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Black Mage
04-06-2004, 02:10 PM
In the game "Risk" you have 5 types of confrotations when a player attacks a territory. They are:

1) One attacker versus one defender
2) Two attackers versus one defender
3) Three attackers versus one defender
4) One attacker versus two defenders
5) Two attackers versus two defenders
6) Three attackers versus two defenders

I assume many people reading this are familiar with the rules but I'll try to explain. For each attacker or defender a player gets to roll s 6 sided die. For multiple dies (say when two attackers versus two defenders) it is the highest numbered die of each player that is looked at, then it is the next highest. If there is a tie then the defender wins. What are the odds for each situtation of the attacker losing?

I calculated #1 as 21/36, but I cannot figure out the others. Any assistance would be welcomed.

:duh:

cubedbee
04-06-2004, 02:17 PM
Here, I cheated: http://www.sciencenews.org/articles/20030712/mathtrek.asp

Gandalf
04-06-2004, 02:43 PM
CubedBee, that table may not be accurate, depending on what question stonybrookcalc is asking.

Consider his question about 3 attackers vs 1 defender.

The table shows .917 for the attackers. That can't be right for the attackers winning that attack. The sole defender will role a 6, and win, 1/6 of the time. The defender can also win with less than a 6, even (exceedingly rarely) with a 1.

What the table must mean is that the 3 attackers eventually prevail with probability .917. Either they win immediately, or they lose then win the 2-1 rematch, or they lose then win the 1-1 rematch. Even that should be .916375, but who's counting?

The 2-2 and 3-2 situations are ambiguous for winning that attack, as one possible outcome of that attack is that each side loses one army.

Macroman
04-06-2004, 08:57 PM
According to Gandalf's explanation we should expand Stonybrook's list of possible number of armies in an attack to any number of attackers and defenders. Though only 3 attackers and 2 defenders can participate in each throw, the ultimate probabilities of success in a series of attacks would be much different if 10 against 2 as opposed to say 3 against 2 or 3 against 10 even though in each case the same number of dice are thrown for the first throw.

There is also more general strategy to the game then the outcome of a single series of attacks. Often a successful attack will weaken the attacker so much that another player will recapture the gains and more.

abacustwo
04-06-2004, 11:33 PM
What the table must mean is that the 3 attackers eventually prevail with probability .917. Either they win immediately, or they lose then win the 2-1 rematch, or they lose then win the 1-1 rematch. Even that should be .916375, but who's counting?

It's actually closer to .9163746356, but who's counting?

abacustwo
04-06-2004, 11:34 PM
Or 1-(7^4*13)/(3^6*2^9) if you want to be exact.

abacustwo
04-07-2004, 12:58 AM
1v1 21/36 = .5833
2v1 91/216 = .4213
3v1 441/1296 = .3403
1v2 161/216 = .7454

Gandalf
04-07-2004, 08:43 AM
What the table must mean is that the 3 attackers eventually prevail with probability .917. Either they win immediately, or they lose then win the 2-1 rematch, or they lose then win the 1-1 rematch. Even that should be .916375, but who's counting?

It's actually closer to .9163746356, but who's counting?
OK, as is your formula. .916375 is accurate to the number of digits shown; .917 was not.