The Gaming Den

For the hundredth time: Probabilities depend on what you already know.

The odds of rolling two threes in a row are low. The odds of rolling a second three, when the first one is already known, is substantially higher (the same as your odds of rolling the first).

The odds of this actually meaning that your actual number of "two 3s in a row" has anything to do with the theoretical mathematical possibility of "two 3s in a row" abstractly = zero.

It's a 50% chance of surviving the second encounter, conditional on your seeing the second encounter, so a total 25% chance of surviving it before the first encounter.

And if you don't survive the first encounter, the probability of -having- a second encounter is zero. (barring resurrection). So this "total of 25%" doesn't actually mean anything.

I think you don't like the probability of a loss occuring, in which case you should simply deny chance for those situations.

No, I don't like people using math in ways that basically add up to statistics saying one thing and whatever reality you play with making those meaningless but interesting hypotheticals.

Until you win one battle, you have zero chance of winning two in a row.

If it's a cute benefit you can never guarantee like picking up the Quad Damage, then people don't care much about it, but if it's something you can reliably get, especially by, say, sleeping 8 hours, you now will be observing quite a few people patterning themselves around that so they have the resource always.

And once again the instinct to be a gamer first and put roleplaying what the character would do a distant second rears its ruggedly handsome* head.

Dude - you can either write a story or make a game.

If you are going to write a story then who frickin cares what the math is - there is no math, it's just the story you are writing.

If you are making a game then you need rules. THe assumption here is that there will be some random number generator used to determine the success or failure of actions... combat being the primary action in your proposed game.

You have 2 stated goals:

1) About 40 combats in the campaign.

2) I would like it to be lethal.

Do you want to make the 40 encounter campaign and then kill the players in the first or 2nd encounter? Is that really what you want to do - I mean, that's lethal for sure. And if each encounter is about as lethal as tails I live, heads I die then the most likely result is campaign over by encounter 2.

OR - would you like it to be lethal so that only about half of the PC's survive the entire 40 encounter campaign... in that case you want to make each encounter have around a 1% or 2% chance of killing the PCs.

This is a very, very basic question that will inform the entire mechanics of your system. AND YOU WONT ANSWER IT - so what are you trying to do?

Elennsar wrote:

If it's a cute benefit you can never guarantee like picking up the Quad Damage, then people don't care much about it, but if it's something you can reliably get, especially by, say, sleeping 8 hours, you now will be observing quite a few people patterning themselves around that so they have the resource always.

And once again the instinct to be a gamer first and put roleplaying what the character would do a distant second rears its ruggedly handsome* head.

If you don't like games where people intensely game the resources, then make a game where the resource cannot be intensely gamed. To do this, you need to understand how the probabilities of your game work, but I can pretty much already tell you, anything that significantly increases survivability has to be either always available or never available in order to have the "bravery" you so desire be commonplace.

1) About 40 combats in the campaign.

2) I would like it to be lethal.

40 battles if it goes as "planned" - depending on what you do, you might be able to reduce that or they might increase, but that's the estimate.

This is a very, very basic question that will inform the entire mechanics of your system. AND YOU WONT ANSWER IT - so what are you trying to do?

What bloody part of "it depends on what you do" is sneaking into evasiveness? If you treat rank and file opponents as capable of killing you and exercise a moderate amount of prudence, they should fear YOU (numbers or not).

If you don't, you -will- get gangfucked into death.

As for a 1-2% chance - yes, by the totally nonapplicable equation that utterly ignores that you don't -have- a battle two if you don't survive battle 1 so you might well be in more danger than that AND OVERCOME IT.

If you don't like games where people intensely game the resources, then make a game where the resource cannot be intensely gamed.

Fuck that. Seriously. If you're incapable of playing a goddamn roleplaying game without intensely gaming the resources whether that involves doing things that are appropriate or not, then you fething suck at roleplaying.

Personally, I'm not even sure how I intend to hand it out - I'd like to have it always available to some extent in any major encounter (said forty battles, for instance), but that could be "you always have at least 1 at the start of an encounter".

Elennsar wrote:Fuck that. Seriously. If you're incapable of playing a goddamn roleplaying game without intensely gaming the resources whether that involves doing things that are appropriate or not, then you fething suck at roleplaying.

You cannot seriously tell me you have never played a warrior who has retreated when wounded, a wizard who has rested when he needs more MP (or whatever), or a rogue who moved to flank an opponent before attacking. You cannot seriously tell me you have never assented to the party stopping because the cleric (or whatever) was out of healing spells. You cannot seriously tell me that you have never been in a party that chose to ambush instead of attack head-on. You cannot tell me that there has never been moments in your gaming history where you have sought to pull together the advantages to win a fight.

The entire procedure of gaming is manipulating the resources you have in such a manner as to achieve the desired outcome. Roleplaying is merely doing so in a manner that coincides with your character's behaviours. While it is understandable that you want to have a high-risk, high-reward system, you cannot create this while pretending you are blind to the basics of how the system operates. Given that you have apparently chosen blindness, I do not see how anyone can help you.

Good day, Sir.

No, what I have chosen is that if wounded but it is important to press on or my character would be confident enough to press on (whether or not I would be), my character will press on.

Your character does not know that if he eats gobstoppers that the improbable will happen more often than it is supposed to. Playing him as if he does is bad.

Elennsar wrote:Giant Frog.

Your right El - the frog is quite giant.

spasheridan wrote:

Elennsar wrote:Giant Frog.

Your right El - the frog is quite giant.

But is it made of Chocolate? Because Wizards love those.

Your right El - the frog is quite giant.

If you don't believe that what you do is going to have a huge role in what happens to you, and that therefore pointing that out is legitimate, then say so and stop sniping.

Otherwise, deal with the fact that I don't -know- whether "shield block" risks your shield arm (even if you block) or not, because I don't know if that would be a good thing or not.

Elennsar wrote:Giant Chocolate Frog

I already agreed that the frog is quite giant, and apparently it is also made up of chocolate.

I eagerly await more information about how giant the frog is - please post it when you have thought on your frog long enough to be able to talk about how chocolate and giant it is.

Elennsar wrote:

For the hundredth time: Probabilities depend on what you already know.

The odds of rolling two threes in a row are low. The odds of rolling a second three, when the first one is already known, is substantially higher (the same as your odds of rolling the first).

The odds of this actually meaning that your actual number of "two 3s in a row" has anything to do with the theoretical mathematical possibility of "two 3s in a row" abstractly = ONE-HUNDERED PERCENT!!!!!.

Fixed it for you.

It's a 50% chance of surviving the second encounter, conditional on your seeing the second encounter, so a total 25% chance of surviving it before the first encounter.

And if you don't survive the first encounter, the probability of -having- a second encounter is zero. (barring resurrection). So this "total of 25%" doesn't actually mean anything.

wait what?! are we back to this? look man, it was funny once, but that time has long since passed. if it's really that hard for you, just think of it this way; If you don't have a second encounter, then your chance of winning the second encounter is zero. (barring Giant Frog). so the fact that dying in one encounter prevents a second actually reinforces the point you think it disproves.

I think you don't like the probability of a loss occuring, in which case you should simply deny chance for those situations.

No, I don't like people using math in ways that basically add up to statistics saying one thing and whatever reality you play with making those meaningless but interesting hypotheticals.

Until you win one battle, you have zero chance of winning two in a row.

good thing nobody has done that then.

seriously man, I want you to do something for me. go look in the mirror. do you see a small child? answer honestly

Fixed it for you.

No, missed my point.

wait what?! are we back to this? look man, it was funny once, but that time has long since passed. if it's really that hard for you, just think of it this way; If you don't have a second encounter, then your chance of winning the second encounter is zero. (barring Giant Frog). so the fact that dying in one encounter

We are back to this because you insist that somehow the first encounter hurts anything that anyone who has encounter #2 cares about simply because it happened.

It doesn't.

If you made it to encounter #2, the odds of you surviving encounter #1 don't matter anymore.

good thing nobody has done that then.

seriously man, I want you to do something for me. go look in the mirror. do you see a small child? answer honestly

No, bad thing that this has been exactly what they are doing. Because the fact that if you fail the first encounter then the second is irrelevant to your experience means that only the people who survive #1 are concerned with the chance of surviving both.

To put it another way, do the people who don't take part in the archery contest count as part of any calculation that someone will hit the bull's eye?

Because I never learn, I'm going to try explaining iterative probability to the spambot again.

The math doesn't care whether you are coming to the enemy, they're coming to you, you and the enemy group are both moving toward eachother and meet in the middle, or you're both following odd paths that happen to intersect at various points, so let's use a simple situation: the PCs are guarding a bridge, and there are a few groups of barbarians trying to cross it.

Each group of barbarians has a 1-in-10 chance of beating the PCs and crossing the bridge, and a 9-in-10 chance of being beaten by the PCs.

The barbarians' chances of crossing the bridge look like this:
First group has 10%
Second group has 19%: a 10% chance of finding the bridge unguarded, and a 10% chance of the remainder (so 9%) that they'll be able to cross.
Third group has 27.1%: 19% of the bridge being unguarded, 8.1% of forcing their way across).
Fourth group has 34.39%
Fifth group has 40.961%
and so on and so forth. If the barbarians send five groups at the PCs bridge in these circumstances, then there's about a 41% chance of the PCs having to regroup to fight the barbarians on the other side.

---

Also, he didn't miss your point. Your point is, quite simply, wrong. Theories of probability are directly applicable here. Yes, probabilities change as you get more information. However, over enough rolls, probabilities apply perfectly to the situations defined by the information used for the calculation.

If you made it to encounter #2, the odds of you surviving encounter #1 don't matter anymore.

You're missing the point here. Yes, the odds that you will face encounter 2 are 100% for all of the situations that do face encounter 2, but not for all of the DMs that plan an encounter 2 even with PCs that are perfectly railroaded.

The odds that you will make it past encounter N from encounter 1 must take into account the odds of even making it to encounter N. There will be players who don't make it to encounter N, and you have to take them into account.

To put it another way, do the people who don't take part in the archery contest count as part of any calculation that someone will hit the bull's eye?

Do the people who get eliminated in round 1 of a tournament not count when you're looking at the odds of winning? That's more like the situation here. The people not in the contest are more like the people who are off playing Warp Cult in the next room. A calculation of the odds of making it through a group of encounters must take into account the odds of getting through each one, not just the last one.

I'm just simplifying it to this to get this answered straight on and not twisted into "it is but it isn't":

Are the odds of me rolling a 4th 3 the odds of rolling one 3 times stuff, or are they the odds of rolling a 3?

Because if I roll a 4th 3 one time in two hundred and sixteen rolls, then the only thing that makes it less likely is not getting there to begin with, which is seperate from the odds of succeeding on the roll IF I do get there.

let me ask you this, do you, or do you not, believe that your odds of rolling a "fourth 3" (defined specifically as a 3 immediately precedded by 3 other 3s) is not better after you've rolled the first 3 3s?

I'm not being rhetorical, I've seen you alternate between blatantly not understanding probability and pretending to understand by repeating your misinterpretation of what we're trying to explain

I believe that my odds of rolling a 3 are one in two hundred and sixteen, assuming 3d6.

I believe that it is improbable that I will roll more or less 3s than one in two hundred and sixteen rolls.

Elennsar,

Let's say you have a 6 sided dice and a spinner with 3 colored sections of equal size, red, blue and yellow. Based on this information, please estimate, to the best of your ability, the probability of the following events:

1] The spinner comes up red and you roll a 6.

2] The spinner comes up blue and you roll an even number.

3] The spinner comes up green and you roll a 2.

4] The spinner comes up yellow and you get a number less than 6 OR the spinner comes up blue and you get a number more than 1.

5] You spin 3 times and the spinner comes up blue three times.

6] You spin 3 times and the spinner comes up blue, then yellow, then red.

7] The spiner produces at least 2 reds AND at least 1 blue over the course of 10 spins.

8] You roll a 1 on your first roll of the dice.

9] You roll a 1 on your next roll, even though your last roll was a 6.

10] You roll a 1 on your next roll, even though your last roll was a 1.

Grek: Please explain to the best of your ability what any of this has to do with me failing or making a given roll.

They are sample "rolls". I want to make sure we are parsing them the same way and arriving at the same results.

What I want to know is what these results are supposed to mean.

If I don't get a red, then the chance of getting a 6 doesn't matter for "red and 6" If I don't get a 6, same with getting a red.

For the 1st question, it's the chance of getting both a red and a 6. If you do not get red, you fail the roll. If you do not get 6, you fail the roll. If you get neither red nor 6, you fail the roll. You only succeed the roll if you get both a red and a 6.

Which means that until one condition is met, the chance of meeting both is 0.

Assume both the dice roll and the spin are done at the same time. What is the chance of getting both a red and a 6?

That's not the until I meant - unless you get X or Y, you are literally incapable of getting XY.

Allow me to rephrase. You are using an ability in a game. In order for the ability to succeed in producing the desired effect, you must spin a spinner with three colored section of equal size (red, yellow and blue) and get red. You must also roll a single d6 and get the number 6. If you do not get red or do not get 6, you fail to get the desired result this round. If you get both red and 6, you get the desired effect.

What is the probability of the ability producing the desired effect in 1 round consisting of 1 person making 1 attempt to use the ability?