I wanted to find a mathematical relationship between the time-length of a relationship or marriage and the likelihood of it ending in breakup or divorce. So I decided to on another forum ask members the time-lengths of all their failed relationships/marriages. This does not include widows, meaning they ended in breakup.
So they would list for example, something like:
3 months
6 months
9 months
21 years and still going/counting
Now I asked for failed relationships, so I wouldn't include the last one. When I decided to analyze the data, I had 58 data points (aka failed relationships and their time-lengths).
I counted this so for example, 3 months would also count for 2 and 1 because it did at one point be that old. So I took all the times and I listed the all the numbers down a line and then I would take one data point and mark tallies up the list. So for example, for 3 months, I marked a tally for 3 months, 2 months, and 1 month and so on. So I had a bunch of tallies, and at 1 month, I had 58 tallies all the way down to 1 tally at 288 months. Yes, one relationship ended in breakup after 288 months.
You might still be wondering why I counted all the ones that ended at 3, 4, 5, etc months for 1 month as well. The reason is because this is finding the mathematical relationship between the time-length of a relationship and the likelihood of it ending in breakup or divorce altogether, not when breakups happen. So a 6-month failed relationship was still a 1-month relationship at one point and it ended in a breakup.
Now blindly interpreting this data would cause one to believe that 100% of one-month long relationships fail! This can't be possible because that would mean that all relationships fail and we know that's not true considering that approximately 50% of first marriages do not end in divorce.
What I did with the tallies is I multiplied them by 100 and then divided them by 58 to yield a percentage. So at 2 months, I had 52 tallies, so I entered 5200/58. So this data point would indicate that at two months into a relationship there is approximately an 89.655% chance that it will end up in breakup or divorce between the status quo and infinite future. I did this for all time-lengths that I had tallies for and it went all the way to approximately 1.724% for a 288-month long relationship.
Now if you have good visualization when it comes to mathematics, you can imagine the end behavior of this graph. For now, let x represent the time-length of a failed relationship and y represent the likelihood of the relationship or marriage ending in breakup or divorce. The limit as x approaches infinity of this unknown (for now) function equals 0, which makes sense because a relationship lasting forever will never end! Now, let's think about x=0. Well, there was never a relationship so we can't really have a data point there, right? But we can have a discontinuity! Think about it. You can't go over 100%. So we will place a restriction on the function (when it's determined) to prevent the y value from exceeding 100. Now if you think about it, after one day of being together, there is almost a 100% chance you will not stay together right. This is what I am saying. If you were to blindly look at my data, then at x=1, y=100, which isn't possible and the reason is exp lained above. That's why I excluded x=1 when I graphed my data points on the calculator. But you can see that as x approaches 0, y is getting closer and closer to 100, which makes sense because chances are slim that you are destined to stay with a person you have only been with for a week. So we can conclude, that the limit as x approaches 0 equals 100, so while we can't have a point at x=0, we can certainly have a "hole" or a discontinuity at the point (0, 100).
Now visualize the graph again. Since we know that the limit as x approaches infinity equals 0, we know that there is a horizontal asymptote at y=0. With a little bit of playing around, I found that this would have to be an exponential function.
So the format for an exponential function is y=ab^x.
Now let's consider the discontinuity (0, 100). It's a discontinuity, but we still have to account for it in the process of writing a function. So:
100=ab^0
b can be any number except 0 and it will equal 1. We'll leave b alone for now and find that later. But if we assume that b is just some number except 0, then:
100=a(1)
So a=100. So now we have y=100b^x and therefore y is destined to equal 100 at x=0. If you forgot why this makes sense, reference the above explanation.
Now a scatter plot won't give you a perfect equation. Of course, data will be all over the place. But it will give you a trend from which you can figure out what type of function it will be. As we already established, it's exponential. The reason I point this out is to remember that the function we find won't necessarily pass through all data points, or even through most. But it should follow the trend of the data points. I figured the best way to find a best-fit curve equation is to take y=100b^x and then use our data point (288, 100/58) (There is a 1/58 chance of a relationship 288 months long ending in breakup or divorce, and to convert it into a percentage, simply multiply 1 by 100), and plug that in and find b. Through the use of logarithms to solve for b, we found that b equals approximately 0.970.
So now to right this as a formal equation, I'll let t represent the time-length of a relationship (in months) and b(t) represent the likelihood of it ending in breakup or divorce (in percentage) (between now and the infinite future), and write:
b(t)=100(0.970)^t ; t must be greater than 0.
The restriction is necessary because a relationship must have some time-length and without this restriction, y will exceed 100 which can't be allowed because y is a percentage.
I graphed this and it looked pretty good (instructions for graphing are at the bottom of the post).
So I decided to do a few plug-and-chugs to test out the function.
So for a 6-month relationship, I get y=83.379%.
For 1-year, I got y=69.38%.
This seemed to fit my data points relatively well. There were some data points that were a little far off, but that's to be expected.
And notice how this function satisfies the original conditions I had placed for this function. Like for the limit as x approaches infinity equaling 0, let's take a look:
b(t)=100(0.970)^inf.
That makes sense because 0.970^inf. equals 0 and then 100*0 equals 0. The reason 0.970^inf. equals 0 (for those of you not so math-savy) is because 0.970 multiplied by itself repeatedly will yield a smaller and smaller number, and it will approach 0 as the number it is being raised to keep getting larger and larger. So when it's raised to the infinity power, it will equal 0.
Also, this function would satisfy the point (0,100).
b(t)=100(0.970)^0
b(t)=100(1)
b(t)=100
However, because of the restriction, on the graph we will mark an open circle.
If you want to see the graph, simply enter "(100)(0.970)^x" for y=___. Then zoom out (using the out 10x button) so the viewing windows is a little past -200 and 200 for x and -100 and 100 for y. This will show the function without the restriction of x having to be greater than 0. But it's still useful, for visualization purposes [just imagine the open circle at (0, 100) and the function not continuing past that] because if you're too lazy, you can enter x values next to "points" below the graph, and it will calculate the y value, or in this case, the "likelihood of the relationship or marriage ending in breakup or divorce between now and the infinite future in percentage". And remember, x values are in months.
If you're curious, you can also calculate for a certain days. The average month is 30.4375 days, so simply take the number of days and divide it by 30.4375 and enter that as your x value. For 1 day, I got approximately 99.90%.
Of course, I don't know how "accurate" this is as it was only done off of 58 data points. I suppose for a really accurate equation, we would want 1,000 or so data points.
So go ahead and tell your boyfriend/girlfriend of 3 years that approximately there is only a 33.403% of chance of you guys ever breaking up now or in the future, at this point! :D
What do you think?
EDIT: for the grapher, visit mathisfun.
Also, time starts from the first date or date or transformation of relationship type to romantic.
So they would list for example, something like:
3 months
6 months
9 months
21 years and still going/counting
Now I asked for failed relationships, so I wouldn't include the last one. When I decided to analyze the data, I had 58 data points (aka failed relationships and their time-lengths).
I counted this so for example, 3 months would also count for 2 and 1 because it did at one point be that old. So I took all the times and I listed the all the numbers down a line and then I would take one data point and mark tallies up the list. So for example, for 3 months, I marked a tally for 3 months, 2 months, and 1 month and so on. So I had a bunch of tallies, and at 1 month, I had 58 tallies all the way down to 1 tally at 288 months. Yes, one relationship ended in breakup after 288 months.
You might still be wondering why I counted all the ones that ended at 3, 4, 5, etc months for 1 month as well. The reason is because this is finding the mathematical relationship between the time-length of a relationship and the likelihood of it ending in breakup or divorce altogether, not when breakups happen. So a 6-month failed relationship was still a 1-month relationship at one point and it ended in a breakup.
Now blindly interpreting this data would cause one to believe that 100% of one-month long relationships fail! This can't be possible because that would mean that all relationships fail and we know that's not true considering that approximately 50% of first marriages do not end in divorce.
What I did with the tallies is I multiplied them by 100 and then divided them by 58 to yield a percentage. So at 2 months, I had 52 tallies, so I entered 5200/58. So this data point would indicate that at two months into a relationship there is approximately an 89.655% chance that it will end up in breakup or divorce between the status quo and infinite future. I did this for all time-lengths that I had tallies for and it went all the way to approximately 1.724% for a 288-month long relationship.
Now if you have good visualization when it comes to mathematics, you can imagine the end behavior of this graph. For now, let x represent the time-length of a failed relationship and y represent the likelihood of the relationship or marriage ending in breakup or divorce. The limit as x approaches infinity of this unknown (for now) function equals 0, which makes sense because a relationship lasting forever will never end! Now, let's think about x=0. Well, there was never a relationship so we can't really have a data point there, right? But we can have a discontinuity! Think about it. You can't go over 100%. So we will place a restriction on the function (when it's determined) to prevent the y value from exceeding 100. Now if you think about it, after one day of being together, there is almost a 100% chance you will not stay together right. This is what I am saying. If you were to blindly look at my data, then at x=1, y=100, which isn't possible and the reason is exp lained above. That's why I excluded x=1 when I graphed my data points on the calculator. But you can see that as x approaches 0, y is getting closer and closer to 100, which makes sense because chances are slim that you are destined to stay with a person you have only been with for a week. So we can conclude, that the limit as x approaches 0 equals 100, so while we can't have a point at x=0, we can certainly have a "hole" or a discontinuity at the point (0, 100).
Now visualize the graph again. Since we know that the limit as x approaches infinity equals 0, we know that there is a horizontal asymptote at y=0. With a little bit of playing around, I found that this would have to be an exponential function.
So the format for an exponential function is y=ab^x.
Now let's consider the discontinuity (0, 100). It's a discontinuity, but we still have to account for it in the process of writing a function. So:
100=ab^0
b can be any number except 0 and it will equal 1. We'll leave b alone for now and find that later. But if we assume that b is just some number except 0, then:
100=a(1)
So a=100. So now we have y=100b^x and therefore y is destined to equal 100 at x=0. If you forgot why this makes sense, reference the above explanation.
Now a scatter plot won't give you a perfect equation. Of course, data will be all over the place. But it will give you a trend from which you can figure out what type of function it will be. As we already established, it's exponential. The reason I point this out is to remember that the function we find won't necessarily pass through all data points, or even through most. But it should follow the trend of the data points. I figured the best way to find a best-fit curve equation is to take y=100b^x and then use our data point (288, 100/58) (There is a 1/58 chance of a relationship 288 months long ending in breakup or divorce, and to convert it into a percentage, simply multiply 1 by 100), and plug that in and find b. Through the use of logarithms to solve for b, we found that b equals approximately 0.970.
So now to right this as a formal equation, I'll let t represent the time-length of a relationship (in months) and b(t) represent the likelihood of it ending in breakup or divorce (in percentage) (between now and the infinite future), and write:
b(t)=100(0.970)^t ; t must be greater than 0.
The restriction is necessary because a relationship must have some time-length and without this restriction, y will exceed 100 which can't be allowed because y is a percentage.
I graphed this and it looked pretty good (instructions for graphing are at the bottom of the post).
So I decided to do a few plug-and-chugs to test out the function.
So for a 6-month relationship, I get y=83.379%.
For 1-year, I got y=69.38%.
This seemed to fit my data points relatively well. There were some data points that were a little far off, but that's to be expected.
And notice how this function satisfies the original conditions I had placed for this function. Like for the limit as x approaches infinity equaling 0, let's take a look:
b(t)=100(0.970)^inf.
That makes sense because 0.970^inf. equals 0 and then 100*0 equals 0. The reason 0.970^inf. equals 0 (for those of you not so math-savy) is because 0.970 multiplied by itself repeatedly will yield a smaller and smaller number, and it will approach 0 as the number it is being raised to keep getting larger and larger. So when it's raised to the infinity power, it will equal 0.
Also, this function would satisfy the point (0,100).
b(t)=100(0.970)^0
b(t)=100(1)
b(t)=100
However, because of the restriction, on the graph we will mark an open circle.
If you want to see the graph, simply enter "(100)(0.970)^x" for y=___. Then zoom out (using the out 10x button) so the viewing windows is a little past -200 and 200 for x and -100 and 100 for y. This will show the function without the restriction of x having to be greater than 0. But it's still useful, for visualization purposes [just imagine the open circle at (0, 100) and the function not continuing past that] because if you're too lazy, you can enter x values next to "points" below the graph, and it will calculate the y value, or in this case, the "likelihood of the relationship or marriage ending in breakup or divorce between now and the infinite future in percentage". And remember, x values are in months.
If you're curious, you can also calculate for a certain days. The average month is 30.4375 days, so simply take the number of days and divide it by 30.4375 and enter that as your x value. For 1 day, I got approximately 99.90%.
Of course, I don't know how "accurate" this is as it was only done off of 58 data points. I suppose for a really accurate equation, we would want 1,000 or so data points.
So go ahead and tell your boyfriend/girlfriend of 3 years that approximately there is only a 33.403% of chance of you guys ever breaking up now or in the future, at this point! :D
What do you think?
EDIT: for the grapher, visit mathisfun.
Also, time starts from the first date or date or transformation of relationship type to romantic.
Put the internet to work for you.
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