Niko Ruf
Germany Schönaich

bestbandis wrote: I then thought about inverting it, and looking at the odds of not rolling a 6. So that would be 5 over 6 times 5 over six times 5 over six, which comes out at 125 over 216. If I then subtract 125 from 216, I should have my answer, which is 91 over 216, or 42.13%.
But this seems a little too pat, and I'd appreciate it if someone could put me straight.
Your solution using the inverse probability is correct. There are 125 ways to roll not a single "1" on 3 dice, so the other 91 outcomes include at least one "1" each.

Walt
United States Orange County California
Saddleback, covered by a blanket of winter snow
Please contact me about board gaming in Orange County.

Yep. Dead on. Ask us a hard one.

Mike K
United States Fairless Hills Pennsylvania

Tall_Walt wrote: Yep. Dead on. Ask us a hard one. OK, here's one:
In the classic game of Risk, assume that the attacker uses 3 dice and the defender uses 2 dice. What are the exact probabilities that (a) the defender loses two armies, (b) they each lose one, and (c) the attacker loses two armies.
(I did this once, programming a TI84 to get the exact odds. Sadly, that calculator was stolen a couple years ago. I forget the exact answers now.)

(The Artist formerly known as) Arnest R
Germany Munich Bavaria
Keep calm and carry on...

Coyotek4 wrote: Tall_Walt wrote: Yep. Dead on. Ask us a hard one. OK, here's one:
I'm afraid you'll have to hand the "Math Geek" badge back, please...

Mike K
United States Fairless Hills Pennsylvania

arnest_r wrote: Coyotek4 wrote: Tall_Walt wrote: Yep. Dead on. Ask us a hard one. OK, here's one: I'm afraid you'll have to hand the "Math Geek" badge back, please... Oh, it's not like I couldn't redo those calculations by hand. I just don't feel like going through the 6^5 permutations.
(And if you think normal people memorize such numbers ... it's a 'Math Geek' badge, not a 'Rain Man' badge.)



Tall_Walt wrote: Yep. Dead on. Ask us a hard one.
There you go (not exactly a probability question). I've been pondering this for quite a while, but have not been able to solve.
Two players, A and B. A deck of n cards, numbered from 1 to n, n even. The cards are shuffled and dealt so that each player gets n/2 cards.
A starts the first trick by playing a card. B plays a card after having seen A's card. Higher card wins the trick, and the winner of the trick gets one point. The game proceeds like this, except that the winner of each trick starts the next one. The game ends when all the cards have been played, and the player with a higher score wins.
Question: What are the optimal strategies for each player? Does there exist a simple algorithm to calculate the scores for each player (assuming they play with optimal strategies), if the algorithm is given the hands of each player as input? ("simple" means easily calculable by hand, not "brute force")

Walt
United States Orange County California
Saddleback, covered by a blanket of winter snow
Please contact me about board gaming in Orange County.

Coyotek4 wrote: In the classic game of Risk, assume that the attacker uses 3 dice and the defender uses 2 dice. What are the exact probabilities that (a) the defender loses two armies, (b) they each lose one, and (c) the attacker loses two armies.
Frankly, 6^5 is only 7776, or maybe more conveniently, 216 x 36, so I'd just brute force it with Excelless chance of a subtle mistake. Maybe someone would like to try it more elegantly?
Spoiler (click to reveal) (a) 2890/7776 (37%), (b)2611/7776 (34%), (c) 2275/7776 (29%)

Walt
United States Orange County California
Saddleback, covered by a blanket of winter snow
Please contact me about board gaming in Orange County.

Punainen Nörtti wrote: Two players, A and B. A deck of n cards, numbered from 1 to n, n even. The cards are shuffled and dealt so that each player gets n/2 cards.
I don't have a proof, just intuition on this one. If you're second player, you win with the lowerst card possible, or if you can't, play your lowest card. On lead, play the lowest card possible: if you've got guaranteed winners, like 100 in a 100 card deck, you'll get them sooner or later. In either case, the idea is to use the lowest card possible to get the effect you want or you're stuck with.

martin sgattoni
Argentina Capital Federal Argentina

The chances of rolling at least one 1 in 3 dices are:
1/6 + 1/6 + 1/6
If you want to know the chances of rolling three 1s, you have to multiply instead of adding:
1/6 * 1/6 * 1/6

Philip Thomas
United Kingdom London London

Martin, that can't be right. By the same logic the chances of rolling at least one one in 6 dice is 1/6+1/6+1/6+1/6+1/6+1/6=1!
The actual chance is 1 (5/6)^3. 5/6 being the chance not to roll a 1 on one dice.



Tall_Walt wrote: I don't have a proof, just intuition on this one. If you're second player, you win with the lowerst card possible, or if you can't, play your lowest card. On lead, play the lowest card possible: if you've got guaranteed winners, like 100 in a 100 card deck, you'll get them sooner or later. In either case, the idea is to use the lowest card possible to get the effect you want or you're stuck with.
At least your leading strategy does not provide optimal scores:
Assume 6 cards, A having 1,3,4 and B having 2,5,6. If A leads 1, he loses all the tricks. If A leads 3 (or 4), B wins, and must lead at A:1,4, B:2,5, which guarantees A at least one trick.
Nice try, anyway :)

Patrick Barringer
United States Rushford New York

martin47 wrote: The chances of rolling at least one 1 in 3 dices are:
1/6 + 1/6 + 1/6
If you want to know the chances of rolling three 1s, you have to multiply instead of adding:
1/6 * 1/6 * 1/6
This is actually the method used to determine expected value. If you roll three dice the expected number of ones is 1/2. Whne you're up to six dice the expected number of ones is one, etc...

James Barton
Australia Melbourne Victoria

I'm in. I'll answer the risk question. First, I need to know the rules of risk combat, though.



jman11 wrote: I'm in. I'll answer the risk question. First, I need to know the rules of risk combat, though.
The highestnumbered dices or each player are matched against each other, as well as the secondhighestnumbered dices or each player. The player having a lower value in a comparison loses one army. If the dice are tied, it is interpreted as the attacker's loss.

Mike K
United States Fairless Hills Pennsylvania

jman11 wrote: I'm in. I'll answer the risk question. First, I need to know the rules of risk combat, though. a) Attacker rolls 3 dice. b) Defender rolls 2 dice. c) Highest A die matches with higher D die. High die wins; D wins on a tie. d) Middle A die matches with lower D die. High die wins; D wins on a tie. e) Lowest A die ignored.
That's it.

James Barton
Australia Melbourne Victoria

Ouch, those rules suck. I'm not sure there is an elegant approach. Brute force is easy, but no fun. It's going to take a little while, though.

Chris B
United States Oxford Mississippi
Hotty Toddy Rebels!
Lets go Blues!

Tall_Walt wrote: Yep. Dead on. Ask us a hard one.
Alright, so your on this gameshow called Lets make a deal...

James Barton
Australia Melbourne Victoria

So I'm not done yet, but I have a really good idea for how I am going to do it. It isn't very elegant sorry, but then again neither is the problem.
There are actually only 21 combinations for the defensive player (who is rolling 2 dice), and 58 for the offensive player. The original estimates of 36 and 216 follow from taking 2^6=36 and 3^6=216. This estimate is too many, because for the defender 1,5 and 5,1 are really the same. However when taking the square you count both as being different. The 21 number follows from taking 6+5+4+3+2+1=21, and 58=21+15+10+6+3+2+1. Thus there are only 1218=21*58 different combinations of rolls on the two sets of dice.
In essence I have just reduced the amount of calculations and found a whole bunch of patterns. I have not yet done 3d vs 2d, but I have done every other permutation. I am now going to use my 2d vs 2d work to get an answer for 3d vs 2d by multiplying by the number of different ways of getting the two highest. It's all a bit tough, but I have not used brute force.
Attack vs Defense:
1v1: 6 x 6 = 36 Defender wins = 6+5+4+3+2+1 = 21 Attacker wins = 3621 = 5+4+3+2+1 = 15 Results: 58% atacker loses 1 guy. 42% defender loses 1 guy.
2v1: 21 x 6 = 126 Defender wins = 1+3+6+10+15+21 = 56 Attacker wins = 12656 = 70 Results: 44% attacker loses 1 guy. 56% defender loses 1 guy.
3v1: 58 x 6 = 348 Defender wins = 1+(3+1)+(6+3+1)+(10+6+3+1)+(15+10+6+3+1) = 70 Attacker wins = 34870 = 278 Results: 20% attacker loses 1 guy. 80% defender loses 1 guy.
1v2: 6 x 21 = 126 Attacker wins = 0+1+3+6+10+15 = 35 Defender wins = 21+(211)+(213)+(216)+(2110)+(2115) = 91 Results: 72% attacker loses 1 guy. 28% defender loses 1 guy.
2v2: 21 x 21 = 441 Attacker wins both = 15 + 10 + 6 + 3 + 1 ~35 (151) + (101) + (61) + (31) ~30 (153) + (103) + (63) ~22 (156) + (106) ~13 (1510) ~ 5 = 105 Defender wins both = 21 + 15 + 10 + 6 + 3 + 1 ~56 (211)+(151)+(101)+(61)+(31) ~50 (213)+(153)+(103)+(63) ~40 (216)+(156)+(106) ~28 (2110)+(1510) ~16 (2115) ~ 6 = 196 One of each = 441105196 = 140 Results: 44% attacker loses 2 guys. 32% each loses 1 guy. 23% defender 2 guys.
If people want to see how I have done all this (beyond the meagre calculations here). Just reply and I'll put together some pictures and stuff. An interesting little combinatorial challenge BTW.



jman11 wrote: There are actually only 21 combinations for the defensive player (who is rolling 2 dice), and 58 for the offensive player. The original estimates of 36 and 216 follow from taking 2^6=36 and 3^6=216. This estimate is too many, because for the defender 1,5 and 5,1 are really the same.
Assume (for illustration) that one of the defender's dice is light blue and another is dark blue. Denote by "probability(p,q)" the probability of "the result of the light blue dice is p, and the result of the dark blue dice is q".
Now probability(5,1)=probability(1,5)=probability(1,1), thus probability(5,1 or 1,5) = 2* probability(1,1). You must take this into the account when you compute the probabilities.
That is, if you reduce the defender to 21 cases, then some of the cases (those where the numbers differ) are more likely than others (those where the numbers are the same.)

Mike K
United States Fairless Hills Pennsylvania

jman11 wrote: There are actually only 21 combinations for the defensive player (who is rolling 2 dice), and 58 for the offensive player. The original estimates of 36 and 216 follow from taking 2^6=36 and 3^6=216. This estimate is too many, because for the defender 1,5 and 5,1 are really the same. However when taking the square you count both as being different. The 21 number follows from taking 6+5+4+3+2+1=21, and 58=21+15+10+6+3+2+1. Thus there are only 1218=21*58 different combinations of rolls on the two sets of dice.
There's a flaw in your logic. While 1,5 and 5,1 are the same, they combine to come up twice as often as, say 1,1. In short, every nondouble combo comes up twice as often as any double combo.
I believe your 1vs1 results are correct. After that, I'm not sure.
I think Tall_Walt posted the correct answers above, but it's too much for me to confirm right now.
(I was just beaten to the punch, I see.)

James Barton
Australia Melbourne Victoria

Damn, you're right. I'm an idiot.



jman11 wrote: Damn, you're right. I'm an idiot.
Even smart people make stupid mistakes



Coyotek4 wrote: In the classic game of Risk, assume that the attacker uses 3 dice and the defender uses 2 dice. What are the exact probabilities that (a) the defender loses two armies, (b) they each lose one, and (c) the attacker loses two armies.
7776 outcomes? Hmm... I'm going to go with:
2890: defender loses two armies (~37%) 2611: they each lose one (~34%) 2275: attacker loses two armies (~29%)

Walt
United States Orange County California
Saddleback, covered by a blanket of winter snow
Please contact me about board gaming in Orange County.

Punainen Nörtti wrote: Assume 6 cards, A having 1,3,4 and B having 2,5,6. If A leads 1, he loses all the tricks. If A leads 3 (or 4), B wins, and must lead at A:1,4, B:2,5, which guarantees A at least one trick.
Very good: quite right. You're setting the cards up so B finesses himself. What I knew, but oversimplified, is that you do not want to be leading, but still you must take a majority of tricks to lead.

Tim Klein
United States Alexandria Virginia

Tall_Walt wrote: Punainen Nörtti wrote: Assume 6 cards, A having 1,3,4 and B having 2,5,6. If A leads 1, he loses all the tricks. If A leads 3 (or 4), B wins, and must lead at A:1,4, B:2,5, which guarantees A at least one trick. Very good: quite right. You're setting the cards up so B finesses himself. What I knew, but oversimplified, is that you do not want to be leading, but still you must take a majority of tricks to lead.
I will wager a guess. The optimal lead is your highest card that your opponent can beat. The optimal follow has been suggested by tall_walt: if you can beat a trick, play the lowest card to do so; if not play your lowest card.
As for the number of tricks you will win, it should be the number of your cards above n/2. However, if the weaker hand is forced to start the lead, this can be off by one trick. An example would be A has 1,2,4 and B has 3,5,6: if A leads, A will lose all the tricks, if B leads, A will win one. So I would play the first trick and then recalculate the number of tricks you will take now that the stronger hand is leading.
Edit: I can't spell.


