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Question for the maths experts
Posted:
Thu Nov 27, 2008 10:18 amby ahunda
If I have a shot at someone in an Escalating game, with a 50 % success chance on 2 fronts each: What is my overall success chance for the 2 fronts combined ? 50 % ? Or 25 % ?
Re: Question for the maths experts
Posted:
Thu Nov 27, 2008 10:21 amby White Moose
Was a couple of years ago i had stuff like that in School, but i'm quite sure it's 25% chance.
Quite funny though, i did math like this when i was 16... and i sure ain't a math expert..
Re: Question for the maths experts
Posted:
Thu Nov 27, 2008 10:29 amby RashidJelzin
Yeah that's easy...
you multiply both strings of possibilities, since it's 2 different fronts and they're not dependent on each other.
So it's (50/100) x (50/100) = (2500/10000) = (25/100) = 25%
Well that's how I remember it... and I'm pretty sure I got that right.
If I dare advise you... don't take 50% shots. That's suicide.
Re: Question for the maths experts
Posted:
Thu Nov 27, 2008 10:32 amby White Moose
RashidJelzin wrote:Yeah that's easy...
you multiply both strings of possibilities, since it's 2 different fronts and they're not dependent on each other.
So it's (50/100) x (50/100) = (2500/10000) = (25/100) = 25%
Well that's how I remember it... and I'm pretty sure I got that right.
If I dare advise you... don't take 50% shots. That's suicide.
Yeah, thats it.
Re: Question for the maths experts
Posted:
Thu Nov 27, 2008 10:38 amby ahunda
RashidJelzin wrote:If I dare advise you... don't take 50% shots. That's suicide.
I usually don´t. But if I have a 50 % overall shot at a game, that I´ll almost certainly lose, if I don´t take said shot, I might go for it ...
[EDIT] And cheers for the answer, of course.
Re: Question for the maths experts
Posted:
Thu Nov 27, 2008 6:28 pmby whitestazn88
technically the two are separate systems, so rashid jezens math is correct. if you had the opportunity to take out 2 separate players, and said you had a 50% chance on each, the chance of you defeating any single player is 50% though.
Re: Question for the maths experts
Posted:
Fri Nov 28, 2008 8:45 amby hecter
Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
Re: Question for the maths experts
Posted:
Fri Nov 28, 2008 11:28 amby White Moose
hecter wrote:Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
By your formula, then you have 50% chance to toss 1000 coins and always get heads. Which i really don't think is correct
Re: Question for the maths experts
Posted:
Fri Nov 28, 2008 4:55 pmby hecter
White Moose wrote:hecter wrote:Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
By your formula, then you have 50% chance to toss 1000 coins and always get heads. Which i really don't think is correct
No, but that's different. You're looking at the same possible thing happening multiple times, whereas we're only looking at multiple possible things happening once. If he had said "I have a 70% chance of winning one way with X armies and an 80% chance of winning another way with X armies, what's my probability of winning if I split X armies and go both ways?" I'd say it would be 75% chance. If he had said "If I run a simulation 10, each with a 75% probability of winning, what's the likelihood that I'll win every time?",
then I'd say about 5.6%.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 5:05 amby White Moose
hecter wrote:White Moose wrote:hecter wrote:Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
By your formula, then you have 50% chance to toss 1000 coins and always get heads. Which i really don't think is correct
No, but that's different.
It's not.
By your forumla... then this would happen... 50% at 10 diffrent places at once. (50%+50%+50%+50%+50%+50%+50%+50%+50%+50%)/10 = 50%
You are just calculating the average number by doing that.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 5:12 amby BaldAdonis
hecter wrote:If he had said "If I run a simulation 10, each with a 75% probability of winning, what's the likelihood that I'll win every time?", then I'd say about 5.6%.
Ay, there's the rub. In order to eliminate the player, he will have to win every time (ie win on both fronts). If he only had to win on one, then like whitestazn mentioned, he'd have a 50% chance of making it. But he needs to do both. It's like running two simulations, each with 50% probability.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 5:32 amby Stroop
hecter wrote:White Moose wrote:hecter wrote:Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
By your formula, then you have 50% chance to toss 1000 coins and always get heads. Which i really don't think is correct
No, but that's different. You're looking at the same possible thing happening multiple times, whereas we're only looking at multiple possible things happening once. If he had said "I have a 70% chance of winning one way with X armies and an 80% chance of winning another way with X armies, what's my probability of winning if I split X armies and go both ways?" I'd say it would be 75% chance. If he had said "If I run a simulation 10, each with a 75% probability of winning, what's the likelihood that I'll win every time?",
then I'd say about 5.6%.
It's exactly the same. Trying to eliminate an opponent has two possible outcomes, failure or success, and tossing a coin has two possible out comes, heads or tails. The only thing that's different is that the chances of taking out an opponent might not be 50%, but the maths work the same.
As mentioned, what you are doing is taking an average. If you take a shot at two opponents, with the chances of success being 80% and 70%, you have a 38% chance of eliminating one of them, and a 56% chance of eliminating them both.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 6:37 amby MrBenn
Stroop wrote:As mentioned, what you are doing is taking an average. If you take a shot at two opponents, with the chances of success being 80% and 70%, you have a 38% chance of eliminating one of them, and a 56% chance of eliminating them both.
I'm not too sure about that... the probability of beating one opponent would be greater than that of beating two...
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 7:00 amby Stroop
MrBenn wrote:Stroop wrote:As mentioned, what you are doing is taking an average. If you take a shot at two opponents, with the chances of success being 80% and 70%, you have a 38% chance of eliminating one of them, and a 56% chance of eliminating them both.
I'm not too sure about that... the probability of beating one opponent would be greater than that of beating two...
The probability of defeating at least one opponent is greater (94%), the probability of beating exactly one isn't in this case.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 7:16 amby Cundy
hecter wrote:Ya, I think it's all your chances added together divided by the number of chances. So you have two shots at 50%, that's (50%+50%)/2, which is 50%.
thats (.5)^2 = .25
its just the probability to the power of how many times. thats common sense but if you want to know. it comes from the binomial distribution formula.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 10:13 amby hecter
White Moose wrote:By your forumla... then this would happen... 50% at 10 diffrent places at once. (50%+50%+50%+50%+50%+50%+50%+50%+50%+50%)/10 = 50%
You are just calculating the average number by doing that.
You're right, that doesn't look correct. So...
I dug out some formula's...
http://www.easycalculation.com/statisti ... bility.phpStroop wrote:If you take a shot at two opponents, with the chances of success being 80% and 70%, you have a 38% chance of eliminating one of them, and a 56% chance of eliminating them both.
No, the probability of either of them happening, by the formula at the bottom, is 94%, 56% of both of them happening. Going back to the original question... If he only had to succeed in one of the directions, the probability of success would be 75%, 25% if he only had to win both.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 10:43 amby Stroop
hecter wrote:Stroop wrote:If you take a shot at two opponents, with the chances of success being 80% and 70%, you have a 38% chance of eliminating one of them, and a 56% chance of eliminating them both.
No, the probability of either of them happening, by the formula at the bottom, is 94%, 56% of both of them happening. Going back to the original question... If he only had to succeed in one of the directions, the probability of success would be 75%, 25% if he only had to win both.
I said 38% to eliminate one of them, not at least one of them... When I say one, I mean exactly one, not one or more...
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 10:55 amby knubbel
How do you guys calculate the propability of killing an enemy? Do you have a programm? I always just try to guess, but sometimes I would like to calculate it exact.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 11:04 amby jpcloet
It's 25%, however, consideration must be made to acquisition of cards, points for terminator etc.....
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 2:11 pmby lancehoch
hecter, stroop is right. If you want to take the odds of eliminating one or the other, but not both (xor/exclusively or) then you take the odds of eliminating one or the other or both (94%) and subtract the odds of eliminating both (56%) and you get the exclusive value (94%-56%=38%).
The 38% is also the sum of .7*.2+.8*.3=.14+.24=.38=38% where .7 is the success of A, .3 is the failure of A, .8 is the success of B, .2 is the failure of B.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 2:33 pmby BaldAdonis
knubbel wrote:How do you guys calculate the propability of killing an enemy? Do you have a programm? I always just try to guess, but sometimes I would like to calculate it exact.
http://gamesbyemail.com/Games/Gambit/BattleOdds
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 4:09 pmby hammocksleeper
I think there have been some gross misunderstandings about what the OP is asking. Let's read it again:
ahunda wrote:If I have a shot at someone in an Escalating game, with a 50 % success chance on 2 fronts each: What is my overall success chance for the 2 fronts combined ? 50 % ? Or 25 % ?
Everyone who has said 25% doesn't understand the question. Likewise, those people who thought he was attacking two different enemies. Let's set up his question with a concrete example so everyone can understand it.
Classic map. Red is on Russia and China. Green (the SINGLE enemy) is on Japan. Red has enough armies on Russia such that Russia->Japan has 50% chance of winning, and also has enough armies on China such that China->Japan has 50% chance of winning. Now the question is, if Red goes at Japan from BOTH directions, what is his probability of winning. Get it?
At least that's how I understood the question, I could be wrong. The OP hasn't posted in here yet. Also you could easily expand "Japan" to mean "a string of 10 countries that you want to sweep" but I simplified it for simplicity's sake.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 4:15 pmby Stroop
hammocksleeper wrote:I think there have been some gross misunderstandings about what the OP is asking. Let's read it again:
ahunda wrote:If I have a shot at someone in an Escalating game, with a 50 % success chance on 2 fronts each: What is my overall success chance for the 2 fronts combined ? 50 % ? Or 25 % ?
Everyone who has said 25% doesn't understand the question. Likewise, those people who thought he was attacking two different enemies. Let's set up his question with a concrete example so everyone can understand it.
Classic map. Red is on Russia and China. Green (the SINGLE enemy) is on Japan. Red has enough armies on Russia such that Russia->Japan has 50% chance of winning, and also has enough armies on China such that China->Japan has 50% chance of winning. Now the question is, if Red goes at Japan from BOTH directions, what is his probability of winning. Get it?
At least that's how I understood the question, I could be wrong. The OP hasn't posted in here yet. Also you could easily expand "Japan" to mean "a string of 10 countries that you want to sweep" but I simplified it for simplicity's sake.
If this is indeed the case, you're looking at a chance of 75%.
Either you make it on your first shot at Japan, which is a 50% chance, or you fail the first and make the second, which is a 25% chance.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 4:18 pmby lancehoch
I agree with the 75%: .5 that you win on the first front, if you fail (.5) you still have a .5 chance of success on the second front: .5+.5*.5=.5+.25=.75=75%
On a side note, it would probably be greater, due to the fact that you are not likely to only have a 50% chance on the second front, it will be greater than 50% if you eliminate even one opposing army.
Re: Question for the maths experts
Posted:
Sat Nov 29, 2008 4:29 pmby Geger
BaldAdonis wrote:knubbel wrote:How do you guys calculate the propability of killing an enemy? Do you have a programm? I always just try to guess, but sometimes I would like to calculate it exact.
http://gamesbyemail.com/Games/Gambit/BattleOdds
Thanks for the link
