## Odds in the game of Risk

## 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.

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

## RE: Odds in the game of Risk

## 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

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

--------------

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: 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

## Hidden:

## 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

## Hidden:

## RE: Odds in the game of Risk

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:

@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

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 readFAQ181-2886: 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

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

## RE: Odds in the game of Risk

## RE: Odds in the game of Risk

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

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

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

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

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

## RE: Odds in the game of Risk

.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

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 readFAQ181-2886: 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

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?

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: 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

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.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: 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

The lines that appear to credit incorrect outcomes are:

## RE: Odds in the game of Risk

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 readFAQ181-2886: 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

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

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

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

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

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.

Good Luck

To get the most from your Tek-Tips experience, please readFAQ181-2886: 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

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

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

## RE: Odds in the game of Risk

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 3.5That'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 13

^{th}roll, but you have to after the 13^{th}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 readFAQ181-2886: 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

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

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

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

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

## RE: Odds in the game of Risk

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

## RE: Odds in the game of Risk

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

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

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

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

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

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

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 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

Here is a link to a Risk odds calculator

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

## RE: Odds in the game of Risk

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

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

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

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.