Re: A certain bias, there
A lot of beers are significantly more than 2 units/pint.
2 units/pint = 3.52% alcohol. Sessions beers were traditionally ~3%, most beers in my local are above 3.5, some considerably above.
Analytical skills are in big demand so it is really important not to make the basic, common, mistakes that show you up as a newbie. For example, probability calculations are often performed on binary outcomes such as "What is the probability that a given policy holder will claim?" The result is binary because they will either …
Correct, kind of. But that 8 hours assumes you drank all 4 pints at once.
units of alcohol is about strength % * volume (in litres)
So a metric pint (half litre) of 4% beer is 2 units.
If you begin at at fairly normal 20:00, drink 4 pints between then and 00:00, that's already 2 of the pints gone from the system, the next 2 by 02:00. Driving to work the next day at 08:00 won't be an issue.
(This is a rough estimate as it takes some time for your liver to start processing the alcohol obv.)
In the UK, the drink driving limit, roughly, for a 12 stone man, is around 4 units, so you would unlikely be over the limit if you waited an hour or so after midnight and then drove (but I wouldn't recommend risking it).
Wasn't aware that beer could have that kind of effect.
So I have to amend my position on alcohol in the morning, since it could simply mean that you had "too much" the night before. I'm okay with that, although I will be encouraging you to drink lots of water during the day.
The quick way is to do 1-(0.95)^52=0.93 (2 sig figs), which gets you almost exactly the right answer. That's how we were taught to do it in school for continuous events and adding probabilities. The extra accuracy you get by withdrawing balls gives you the correct answer, but the short form above works well, if the testing is still relatively infrequently.
@DavCrav:
> The quick way is to do 1-(0.95)^52=0.93 (2 sig figs),
> which gets you almost exactly the right answer.
No, it gets you exactly the right answer. The use of a hypergeometric distribution is technically correct (but only if the sot keeps track of how many times he's been drunk and stops after twelve such days); personally, I would apply a binomial distribution with a fixed probability of drunkenness on any one day of p=12/260, just as you did.
The results are: P(X>=1) = 0.935646978954 for the hypergeometric case, P(X>=1) = 0.914321213168 for the binomial distribution. The commenter that decided that the result must round to 0.90 is living in a state of sin.
"No, it gets you exactly the right answer."
It gets you the right answer, but to a different question. It gets you the right answer to the question "what is the chance of getting caught given that each day has a fixed probability of you being drunk", whereas the question was that in a given year there are exactly twelve drunk days, so it becomes a finite problem which needs finite methods rather than a continuous probability.
I think your rough calculation is actually closer to correct, while the one given in the article is slightly wrong. This is not a case of "removing one ball from the bag" each time as the writer has assumed. In actuality it is likely that the probability of the employee being drunk on any given day is always 12/260, independent of any previous day. In other words, the employee decides on any given day "do I need a drink today?" depending on his or her current circumstances, and is not thinking "wait, I've already gone in to work drunk 5 times this year, so maybe not today.".
I think we accidentally get hung up on the pertinent request, "If I do something X times, what are the facts about X?". In this case we get caught up on getting to that magic fact being driven by the 52 times/year, the 4 times/year or twice a year and start from there. Just human nature I suppose to look for the quickest route through a problem rather than the correct logical route.
The "bag of marbles" visual idea for probabilities takes me back to my school days and later on working through maths homework with my own kids. It's such a great little visual image that immediately registers with the imagination. Don't try to imagine an abstract idea of 12 days over 260 because you cannot be two mutually exclusive things at the same time ( drunk and sober ) , but imagine it's ( 260 = 12 + 248 ) , 12 red marbles and 248 green marbles. Immediately most people, even those who really struggle with maths, can visualise instantly and start to get a handle on probabilities. Oddly my imagined bag of marbles is always dark grey felt with a black draw cord, ha ha!!
I've long tended to grab two socks from the drawer in the morning without really checking to see if they match. SWMBO objected to this behavior. I caved in (always a good idea on minor issues; choose one's battles carefully, etc.) and have since taken a moment or two to ensure a match. But I did explain to her that my previous behavior was just in memory of (seemingly countless) probability problems as a lad such as:
Eddie has five red socks, three black socks, and two white socks. If Eddie grabs two socks in the morning at random, what's the probability that he'll grab two socks that match?
If one of my employees shows up obviously under the influence, I'll show them the door. It's that simple, and they know it. They have the option of taking a blood test immediately after being fired, and I'll deal with the consequences if I'm wrong. Thankfully I have never needed to resort to this ... But that's not to say I'll fire 'em if they have booze on their breath. I operate a small brewery, winery, and fledgling distillery. Almost all of my employees have access to, and in some cases are required to sample the product ... Either you trust your employees, or you don't. If you don't, get rid of them. Makes life much easier all around.
Buzzword, I'm fairly certain I can tell the difference. My velcro-whippet certainly can. I'll make the call if/when. And as I said, if I'm wrong I'll face the consequences.
Hasn't happened yet. Somehow, I doubt it will ever be an issue. Not with this crew; they've all been with me for nearly two decades, and all have a piece of the holding company.
Can you tell the difference between someone who is still drunk from the night before, and someone who has a cold? What if they simply haven't slept properly because of external factors (heat wave, noisy neighbours, etc.)?
Yes, because of the smell of ethanol coming from their skin pores.
"If one of my employees shows up obviously under the influence, I'll show them the door. It's that simple, and they know it. They have the option of taking a blood test immediately after being fired, and I'll deal with the consequences if I'm wrong."
If you are in the lovely US in one of those at will states then fine, but in the UK you would be in court pretty quickly with that kind of approach.
Instant dismissal isn't really necessary. When coming back from a liquid lunch, and having been judged to have forgotten how to count above two pints, my manager would meet me at the office door and offer me a pen and an instant, one day, holiday request form. Then off home. It was usually worth it. Fighting in the car park optional.
I've been at places where when we know we are doing a lunchtime celebration then the network access is revoked in advance. I've also been on lunch celebrations where the boss decided it would be better for people to go straight home from the pub rather than be let loose on the IT systems.
How many people do you have to have in a room before there is a greater than 1 in 2 chance that 2 of them have the same birthday?
People tend to guess a number around 183, but the same logic as the drunk testing applies and the answer is in the low 20s. (It is not exactly as calculated because there tend to be more births at certain times of year.)
Hmm. Overthinking it a bit.
My one and only experience of a drunk at work goes back to my first job in the early 90s.
The 'drink because my wife and kids have left me because I moved the family from <somewhere nice> to Bracknell' rolled in at 3pm, fell asleep, then stood up and peed in his top desk drawer.
Sitewide no boozing at work policy the next week.
"The 'drink because my wife and kids have left me because I moved the family from <somewhere nice> to Bracknell'"
Bracknell = RACAL perchance.
"rolled in at 3pm, fell asleep, then stood up and peed in his top desk drawer." = https://www.youtube.com/watch?v=oNFz0hkPXvA
There was a similar sort of Christmas party at RACAL Seaton (Before I joined), there was sex under benches & all sorts of celebrations on return from the pub\in house supplied booze, following year Management wizened up & no such celebrations took place, they just kicked us out at lunchtime & we went straight to The George instead.
We know they test once a week. Wait sober until they test, and drink until the weekend. Probability of being caught 0!
Alternatively, it's almost certain that the boss will have been seduced by sales spiel into buying the latest fanciest internet connected alcohol testing machine. Even the greenest PFY should be able to insert the 'if dept = IT then alcohol = random number between totally sober and reasonably sober' into the code. Or maybe the simpler 'if dept = HR then alcohol = embarrassingly pissed'
And pedantically, assuming that the employee's drinking is reasonably spread your green and red balls are a bit skewed - (I think - of course last night's bender might be clouding my judgement), because while in week 1 you may have 248 red 12 green, but week 2 we should be calculating 243 red 11.75 green, - the odds of being caught aren't changing any more than the odds of throwing a coin change? Your numbers assume that if I was sober on test 1 the chances of me being sober on test 2 are slightly less, effectively assuming that there are still 12 drunk balls, but now 247 sober ones.
I think it was just a poorly labelled table and the best way of thinking about it is there's a 12/260 chance that I'm drunk on a given day and a 52/260 chance that I will be tested on a given day
so for a given day the probability of getting away with it = 0.04 * 0.2 = 0.009 as it was in the table
They just forgot that I come to work 260 days a year (for which I'd need to be on something) so
260*0.009 = 2.4, a 240% chance of being busted. That empirically feels right, If I do something where i have a 20% chance of being caught 12 times in a row, I'm going to get caught.