Conquer Club

i was wondering about the probability of rolling 3 6's on a 6 sided die. it came up in a character generating session of a d&d game and i was thinking about how it applies to risk as well. here's what i found out on wikipedia;


The probability of throwing several dice (or one die several times successively) and getting the same number on each roll is computed by multiplying by 1/6 for each additional die:
No. of Dice 2 3 4 5
Probability 1/6 1/36 1/216 1/1296

The above list only applies when any roll is acceptable provided it's the same on all dice. To throw a particular chosen number more than once in a row, the probability is lower:
No. of Dice 2 3 4 5
Probability 1/36 1/216 1/1296 1/7776

That means nothing its just fucking numbers.....how am i ment to read that ?

3 sixes in a row = 1/6*6*6 or 1/216 (the hypothetical odds)

this means that with an infinate ammount of 3 dice rolled, 1 out of every 216 throws will give you triple 6's.

If the program controlling the randomness of the dice is accurate, then 1/216 will be the odds for EVERY ONE out of all of CC, which means that if you dont get a triple 6, someone else might get two triple 6's.

the only way to accurately test the randomness of the dice would be to gather all of the dice rolls from everyone that has a game on CC, which would most likely be very impractical to attempt due to the vast ammounts of games and members on CC.

Edit: does this help you understand?

I've done it 4 times in one risk game in real life.

Attacker rolls 6 6 6

Defender rolls 6 6

Happened twice in a row once here. Was semi unhappy.

each fraction should be under a number. i can't editit to make it look right
here's a link where they show the probability numbers more clearly
http://en.wikipedia.org/wiki/Dice

What's more useful is figuring out what are the odds that your three dice will beat their two dice. Now that's some heavy math.

mybike_yourface wrote:each fraction should be under a number. i can't editit to make it look right
here's a link where they show the probability numbers more clearly
http://en.wikipedia.org/wiki/Dice

yeah - but you should have the numbers 1 to 4 for the fractions you've shown - not 2 to 5.

wcaclimbing wrote:3 sixes in a row = 1/6*6*6 or 1/216 (the hypothetical odds)

this means that with an infinate ammount of 3 dice rolled, 1 out of every 216 throws will give you triple 6's.

If the program controlling the randomness of the dice is accurate, then 1/216 will be the odds for EVERY ONE out of all of CC, which means that if you dont get a triple 6, someone else might get two triple 6's.

the only way to accurately test the randomness of the dice would be to gather all of the dice rolls from everyone that has a game on CC, which would most likely be very impractical to attempt due to the vast ammounts of games and members on CC.

Edit: does this help you understand?

Not that i asked but yes, it was.

x wrote:What's more useful is figuring out what are the odds that your three dice will beat their two dice. Now that's some heavy math.

I have those odds. I managed to find the equation for 2v1 throws, but for 3v2 throws I wrote a python script to brute-force every possible combination (all 6^5/6**5/7776, of them). I now have a file on my computer with approximate odds of each occurance of the 6 different dice combinations (3v1, 3v2, 2v1, 2v2, 1v1, 1v2). If anyone wants it and pays me $5, I'll give you my numbers.

I will tell you this: in an even distribution of 100 throws of 3v2, odds are that the defender will lose a few more armies than the attacker. Also, the odds of each person loses an army is between the two numbers. So if A = Times attacker "wins"; D = Times defender "wins"; T = Times each loses one army:

A > T > D

But I will not tell you guys the actual distribution numbers. What I have said is absolutely true though. And don't attack a country with 2 armies on it if you only have two armies. You'll lose your army nearly 3/4 the time. And you only win approximately 2/3 of the time against a single army with 4+ armies attacking.

kc-jake wrote:I have those odds. ... If anyone wants it and pays me $5, I'll give you my numbers.

Hmm, just open the grade 12 math book and with simple math you can have it, you don't need a program to do this.

kc-jake wrote:

x wrote:What's more useful is figuring out what are the odds that your three dice will beat their two dice. Now that's some heavy math.

I have those odds. I managed to find the equation for 2v1 throws, but for 3v2 throws I wrote a python script to brute-force every possible combination (all 6^5/6**5/7776, of them). I now have a file on my computer with approximate odds of each occurance of the 6 different dice combinations (3v1, 3v2, 2v1, 2v2, 1v1, 1v2). If anyone wants it and pays me $5, I'll give you my numbers.

I will tell you this: in an even distribution of 100 throws of 3v2, odds are that the defender will lose a few more armies than the attacker. Also, the odds of each person loses an army is between the two numbers. So if A = Times attacker "wins"; D = Times defender "wins"; T = Times each loses one army:

A > T > D

But I will not tell you guys the actual distribution numbers. What I have said is absolutely true though. And don't attack a country with 2 armies on it if you only have two armies. You'll lose your army nearly 3/4 the time. And you only win approximately 2/3 of the time against a single army with 4+ armies attacking.

2v2 or 3v2 is just stupidity. 2v2 is 1 dice vs 2, and they win on ties. 3v2 is 2 dice v 2, and they win on ties. Only when you get to 4v2 and 4v1 do you stand any chance because you get to roll all 3 dice.

4v1 is very good odds, you may lose an army or two, but you'll almost always win.

4v2 isn't as good, but still gives you decent odds.

100 armies vs. 100 armies gives the attacker a sleight advantage, but not a great one. It's worth doing in certain circumstances, but you really only want to attack when you have the overwhelming advantage.

kc-jake wrote:

x wrote:What's more useful is figuring out what are the odds that your three dice will beat their two dice. Now that's some heavy math.

I have those odds. I managed to find the equation for 2v1 throws, but for 3v2 throws I wrote a python script to brute-force every possible combination (all 6^5/6**5/7776, of them). I now have a file on my computer with approximate odds of each occurance of the 6 different dice combinations (3v1, 3v2, 2v1, 2v2, 1v1, 1v2). If anyone wants it and pays me $5, I'll give you my numbers.

I will tell you this: in an even distribution of 100 throws of 3v2, odds are that the defender will lose a few more armies than the attacker. Also, the odds of each person loses an army is between the two numbers. So if A = Times attacker "wins"; D = Times defender "wins"; T = Times each loses one army:

A > T > D

But I will not tell you guys the actual distribution numbers. What I have said is absolutely true though. And don't attack a country with 2 armies on it if you only have two armies. You'll lose your army nearly 3/4 the time. And you only win approximately 2/3 of the time against a single army with 4+ armies attacking.

Sorry to ruin your little fundraiser, but info is not very hard to find on the internet. Especially when this image to the wikipedia dice odds has been posted many times.

x wrote:

kc-jake wrote:

x wrote:What's more useful is figuring out what are the odds that your three dice will beat their two dice. Now that's some heavy math.

I have those odds. I managed to find the equation for 2v1 throws, but for 3v2 throws I wrote a python script to brute-force every possible combination (all 6^5/6**5/7776, of them). I now have a file on my computer with approximate odds of each occurance of the 6 different dice combinations (3v1, 3v2, 2v1, 2v2, 1v1, 1v2). If anyone wants it and pays me $5, I'll give you my numbers.

I will tell you this: in an even distribution of 100 throws of 3v2, odds are that the defender will lose a few more armies than the attacker. Also, the odds of each person loses an army is between the two numbers. So if A = Times attacker "wins"; D = Times defender "wins"; T = Times each loses one army:

A > T > D

But I will not tell you guys the actual distribution numbers. What I have said is absolutely true though. And don't attack a country with 2 armies on it if you only have two armies. You'll lose your army nearly 3/4 the time. And you only win approximately 2/3 of the time against a single army with 4+ armies attacking.

2v2 or 3v2 is just stupidity. 2v2 is 1 dice vs 2, and they win on ties. 3v2 is 2 dice v 2, and they win on ties. Only when you get to 4v2 and 4v1 do you stand any chance because you get to roll all 3 dice.

4v1 is very good odds, you may lose an army or two, but you'll almost always win.

4v2 isn't as good, but still gives you decent odds.

100 armies vs. 100 armies gives the attacker a sleight advantage, but not a great one. It's worth doing in certain circumstances, but you really only want to attack when you have the overwhelming advantage.

For one thing, the other poster meant 3 dice vs 2 dice when he said 3v2, not 3 men vs two men.

Also, the 3v1 odds are not that great. You win about 2/3 of the time and lose about 1/3 of the time.

sully800 wrote:For one thing, the other poster meant 3 dice vs 2 dice when he said 3v2, not 3 men vs two men.

Also, the 3v1 odds are not that great. You win about 2/3 of the time and lose about 1/3 of the time.

Probably, but it's worth clarifying because this is something I'd ignored for a bit when I first started playing here - it'd been a whlie since I played real tabletop risk and I forgot how many dice you were allowed to roll for how many attacking/defending armies.

Your calculations are also PER ROLL, not "odds of taking the territory."