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Math Help?

JoeBobMack ๐Ÿšซ
Updated:

I feel like I ought to be able to figure this out, but my decision not to pursue any math-based career in college is coming home to roost. A little help?

Assume a population of people that is 15% Green, 30% Red, and 55% Yellow. If all individuals are randomply assigned to a partner, what percentage will be in each possible pairing, GG, GR, GY, RR, RY, and YY? For bonus points, what's the name of this type of problem and the formula for solving? Please feel free to dump on my lack of mathematical insight... I'm right there with you.

LupusDei ๐Ÿšซ
Updated:

@JoeBobMack

You're looking at probability of two independent events. Basically you have a bag of peas with are 15% Green, 30% Red, and 55% Yellow and blindly draw two. Thus, you can simply multiply the probabilities of each event.

Chance to draw two green is 0.15 * 0.15 = 0.0225 or two and a quarter percent, 2.25%
Green/Red 0.15 * 0.30 = 0.045 = 4.5%
Green/Yellow 0.15 * 0.55% = 0.0825 = 8.25%

Red/Green 4.5%
Red/Red 9%
Red/Yellow 16.5%

Yellow/Green 8.25%
Yellow/Red 16.5%
Yellow/Yellow 30.25%

Yes, you get nine combinations and not just six because order matters. If it doesn't and you just want to know how many pairings where, say, one is Green and one Red you add GR + RG = 9%

Note that probability distributions for first and second event not necessarily need to be identical as we have in this example. If you, for example, had a strict distinction between two genders with different "color" distributions and paired one of each, you would multiply percentage of a male color with percentage of a female color and GR probability would differ from RG probability.

Replies:   Dominions Son
Dominions Son ๐Ÿšซ

@LupusDei

Except that only works if every draw is fully independent, which means to use your example, you have to put the peas back in the bag and mix them back up.

The op is talking about a human population forming couples. Once a couple forms, those two are no longer available to form new couples.

So to go back to your example of peas in a bag, after each pair is drawn, you have to throw those two peas away, there are now one fewer pea of each color drawn (or two fewer of one color if you draw two peas of the same color).

This means that since each draw reduces the remaining population, the draws are not fully independent as each draw has some small effect on the probability of the next draw.

A better way to look at it is in raw numbers.

150 Green, 300 red and 550 Yellow.

if they form couples at random, how many couples will have a green person, ~150. It won't be exactly 150 because you will get some GG couples, but I don't know of any good way to estimate how many.

So you start with 150/1000 green people = 15%

After forming couples, you effectively cut the population in half to 500 couples.

You will get a minimum of 75 green couples if all greens pair with another green and a maximum of 150 green couples if no you get no GG couples.

That's a minimum of 75/500=15% and a maximum of 150/500=30%

If you average the min and max, that gives 112.5 rounding up since you can't have half a couple and the extra green person goes somewhere, 113/500 = 22.6%

It would work out a bit differently on the other end, since Yellow has the highest population, you would get a minimum of (550-300-150)/2 = 50 YY couples

Replies:   LupusDei
LupusDei ๐Ÿšซ
Updated:

@Dominions Son

Except that only works if every draw is fully independent, which means to use your example, you have to put the peas back in the bag and mix them back up.

If a population is sufficiently large it doesn't matter. If the population is very small, you're right, the events wouldn't be independent and can lead to different outcomes depending on the sequence of events.

In your example with just a thousand, there's a nonzero chance no green-green or red-red pair would form at all, and only few yellow-yellow pairs are guaranteed to form because all yellows can't be exhausted by other color pairings.

But that is complexity that is unnecessary for global averaging, unless, it is indeed stated this happens within confines of a single school or very small village, and even then the independent averages are informative.

Replies:   Dominions Son
Dominions Son ๐Ÿšซ

@LupusDei

If a population is sufficiently large it doesn't matter.

That depends on how many "draws" you intend to make.

If you start with a large population, the effect of individual draws will initially be insignificant. However, after enough draws, the cumulative effect will be significant.

Replies:   Dominions Son
Dominions Son ๐Ÿšซ

@Dominions Son

In your example with just a thousand, there's a nonzero chance no green-green or red-red pair would form at all, and only few yellow-yellow pairs are guaranteed to form because all yellows can't be exhausted by other color pairings.

Actually, the second half of that would hold true no matter how large the population is. As long as the relative percentages are held constant, the majority group can't be exhausted by pairings with the other two groups. It doesn't matter if the population is 1,000, 1,000,000, or 100,000,000,000.

And as long as pairings are truly random, there will be a non zero probability of no in sub-group pairings for the smallest sub-group. The probability of such an out come might approach zero as the population increases, but it will never get there.

Replies:   LupusDei
LupusDei ๐Ÿšซ

@Dominions Son

Sure, for any N there would be 1/N! chance of any specific combination. Only that's a rapidly shrinking number as N grows.

Replies:   Dominions Son
Dominions Son ๐Ÿšซ

@LupusDei

Only that's a rapidly shrinking number as N grows.

Yes, but it never quite reaches zero, so it's always non-zero as long as N is finite.

Replies:   LupusDei
LupusDei ๐Ÿšซ

@Dominions Son

Yes, but we're not talking about exact outcomes, but about likelihood of any given color combination to exist.

I'm not saying exactly 4.5% of all couples must necessarily be Red-Green, I'm saying the unbiased chance to roll a Red-Green couple is 4.5%. For that chance to realize you would necessarily need quite a few rolls.

Quasirandom ๐Ÿšซ

@JoeBobMack

Poking at the other half of your question, the branch of math involved is called combinatorics.

The exact formula is debatable (see above) because of an ambiguity in your question. So for clarification, should we assume that, once two people are paired off, they are out of the pool of available for assignment? Also, how large a pool are we talking โ€” a few dozen or a few million, or what?

JoeBobMack ๐Ÿšซ

@JoeBobMack

Thanks, folks! Very helpful. And, thankfully, I wasn't as far off as I feared. Basically had it, just screwed up the more simple arithmetic! Just to fill in the gaps, my story has magic flowing into the world. About 15% of people are attracted to it (some for good, a few for ill), about 30% are repelled and just don't want to be around it, and the remainder (55%) are inert - don't react much either way. I haven't fully decided if there might be some aspects of pair-bonding that would cause the numbers to skew from pure random and, in any event, the population is on a continuum, so it's just rough numbers. But, I need to think about the fact that roughly one couple in 10 might be repelled + repelled and run for the hills pretty quickly. Another 40%, (roughly) would have a repelled and an inert, so the tendency would be to move away, though ties to the area might slow that. Plus, I wanted to make sure I took into account that only a small percentage of already married couples would be made up of two who were both attracted.

So - y'all have been a great help!! Thank you.

Uther Pendragon ๐Ÿšซ

@JoeBobMack

And, of course, some couples are already formed. Those will presumably stick together however their responses to magic conflict. Others will be in the pair-forming stage; some of the will say. "Ewww! You're one of them."

Replies:   JoeBobMack
JoeBobMack ๐Ÿšซ

@Uther Pendragon

And, of course, some couples are already formed. Those will presumably stick together

Not necessarily. Depends on the relationship and, apparently, how magic is presented to the resisting spouse. I say apparently because the first couple to face this split up, but the circumstances were extreme. As one of my characters keeps suggesting, they have to avoid drawing conclusions until they have a larger sample size.

Uther Pendragon ๐Ÿšซ

@JoeBobMack

And there are other relationships, parents, siblings, employers.
it's easy to say, "Head for the hills," but few of us could support ourselves in the hills, and fewer still could support ourselves comfortably.

Replies:   JoeBobMack
JoeBobMack ๐Ÿšซ

@Uther Pendragon

And there are other relationships, parents, siblings, employers.
it's easy to say, "Head for the hills," but few of us could support ourselves in the hills, and fewer still could support ourselves comfortably.

Yup. The original entry point is a small town near Huntsville, Alabama in 1973. I'm expecting a significant turnover in the town's population. And living in the hills is different as humanity's control over electricity and explosions fails. Of course, the concentration in urban areas of those who are attracted to using magic for power over others could make cities into post-apocalyptic horrors. Depends on how well the team does in carrying out their guidance.

Sorry. Couldn't resist a little teaser. (At least I hope it's a teaser. If you hate it, please be kind and don't tell me now! I'm not to that point yet.) Now, I need to get back to working on the story to see if I can get it to a point where I'd be willing to post it.

steeltiger ๐Ÿšซ

@JoeBobMack

It's a Punitt square

"The Punnett square is a square diagram that is used to predict the genotypes of a particular cross or breeding experiment. It is named after Reginald C. Punnett, who devised the approach in 1905. The diagram is used by biologists to determine the probability of an offspring having a particular genotype. Wikipedia"

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