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-   -   Pre-ordering Meals on Delta, the Definitive Thread (https://www.flyertalk.com/forum/delta-air-lines-skymiles/1754931-pre-ordering-meals-delta-definitive-thread.html)

StayingHomeIsBetter Apr 3, 2016 11:48 am


Originally Posted by WWads (Post 26429784)
... Being able to preorder your food would fix this "problem" anyway, which is what this thread is supposed to be about.

+1 ^

The focus of the thread has gone from the benefits of being able to pre-order up to 24 hours before flight time... to who's most worthy after 24 hours before flight time.

And, as typical, the "correct" answer is "the one that most closely matches my usage patter."

But, this is FT, after all. :D

sethb Apr 3, 2016 5:12 pm


Originally Posted by Bicostal (Post 26427974)
But your experiment is based on numbers, not probabilities. Play again?

No, my experiment's expected results are based on probabilities. What do you think probabilities are?

sethb Apr 3, 2016 5:14 pm


Originally Posted by Bicostal (Post 26429156)
And just to be clear, the experiment should reflect the proposition that with preordering available, those who do not are worse off because the chance of not getting their meal choice goes up as the number of passengers on that flight goes down.

It's about probability on a flight, one flight different numbers of passengers, not different number of flights with the same cumulative number of passengers.

It's the 10 and 100 passenger flight, not the 1 and 10 flights with a total of 100 passengers.

The probability has to be scaled to the total number of passengers. We're looking at the expected number of unhappy passengers per (say) 100; that's the probability of any given passenger being unhappy.

Bicostal Apr 5, 2016 10:11 am


Originally Posted by sethb (Post 26431338)
The probability has to be scaled to the total number of passengers. We're looking at the expected number of unhappy passengers per (say) 100; that's the probability of any given passenger being unhappy.


Originally Posted by sethb (Post 26385440)
I have a little wager to offer you . . .If the number of heads isn't 50, I pay you $10 each for the difference

EACH is NOT a probability, it is a number.

1 in 5 or 10 in 50 or 20 per hundred - all the same probability, all with different payouts....not an EACH in sight.

Your experiemnt is all about EACHes. Not ratios.

Try again?

sethb Apr 5, 2016 4:48 pm


Originally Posted by Bicostal (Post 26440238)
EACH is NOT a probability, it is a number.

1 in 5 or 10 in 50 or 20 per hundred - all the same probability, all with different payouts....not an EACH in sight.

Your experiemnt is all about EACHes. Not ratios.

Try again?

Experiment:

(1) One flight with 20 passengers, no pre-selection, 10 of each of 2 meals boarded. Toss coins to see what people want, count the number of disappointed people. Repeat many times to get the expected number of disappointed people. (Or do some arithmetic, but the coin-tossing makes for a better bet.) The expected number of disappointed people, divided by the total number of people, is the probability of disappointment.

(2) One flight with 6 passengers (same flight, only 14 pre-selected so they don't count). 3 of each meal boarded. Toss coins to get the number of disappointed people. Repeat many times to get the expected number of disappointed people.

If you can't understand it, take a course in probability theory.

Bicostal Apr 6, 2016 5:50 am


Originally Posted by sethb (Post 26442138)
Experiment:

(1) One flight with 20 passengers, no pre-selection, 10 of each of 2 meals boarded. Toss coins to see what people want, count the number of disappointed people. Repeat many times to get the expected number of disappointed people. (Or do some arithmetic, but the coin-tossing makes for a better bet.) The expected number of disappointed people, divided by the total number of people, is the probability of disappointment.

(2) One flight with 6 passengers (same flight, only 14 pre-selected so they don't count). 3 of each meal boarded. Toss coins to get the number of disappointed people. Repeat many times to get the expected number of disappointed people.

If you can't understand it, take a course in probability theory.

Just so we are clear...you think that the chance of disappointment on the 20 passenger flight is less than the 6 passenger flight. Right?

It's a binomial distribution. im not a big fan of chunky distributions but there are mathematical ways to smooth curves. Nonetheless, you still think you are more likely to be happy as the number of passengers goes up, right?

MSPeconomist Apr 6, 2016 5:54 am

Another way to think about it is to imagine 8 people in FC on a single flight without preselected meals, with iid equal probabilities of wanting A over B versus B over A, and 4 A meals and 4 B meals loaded. Now split them into two flights, each having 4 FC passengers who didn't perorder meals and 2 A meals and 2 B meals on each flight. Due to the constraints imposed by splitting the larger group into two halves, those on the smaller flights are more likely to be disappointed because there's no possibility to have, for instance, "abnormally" (i.e., somewhat unlikely) high preference for one meal on one flight to be balanced out by "abnormally" high preference for the other meal on the other flight. FAs can't exchange the meals with another flight up in the air, even though there are times that they might want to do so.

sethb Apr 6, 2016 10:02 pm


Originally Posted by Bicostal (Post 26444301)
Just so we are clear...you think that the chance of disappointment on the 20 passenger flight is less than the 6 passenger flight. Right?

It's a binomial distribution. im not a big fan of chunky distributions but there are mathematical ways to smooth curves. Nonetheless, you still think you are more likely to be happy as the number of passengers goes up, right?

The expected difference from the center increases as SQRT(N), so the probability of disappointment decreases as SQRT(N)/N.

With 2 passengers, the probability they both want the same meal (50% disappointment) is 50%. With 10 passengers, the probability they all want the same meal (the only way to get 50% disappointment) is about 0.02%.

10 flights with 2 passengers each will have, on average, about 5 disappointed passengers. 2 flights with 10 passengers each (same total number of passengers, note) will have, on average, around 2 disappointed passengers.

Bicostal Apr 7, 2016 4:59 am


Originally Posted by sethb (Post 26448871)
The expected difference from the center increases as SQRT(N), so the probability of disappointment decreases as SQRT(N)/N.

With 2 passengers, the probability they both want the same meal (50% disappointment) is 50%. With 10 passengers, the probability they all want the same meal (the only way to get 50% disappointment) is about 0.02%.

OK - and so what? The probability that everyone gets their meal choice is about 25% in the 10 passenger case and 31% in the 6 passenger case. What do you think you have proved?

Certainly not that as the number of passengers goes up the probability of a passenger being disappointed goes down.


Originally Posted by sethb (Post 26448871)
10 flights with 2 passengers each will have, on average, about 5 disappointed passengers. 2 flights with 10 passengers each (same total number of passengers, note) will have, on average, around 2 disappointed passengers.

Again with the "you take 10 flights and i'll take 2" scenario.

You said that if I get on a plane with 20 FC seats and no one has preordered a meal, I stand a better chance of getting what I want than if I get on that same plane where 10 people (but not me) have preordered a meal. I called BS.

Here's why, I fully agree, that the number of unhappy passengers is different, but so is the denominator. More people, more unhappy people.

Second, there are more than 2 states of nature - the all happy or all unhappy states. There are states where one passenger is unhappy, there are probabilities of 2, or 3, or 4 being unhapp - all the way to 5 being unhappy in the 10 passenger case. The probability of each of these states is different - but each contribute to the cummulative probability of me being unhappy. I don't know which plane I am getting on, but I could be on that plane where there is a 7/3 distribution of preference meaning that 2 people will be unhappy - or 20%. The chance of being on a 7/3 plane is lower (12%) than say a 6/4 plane (20%), but the number of chances of being unhappy (2 instead of 1) is higher.

Note - I did NOT change the number of passengers on a flight, just the number of times I do the experiment.

Now do the math - cacluclate the cummulative probability and if you think that you are still better off with more passengers not pre-selecting, then you feel free to think that.

DL2SXM Apr 7, 2016 6:48 am


Originally Posted by MSPeconomist (Post 26427817)
There's a good chance that the elite with the free upgrade doesn't ahve a sLUT(XV) fare but rather has paid more for his/her ticket than the person who "paid" for FC. Compare G A P fares to Y B M.

Than that person, if a comp upgrade has been received, can go into the Trip Extas section of Delta.com and pay for a decent meal if they so decide that they want one. I think this is something delta needs to 'enhance'.

MSPeconomist Apr 7, 2016 8:21 am

Be careful what you suggest. DL could just as easily "enhance" the system so that everyone pays for their meal in domestic FC.

DL2SXM Apr 7, 2016 9:08 am


Originally Posted by MSPeconomist (Post 26450447)
Be careful what you suggest. DL could just as easily "enhance" the system so that everyone pays for their meal in domestic FC.

I don't see whats so terrible about shelling out a few bucks via the Trip Extra's section on the delta website to upgrade your meal experience. Delta wouldn't be forcing you to buy into a better meal deal. Its very simple. Buy an F seat, get a decent meal. Buy a coach seat and wind up in F then you get whatever garbage is loaded on the food cart or use the Trip Extra's section to buy up your meal experience. I like the idea and on a route like JFK-SEA for which I can buy a sLUT fare for $300.00, I would most likely easily spend a few more dollars for an enhanced meal service.

mpheels Apr 7, 2016 10:39 am

I think DL is far more concerned with the number of passengers getting their meal of choice than they are in the probability that any given passengers ends up with a meal the dislike. By letting passengers choose, they effectively remove them from the pool. We can assume that those passengers get their meal of choice (save for especially difficult people who change their mind), and DL reduces the number of people let to chance. So if you have 10 F passengers, and 6 pre-select, then you only have 4 left to choice/chance. If DL loads two of each option for the 4 who didn't pre-select, then at least two will get their preferred meal, and at a most two are disappointed.

rylan Apr 7, 2016 11:44 am


Originally Posted by MSPeconomist (Post 26450447)
Be careful what you suggest. DL could just as easily "enhance" the system so that everyone pays for their meal in domestic FC.

Once the pre-selection infrastructure is in place, it is only a small step to add a purchasing component.

bubbashow Apr 7, 2016 11:49 am


Originally Posted by DL2SXM (Post 26450055)
Than that person, if a comp upgrade has been received, can go into the Trip Extas section of Delta.com and pay for a decent meal if they so decide that they want one. I think this is something delta needs to 'enhance'.

A harsher stance than ME! I want to be your friend. :D


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