Email configuring frustration. What I want is this:

A family email account (call it [email protected] gmail.com) we can use for kid program registration etc. It receives and then forwards on messages to my wife's and my own email inboxes (so that we don't have to remember to check the new inbox). We can reply to the emails from our own inboxes, but as [email protected] gmail.com.

All of this, I can do. Here's the clincher:

I want to set some kind of configuration so that if I reply to that email, my wife is aware that the email has been dealt with. You know, automatically.

I can't, for the life of me, figure out if this is possible. We both use gmail for our inboxes, so that's the system we need. I am aware that we each can cc or bcc the reply to the other, but that requires paying attention and remembering to do it. Is there some kind of outgoing rule that I can set up in gmail, so that "If [email protected] gmail.com then cc:= [email protected] gmail.com?" Or is there a better way to manage this that I don't know about?

Email configuring frustration. What I want is this:

A family email account (call it FeegleFa[email protected] gmail.com) we can use for kid program registration etc. It receives and then forwards on messages to my wife's and my own email inboxes (so that we don't have to remember to check the new inbox). We can reply to the emails from our own inboxes, but as [email protected] gmail.com.

All of this, I can do. Here's the clincher:

I want to set some kind of configuration so that if I reply to that email, my wife is aware that the email has been dealt with. You know, automatically.

I can't, for the life of me, figure out if this is possible. We both use gmail for our inboxes, so that's the system we need. I am aware that we each can cc or bcc the reply to the other, but that requires paying attention and remembering to do it. Is there some kind of outgoing rule that I can set up in gmail, so that "If [email protected] gmail.com then cc:= [email protected] gmail.com?" Or is there a better way to manage this that I don't know about?

You can add the family account as an additional inbox to your main account. You can do this with both the web version and the phone apps.

You can add the family account as an additional inbox to your main account. You can do this with both the web version and the phone apps.

Can you clarify what you mean? The way I do it now, I have FeegleFamily forwarding mail to my account, and then add a "Send Mail As:" option to allow me to send as FeegleFamily. Is that what you mean?

My experiments don't suggest that setting that up will inform my wife when I've sent an email. Am I missing a configuration option?

Or are you suggesting having multiple gmail accounts "logged in" to the same device/app? I'd prefer to avoid that, if I can come up with a different option.

Math fail time!

Say there are 9 siblings, each of whom has a 1-in-3 chance of inheriting a particular trait.

What's the formula for figuring out the probability that 3 siblings has that trait?

Is it (1/3)^3 . (2/3)^6 . #combinations

Where # combinations is given by n! / (n-k)!k!

Rykin wrote:You can add the family account as an additional inbox to your main account. You can do this with both the web version and the phone apps.

Can you clarify what you mean? The way I do it now, I have FeegleFamily forwarding mail to my account, and then add a "Send Mail As:" option to allow me to send as FeegleFamily. Is that what you mean?

My experiments don't suggest that setting that up will inform my wife when I've sent an email. Am I missing a configuration option?

Or are you suggesting having multiple gmail accounts "logged in" to the same device/app? I'd prefer to avoid that, if I can come up with a different option.

I was talking about the second option. On the iOS app it does combined inboxes and I thought there was a way to do that on the web as well but I can't find anything about it other than what you have setup by forwarding. Your best bet is probably to use the BCC to let your partner know it has been handled and so they can know how you handled it.

Math fail time!

Say there are 9 siblings, each of whom has a 1-in-3 chance of inheriting a particular trait.

What's the formula for figuring out the probability that 3 siblings has that trait?

Is it (1/3)^3 . (2/3)^6 . #combinations

Where # combinations is given by n! / (n-k)!k!

Is that the binomial formula? If so then yes assuming you want exactly 3 trial successes (siblings get the trait).

Jonman wrote:Math fail time!

Say there are 9 siblings, each of whom has a 1-in-3 chance of inheriting a particular trait.

What's the formula for figuring out the probability that 3 siblings has that trait?

Is it (1/3)^3 . (2/3)^6 . #combinations

Where # combinations is given by n! / (n-k)!k!

Is that the binomial formula? If so then yes assuming you want exactly 3 trial successes (siblings get the trait).

The combination equation I learned during high school maths bit on "permutations and combinations", which I entirely had to refresh myself about courtesy of Professor Google.

Do you want exactly three siblings to have that trait? Or at least 3 siblings? Any any three siblings, or the first 3?

I need to ponder this for a bit, but I don't think you need the # of combinations.

The number of combinations factor made me think the goal is to have exactly 3, but you don't care which.

Jonman wrote:Math fail time!

Say there are 9 siblings, each of whom has a 1-in-3 chance of inheriting a particular trait.

What's the formula for figuring out the probability that 3 siblings has that trait?

Is it (1/3)^3 . (2/3)^6 . #combinations

Where # combinations is given by n! / (n-k)!k!

Is that the binomial formula? If so then yes assuming you want exactly 3 trial successes (siblings get the trait).

Gravity got it. And Jonman's instinct was right.

We're looking for 3 successes on 9 trials with p = 1/3. The formula is nCx * p^x * (1-p)^(n-x).

9C3 * (1/3)^3 * (2/3)^6 = 0.273

So, 27.3%.

Thanks math wizards.

Thanks

~~math wizards.~~

Mathemagicians!

Do you want exactly three siblings to have that trait? Or at least 3 siblings? Any any three siblings, or the first 3?

To expand on this, the 27.3% was for *exactly* 3 successes (i.e., siblings having the trait). If you wanted *at least* 3 or *at most* 3 you'd expand the formula for the other appropriate cases and sum together. So, P(x <= 3) = P(x = 0) + P(x =1) + P(x = 2) + P(x =3).

Thanks math wizards.

I'll admit you nerdsniped me a bit. Putting my four stats classes to some use!

Jonman wrote:Thanks

~~math wizards.~~

~~Mathemagicians!~~

NERDS!

NERDS! NERDS! NERDS! NERDS! NERDS!

ONE OF US. ONE OF US. GOOBLE GOBBLE, GOOBLE GOBBLE!

Tanglebones wrote:Jonman wrote:Thanks

~~math wizards.~~

~~Mathemagicians!~~NERDS!

Okay, here's what I've always wondered. When a poll has a margin of error, say +/- 5%. What is the 5% added to or subtracted from. Let's say candidate A is a 55% to 45% favorite over candidate B. Is the margin of error added to the difference - so instead of a 10 point favorite A is a 5-15 point favorite? Is it applied to either or both of the candidates polling number - so that candidate A is really polling 50%-60% and candidate B is at 40%-50% (i.e. it's potentially a dead heat)?

If the margin of error is +/- 5%, then wouldn’t candidate A be a 50% favorite over candidate B? The range of 45%-55% simply reflects the margin of error as it is applied to candidate A. Right?

Don’t listen to me. I’m a lay person.

Clumber, I think it's your second one -- that the margin of error is applied to either / both of the candidates' polling number. But again, I'm a lay person here.

Okay, here's what I've always wondered. When a poll has a margin of error, say +/- 5%. What is the 5% added to or subtracted from. Let's say candidate A is a 55% to 45% favorite over candidate B. Is the margin of error added to the difference - so instead of a 10 point favorite A is a 5-15 point favorite? Is it applied to either or both of the candidates polling number - so that candidate A is really polling 50%-60% and candidate B is at 40%-50% (i.e. it's potentially a dead heat)?

(With apologies in advance to any real statisticians present who are wincing at the oversimplifications I'm going to make here.)

For a simple two-choice situation like that, the two values are not independent; B's vote is always 100% minus A's vote. In a case like this you can think of the margin of error as applying to the point along the line (from 0 to 100) where the division is drawn. In reality, though, there are nearly always at least three possible results - there were probably a certain proportion of "don't know" answers in that poll; the real results would have looked something like, say, 45% A, 35% B, 20% Don't Know. The relationship between the errors in that case gets complicated, but to a very rough approximation you can think of the margin of error as applying independently to each of A and B. This becomes closer to the truth the more choices there were.

But it's important to keep in mind that the so-called "margin of error" isn't some kind of absolute limit: it's a standard deviation. If candidate A gets 55% "with a 5% margin of error", that doesn't mean their actual popularity is definitely between 50% and 60%. When something follows a normal distribution (as this sort of thing usually does), the rule of thumb is that you have 2/3 confidence that the result was off by less than one standard deviation (50-60%), 95% confidence that it's within two SDs (45-65%), and 99% confidence that it's within three SDs (40-70%).

Another way of looking at it is to think about the probability that the real outcome will be different to what the survey suggested. (I'm cheating a bit here because, in reality, the final election represents the real underlying distribution and the sample's outcome had a random distribution, not the other way around, but we can use this as a rough approximation.) In your original example, the result would be reversed (B would win) if the survey was wrong by more than 5% (one SD) in one direction. The probability of that is about 16%.

In my three-way (A/B/don't know) example, we can get a rough (i.e. even rougher) answer by treating each candidate's popularity as an independent normal variable. The difference between two normally distributed variables with the same SD has an SD equal to sqrt(2) times the SD of each one, so the difference between A's vote and B's vote has (very approximately) a mean of 10 (=45-35) and a standard deviation of 7.1 (=sqrt(2)x5). For B to win requires this difference to go below 0, i.e. to be off by at least 1.4 SDs. The probability of this is about 8%.

Crowbar, also a layman here but I think your numbers are a ways off. When a poll reports a margin of error it's not a standard deviation, it's typically something like a 95 confidence interval (so more like two SDs). Also poll results aren't necessarily normally distributed (for numbers near 50% they may be, but higher or lower values obviously won't).

I'll write up a fuller answer to Clumber's question.

When a poll has a margin of error, say +/- 5%. What is the 5% added to or subtracted from.

The "I don't care about statistics" answer:

- You add margin of error to each result, so if the poll says "55% to 45% with a 5% margin of error", then margin of error runs from "60-40" to "50-50".

The "I care a little about statistics" answer:

- Usually for something like this the margin of error represents a confidence interval. Often (though not always) polls use a 95% confidence interval, which
*approximately*means "we're 95% confident that the poll results are correct within the margin of error".As such, you could squint and say "well if the poll uses a 95% CI then there's a 5% chance it's wrong, and so a 2.5% chance it's wrong in a particular direction. So it's 2.5% likely that candidate B is actually more popular than A". (None of that is technically correct, but it's good enough for spherical cows.)

The "I care a LOT about statistics" answer, spoilered because nobody cares a lot about statistics:

The precise meaning of a confidence interval is surprisingly unintuitive. If you see a poll that says "50% of americans like cheese, plus or minus 5%" or whatever, and you want to know how accurate it is, it's natural to ask a question like "Okay, so what is the probability that the

actualpercentage is between 45% and 55%?"However, according to traditional statistics (i.e. the kind most of us learn in school), that question is technically unanswerable. Traditional statistics says: "the

actualpercentage is a fixed number - it has no uncertainty or distribution, so one cannot meaningfully talk about the probability of it being this or that. The only thing we can talk about is the probability of whether the random group of people we polled was representative of the population."As such, traditional statistics says "if this poll was repeated lots of times, and we calculated a confidence interval for each set of results, on average 95% of those intervals would included the true value". But that's subtly different from saying "we're 95% confident that the true value is within this particular interval".

There is, incidentally, a whole other school of statistics (called Bayesianism) which is perfectly happy to answer questions about the probability of the true value being between X and Y. The difference is that where traditional ("Frequentist") statistics considers the true value to be fixed and the sampled data to be variable, Bayesian analysis considers the evidence (the poll data) to be fixed and the true value (the actual percentage of people who like cheese) to be variable. The distinction is subtle, but it turns out to be important in various contexts, e.g. in scientific research when you're trying to determine how big a correlation should be considered significant.

But I guess that's enough detail for now

Anybody using EA Access? They're launching it for PSN and I'm looking for advice to see if it's worth it.

I did use it on Xbox.

It's fine if you have an interest in EA's back catalog, although that part is probably hamstrung a bit on PS4 because they can't offer PS3-era games. Personally, without much interest in sports games, there wasn't a lot there for me, but your mileage might vary.

Now, if they changed EA Access and made it more like Origin Access, which offers a much broader catalog including third-party games, I'd sign up. Or if they made it more like Origin Access Premier, which gives you access to EA games when they release, I'd sign up.

But as it is? It seems even less valuable on PS4 than it is on Xbox One, and I don't feel like it's a great value on Xbox.

I tend to sign up for a single month of EA Access right before they release a new game that interests me (i.e. a new Bioware game). I do this because it lets you play the game sooner and they discount the price of the game to the point that the discount price plus the one month subscription price is about the same as the full game price. But really I don't get the point of the single publisher subscription services. There is not a single one of them that releases enough games per year to make it worth it IMHO. I think a lot of them are going to be disappointed when a lot of people just end up signing up for a month or two instead of paying full price for their games or being a continuous subscriber.

Anybody using EA Access? They're launching it for PSN and I'm looking for advice to see if it's worth it.

What you really want is Game Pass. Oh wait. That’s an Xbox exclusive. Sorry bro.

That’s the first time I’ve ever been able to gloat about owning an Xbox rather than PS4.

What are some good, cheap(ish) house warming gifts for a 20 year-old moving out for the first time, into a share house? The usuals may not apply as utensils, tools etc. will likely already be available at their new home.

Is it weird to just gift some nice food? I'm at a bit of a loss.

Edit: I just wanted to clarify that "cheapish" is only because I'm currently studying thus have little income, not because I don't care!

Towel, frying pan.

Towel

They're moving into a house, not a spaceship.

Towel, frying pan.

Vogon poetry?

## Pages