9 minutes ago, Kaptin Krunch said:

say the bullet is in gun A, and after the first trigger pull, there are 3 different outcomes depending on what happened on the first pull.

- Gun A was pulled, if gun A is pulled again (1/3) it has a 1/5 chance to fire

- Gun B was pulled, if gun A is pulled(1/3), it has a 1/6 chance to fire.

- Gun C was pulled, if gun A is pulled(1/3), it has a 1/6 chance to fire.

This is a 1/3 chance of 1/15, and 2 instances of a 1/3 chance of 1/18. Taking the average of the three results in 4/67. This is basically the same, but sllightly different.

because the odds of firing only increase one third of the time (but increase more than 1/18 when they do so) the odds differ slightly.

Okay I think I get what you are saying. So to break down all of the probabilities for the following shots is more work than I actually want commit to.

11 minutes ago, Boom Owl said:

Merry Christmas everyone. Hope you enjoy doing some not X-Wing things today!

Obviously I am spending the day being reminded why I stopped taking math classes.

I feel mean letting the math folks squirm. I want to tell them.

I want to save them.

12 minutes ago, Velvetelvis said:

I feel mean letting the math folks squirm. I want to tell them.

I want to save them.

"On the battlefield, there's no such thing as luck"?

Do you just want to start quoting MGS at us?

Why are we still here? Just to suffer?

2 hours ago, Scott Pilgrim2 said:

EDIT2: I was a PoliSci major, so applying math wrong is almost a requirement for the field.

me too - the only real math I learned was through electives outside my major 😐

2 hours ago, Kaptin Krunch said:

This is close, but you have the odds of a single gun with 18 chambers.

He could theoretically pull 100 triggers and get no bullets.

but he's not. in the problem presented, he's taking 6 pulls of the trigger, which is guaranteed to not recycle any chambers, so it's just firing 6 of 18 chambers, one of which can generate a success, giving him a 1 in 3 chance of success by the end of the exercise (17/18 * 16/17 * 15/16 * 14/15 * 13/14 * 12/13 chance of failure, which is conveniently .666 repeating, but also works out to be the same as the back-of-the-handkerchief "six out of 18 is one third" math)

flip that 12/13 to 1/13 and you get the odds of that particular shot being the successful shot, at .0555 repeating.

31 minutes ago, skotothalamos said:

but he's not. in the problem presented, he's taking 6 pulls of the trigger, which is guaranteed to not recycle any chambers, so it's just firing 6 of 18 chambers, one of which can generate a success, giving him a 1 in 3 chance of success by the end of the exercise (17/18 * 16/17 * 15/16 * 14/15 * 13/14 * 12/13 chance of failure, which is conveniently .666 repeating, but also works out to be the same as the back-of-the-handkerchief "six out of 18 is one third" math)

flip that 12/13 to 1/13 and you get the odds of that particular shot being the successful shot, at .0555 repeating.

You oversimplified the math. The person could never select the gun with the bullet. He could pull the trigger 100 times never finding the right gun, let alone the right chamber.

4 minutes ago, LagJanson said:

You oversimplified the math. The person could never select the gun with the bullet. He could pull the trigger 100 times never finding the right gun, let alone the right chamber.

Here's the thing... You are only modeling the first six shots. You can create an accurate model for the first six that doesn't work for infinite shots.

25 minutes ago, Scott Pilgrim2 said:

Here's the thing... You are only modeling the first six shots. You can create an accurate model for the first six that doesn't work for infinite shots.

Yeah, I’m on board. As soon as the seventh time a trigger gets pulled this isn’t accurate (because you could hit a single chamber twice in two discrete outcomes) but because it is six chambers per gun and six pulls this should work.

The 1/18, 1/17 thing is *close* but not accurate.

If you want to see an example of it differentiating as fast as possible, do the odds of

- There are now 2 guns

- Each gun can now only hold 2 bullets

- One bullet in one gun, the other is empty.

The first shot is 0.25 to fire.

The second shot is a 0.375 to fire, not 0.333 (there was a 50% chance that the pull was in a no-bullets gun the previous time, so the odds are the average of 1/2 and 1/4)

to build on this, the odds of each pull will approach (1/(Guns)), not 100% with each shot, but they approach the number faster.

The third shot is a 75% chance of having there been a pull in the bullets gun previously, which means it's
- 50% of pulling no bullets gun
- If you do pull Bullets gun (50%)

- 25% chance of 50% chance

- 75% chance of confirmed bullet.

That's 50% chance of 87.5, or a 43.75% chance to have it come out on the third pull, given that there were no shots previously

If Mr. Ocelot continues to pull triggers in these 2 shot revolvers and no bullet is fired, the odds of there having been no pulls in a partially loaded gun approaches 0, which means it will approach 50%- a 50/50 odds of pulling the loaded gun with each shot.

Since we are already arguing about math, who wants to hear about a different riddle involving a car and a goat?

6 hours ago, Kaptin Krunch said:

The 1/18, 1/17 thing is *close* but not accurate.

If you want to see an example of it differentiating as fast as possible, do the odds of

- There are now 2 guns

- Each gun can now only hold 2 bullets

- One bullet in one gun, the other is empty.

The first shot is 0.25 to fire.

The second shot is a 0.375 to fire, not 0.333 (there was a 50% chance that the pull was in a no-bullets gun the previous time, so the odds are the average of 1/2 and 1/4)

I get it now, but I think that for most purposes close is good enough. Knowing your expected variance within 5% is good enough for X-Wing. It's a fun mental exercise and an interesting way to evaluate how to set up a problem, but I would be hard pressed to think of an example where knowing the exact percentage would change a decision.

Just now, Scott Pilgrim2 said:

I get it now, but I think that for most purposes close is good enough. Knowing your expected variance within 5% is good enough for X-Wing. It's a fun mental exercise and an interesting way to evaluate how to set up a problem, but I would be hard pressed to think of an example where knowing the exact percentage would change a decision.

The secret is that it doesn't matter what the answer is. What matters is learning how the math works to produce the correct answer.

I'll think of some way to write this up for X-wing that isn't simple enough to be plugged into the dice calc.

Unless people want to hear more unrelated-to-x-wing stuff.

I’m trying out a new way of looking at squads. I’m adding up the points spent on chassis (at lowest to get it on the board) compared to points spent upgrading pilots v buying upgrades, and bid. For example:

squad of legend.

Chassis: 114

Pilots: 34

Upgrades: 35 + Soontir’s Talent

Bid: 17 - Soontir’s Talent

In theory, the more points you spend away from chassis the more you have to combo or leverage in flying to (on average) beat a pure efficiency list.

It might not be fully refined, but maybe it’ll help some of my locals that insist on listbuilding something fresh every time get a new perspective on things.

If Soontir never arc dodges or initiative kills anyone, if redline doesn’t get a torpedo off, and if Whisper doesn’t use Vader or those extra evades, you are only getting ~55% of your squad worth. But if your seven TIE Swarm never stays in the bubble so howl and Iden don’t do anything, you still get 161 points worth of TIE chassis.

47 minutes ago, AEIllingworth said:

I’m trying out a new way of looking at squads. I’m adding up the points spent on chassis (at lowest to get it on the board) compared to points spent upgrading pilots v buying upgrades, and bid. For example:

squad of legend.

Chassis: 114

Pilots: 34

Upgrades: 35 + Soontir’s Talent

Bid: 17 - Soontir’s Talent

In theory, the more points you spend away from chassis the more you have to combo or leverage in flying to (on average) beat a pure efficiency list.

It might not be fully refined, but maybe it’ll help some of my locals that insist on listbuilding something fresh every time get a new perspective on things.

If Soontir never arc dodges or initiative kills anyone, if redline doesn’t get a torpedo off, and if Whisper doesn’t use Vader or those extra evades, you are only getting ~55% of your squad worth. But if your seven TIE Swarm never stays in the bubble so howl and Iden don’t do anything, you still get 161 points worth of TIE chassis.

Have you heard my rants on Words being the strongest thing in x-wing?

Because it applies here.

9 hours ago, LagJanson said:

You oversimplified the math. The person could never select the gun with the bullet. He could pull the trigger 100 times never finding the right gun, let alone the right chamber.

sure, but he only asked for a series of 6 shots, where that doesn't matter.

If he'd said the ocelot fired 7 or more shots, then the vague math I remember from getting my liberal arts education is useless, but he didn't so i'm fine.

Must......not......tell....them...

I'm not the only one that sees this right?

Has nobody else lost entire collections of vintage single action firearms trying to get across a lake somewhere?

3 hours ago, Velvetelvis said:

Has nobody else lost entire collections of vintage single action firearms trying to get across a lake somewhere?

Dude, I’m Canadian, eh? I’ve only fired an old rifle in cadets decades ago. You’re lucky I didn’t think these were battery powered and lobbed potatoes.

6 hours ago, skotothalamos said:

sure, but he only asked for a series of 6 shots, where that doesn't matter.

If he'd said the ocelot fired 7 or more shots, then the vague math I remember from getting my liberal arts education is useless, but he didn't so i'm fine.

It does produce a different result, even with 6 or fewer shots.

15 hours ago, Kaptin Krunch said:

to build on this, the odds of each pull will approach (1/(Guns)), not 100% with each shot, but they approach the number faster.

The third shot is a 75% chance of having there been a pull in the bullets gun previously, which means it's
- 50% of pulling no bullets gun
- If you do pull Bullets gun (50%)

- 25% chance of 50% chance

- 75% chance of confirmed bullet.

That's 50% chance of 87.5, or a 43.75% chance to have it come out on the third pull, given that there were no shots previously

If Mr. Ocelot continues to pull triggers in these 2 shot revolvers and no bullet is fired, the odds of there having been no pulls in a partially loaded gun approaches 0, which means it will approach 50%- a 50/50 odds of pulling the loaded gun with each shot.

Since we are already arguing about math, who wants to hear about a different riddle involving a car and a goat?

The odds are 50/50, since he either pulls the trigger on a loaded barrel, or he doesn't

On 12/23/2018 at 10:44 PM, Kaptin Krunch said:

Revolver Ocelot takes 3 colt single action revolvers, and loads a single bullet into one of them and spins the cylinder (all 3 were unloaded prior). He then begins juggling the three guns, until he has lost track of which one is which.

Revolver will pull 6 triggers at random from the three guns.

What are the odds that the bullet is fired? What are the odds that the bullet is fired on exactly the 6th trigger pull?

For those in bongland/don't know how guns work, the gun in question holds 6 bullets, and rotates through the chambers sequentially.

I’m assuming @Velvetelvis is saying the answer is 0 since a single action revolver doesn’t fire or rotate the cylinder unless you manually cock the hammer each time?

2 hours ago, Transmogrifier said:

I’m assuming @Velvetelvis is saying the answer is 0 since a single action revolver doesn’t fire or rotate the cylinder unless you manually cock the hammer each time?

Yeah, the secret is despite being a huge inverse-weeb, Kojima doesn't actually know how guns work. He got better at figuring it out as the series went on.

1 minute ago, Kaptin Krunch said:

Yeah, the secret is despite being a huge inverse-weeb, Kojima doesn't actually know how guns work. He got better at figuring it out as the series went on.

That's pretty much an industry-wide thing though - very few video game developers bother with going through the effort to have accurate animations etc. Ironically, AAA studios seem to care the least even though they have the most available resources.

9 minutes ago, Transmogrifier said:

That's pretty much an industry-wide thing though - very few video game developers bother with going through the effort to have accurate animations etc. Ironically, AAA studios seem to care the least even though they have the most available resources.

A few years ago I played Wasteland 2. One of the first heavy weapons you get it an M2 heavy machine gun, modified to fire 5.56 mm ammunition. Talk about a waste of technical expertise.

As this thread is in the main X-Wing forum, please bring the topic back to X-Wing. Thank you.