker must have, so that even if the defender has one more army (A + 1), the odds still favor the attacker?

It appears that that number is 19 (to defenders 20)

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

### RE: Odds in the game of Risk

(OP)
In order to get precise results for an n vs. n battle, we need to calculate the exact probability of each outcome for each possible dice combination.  I have done this, although I have not yet double-checked my results.  The numbers are:

for three attacking dice vs. two defending dice
2-0 win for attacker - 2890/7776
1-1 split            - 2611/7776
0-2 win for defender - 2275/7776

for two attacking dice vs. two defending dice
2-0 win for attacker - 295/1296
1-1 split            - 420/1296
0-2 win for defender - 581/1296

for three attacking dice vs. one defending die
1-0 win for attacker - 855/1296
0-1 win for defender - 441/1296

for two attacking dice vs. one defending die
1-0 win for attacker - 125/216
0-1 win for defender - 91/216

for one attacking die vs. two defending dice
1-0 win for attacker - 55/216
0-1 win for defender - 161/216

for one attacking die vs. one defending die
1-0 win for attacker - 15/36
0-1 win for defender - 21/36

### RE: Odds in the game of Risk

Confirming I get the same answers for:

3 vs 2 - Yes
3 vs 1 - Yes
2 vs 2 - Yes
2 vs 1 - Yes
1 vs 2 - Yes
1 vs 1 - Yes

Double check completed

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

### RE: Odds in the game of Risk

(OP)
Ok, using the above probabilities, I get the following inflection points where the advantage switches from the defender to the attacker:

equal armies:
11 vs. 11  -  49.40%
12 vs. 12  -  50.65%

attacker has one fewer army:
17 vs. 18  - 49.42%
18 vs. 19  - 50.39%

attacker has two fewer armies:
23 vs. 25  -  49.87%
24 vs. 26  -  50.69%

attacker has three fewer armies:
29 vs. 32  -  49.82%
30 vs. 33  -  50.54%

attacker has four fewer armies:
34 vs. 38  -  49.46%
35 vs. 39  -  50.13%

attacker has five fewer armies:
40 vs. 45  -  49.45%
41 vs. 46  -  50.01%

attacker has six fewer armies:
46 vs. 52  -  49.74%
47 vs. 53  -  50.30%

### RE: Odds in the game of Risk

(OP)
By the way, these results, although unverified, were actually quite easy to calculate.  That's because the possible states of a Risk battle comprise a Markov chain.  Suppose that P(i,j) denotes the probability that the attacker will win a battle starting with the attacker having i armies and the defender having j armies.  If I want to calculate P(12, 12), for example, I need to have only a few facts at hand.  The only possible results of a 3-attack vs. 2-defend dice roll are 10 vs. 12, 11 vs. 11, and 12 vs. 10.  So, suppose I'm lucky enough to know the actual values of P(10, 12), P(11, 11) and P(12, 10).  I have already calculated the odds of getting from 12 vs. 12 to 10 vs. 12, 11 vs. 11, and 12 vs. 10, so

P(12, 12) = 2275/7776 * P(10, 12) + 2611/7776 * P(11, 11) + 2890/7776 * P(12, 10)

In other words, I know P(12, 12) in terms of the odds of various outcomes of a dice roll, combined with the outcomes of a few other battles.  This type of calculation lends itself extremely naturally to a spreadsheet, and in fact I calculated the above odds using an Excel spreadsheet.  Roughly speaking, I calculated the contents of row 12, column 12 of the spreadsheet by multiplying the contents of row 10, column 12 by 2275/7776, then adding the contents of row 11, column 11 multiplied by 2611/7776, and finally adding the contents of row 12, column 10 multiplied by 2890/7776.

This technique of determining an unknown probability by following a Markov chain down through simpler cases is widely used in probability theory.  It could be used, I believe, to calculate the exact odds of getting a multiple of nine in SidYuca's "guess the missing digit" puzzle.  The Markov chains in that puzzle are more complicated and lengthy than those in the game of Risk, but the same type of calculation would work.

### RE: Odds in the game of Risk

The Even Armies 12 vs 12 agrees with my simulator (or the other way around)

The simulator said 19 vs 20 was required for 1 less army.

50.1%

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

### RE: Odds in the game of Risk

(OP)

#### Quote (kwbMitel):

The simulator said 19 vs 20 was required for 1 less army.
Yes, I noticed the same discrepancy when I posted my results.  I suspect that my results are accurate, mostly because I have taken the liberty to do an internet search and found a site that confirms my calculations.  One also has to take into consideration that simulations have an inherent margin for error, since they only claim to approximate the true odds, not to compute them exactly.
Here is a link to a Risk odds calculator

http://www.dandrake.com/risk.html

### RE: Odds in the game of Risk

Regarding approximation in simulators. This is why I run so many trials. 20000 in the last case. I find the variation cancels out and you end up with a number you can be confident in. As long as you do enough trials anyway.

In this case, 20000 may just not be enough.

I'll try to make my simulator more efficient so that I can manage more trials more quickly.

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

### RE: Odds in the game of Risk

(OP)
It's possible to glean all sorts of other interesting tidbits from my spreadsheet.  Suppose, for example, the attacker is unwilling to attack without a 90% probability of success.  How big of an invading force is needed for various sized defending forces?  I get the following results:

three army advantage required
4 vs. 1

five army advantage required
7 vs. 2
8 vs. 3

six army advantage required
10 vs. 4
11 vs. 5
12 vs. 6

seven army advantage required
14 vs. 7
15 vs. 8
16 vs. 9
17 vs. 10
18 vs. 11

eight army advantage required
20 vs. 12 through 31 vs. 23

nine army advantage required
33 vs. 24 through the limit of my spreadsheet at 50 vs. 41

### RE: Odds in the game of Risk

Ok, I made the simulator more efficient.

I can now run a million trials in less time that the previous 10,000.

It did make a difference.

18 vs 19 now comes in @ 50.3259%

Close enough for me.

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

### RE: Odds in the game of Risk

(OP)
Thank you very much for being so willing to run your simulations, kwbMitel.  Now that I've calculated the exact odds, the simulations provide mostly just reassurance that there's no glaring error in my calculations.  But earlier in this thread they provided a major motivation to me to go out and compute the exact odds.  I knew that my original reasoning that the advantage switches to the attacker at 16 vs. 16 wasn't precise enough to rule out the findings of your simulator.  But, I would have guessed that your estimate of a switch-over at 12 vs. 12 was way too low.  So I knew I needed to get more accurate odds to be able to confirm whether your simulations were accurate.  As it turned out, you were 100% correct on that one.

It looks to me as if the number of different ways to to toss dice grows exponentially as the number of armies increases.  That's probably why you were able to get accurate results with 10,000 trials at 12 vs. 12, but needed 1,000,000 trials at 18 vs. 19.

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