Probability question help

[stocks]

I listed all the years we started an economic recession for the last 2 centuries and sorted them by the last number of the year, 0 thru 9, as follows:

# of events in years ending with number:

0 - 6
1 - 3
2 - 0
3 - 7
4 - 1
5 - 0
6 - 1
7 - 5
8 - 1
9 - 4

28 events

No events in years ending 2,5
18 total in just three years 0,3,7



I have 2 mult-part questions:

1. What is the probability that if you randomly drop 28 unique events into 10 different slots,

2 slots would have 0 events,
3 slots would have 0 events,

2. What is the probability that if you randomly drop 28 unique events into 10 different slots,

3 slots would have a total of 18 or more events,
2 slots would have a total of 18 or more events,

 

[Paul C. Anagnostopoulos]

I believe the answers to the first question are .002 and .000046.

After that it gets hard. Hell, I might be wrong about this.

You may not have enough points for this to be significant. Also, don't forget to try the tens digit of the year.

~~ Paul

 

[Beth]

I

I have 2 mult-part questions:

1. What is the probability that if you randomly drop 28 unique events into 10 different slots,

2 slots would have 0 events,

I compute it as .1741

3 slots would have 0 events,

I compute it as .0331

2. What is the probability that if you randomly drop 28 unique events into 10 different slots,

3 slots would have a total of 18 or more events

I compute it as .1321

2 slots would have a total of 18 or more events,

I compute it as .00004

I'm assuming 0.1 probability for each slot.

 

[Beth]

I believe the answers to the first question are .002 and .000046.

After that it gets hard. Hell, I might be wrong about this.

You may not have enough points for this to be significant. Also, don't forget to try the tens digit of the year.

~~ Paul

Those are the probabities for a given set of 2 or 3 slots, but since the particular slots weren't specified in advance, we need to multiply by all the number of possible combinations of 2 or 3 slots respectively.

 

[Paul C. Anagnostopoulos]

Yup, Beth is correct.

Wait a minute, your numbers still seem too high. Aren't there 45 ways to choose 2 out of 10? So .002 x 45 = .09

~~ Paul

 

[Rodney]

Yup, Beth is correct.

Wait a minute, your numbers still seem too high. Aren't there 45 ways to choose 2 out of 10? So .002 x 45 = .09

~~ Paul

I believe that is the correct answer to the first question, rather than the .1741 calculated by Beth. I also differ with Beth on the other three answers. I get, respectively, .0055 (not .0331); .0101 (not .1321); and .00002 (not .00004).

 

[Just thinking]

I don't believe we can look at all calendar years as having an equal likelihood in starting a recession. Why? Because major elections, which can highly influence economic directions, do not occur on every year ending 0 - 9. Therefore the questions regarding the probability of so many slots with no events may be moot.

 

[rockoon]

I don't believe we can look at all calendar years as having an equal likelihood in starting a recession. Why? Because major elections, which can highly influence economic directions, do not occur on every year ending 0 - 9. Therefore the questions regarding the probability of so many slots with no events may be moot.

Of course that could be the point of calculating the probability.

 

[Roboramma]

I don't believe we can look at all calendar years as having an equal likelihood in starting a recession. Why? Because major elections, which can highly influence economic directions, do not occur on every year ending 0 - 9. Therefore the questions regarding the probability of so many slots with no events may be moot.

Also, having a recesion on year X may make a recession on year X+3 more or less likely. If so, that might skew the results as well. If recessions tend to cluster, that is, if a recession on year X makes a recession on year X+1 much more likely, but a recession on year X+2 less likely, for instance, then I think we'd need a lot more data points to see things start to average out.
I'm sure there are other things that might skew the results as well.

Interesting question.

 

[Kevin_Lowe]

Isn't this a case of drawing the target after the shot has been fired?

 

[T'ai Chi]

Isn't this a case of drawing the target after the shot has been fired?

No, else everytime one collects data and analyzes it, you'd be screaming that they should have analyzed the data before they collected it.

 

[Kevin_Lowe]

No, else everytime one collects data and analyzes it, you'd be screaming that they should have analyzed the data before they collected it.

It seems to me that the OP has noticed some oddities in the distribution and is asking what the odds are that those particular oddities would be present in a randomly distributed set.

That's the same as looking at a string of random numbers, seeing four nines in a row, and asking "What are the odds of four nines in a row cropping up in a string of random numbers of that length?".

A better question would be "What are the odds that some eye-catching oddity would crop up in a string of random numbers of that length?".

Alternatively you could hypothesize that the random number generator had a bias towards producing strings of nines and see if future data bore that out.

 

[T'ai Chi]

It might be more useful instead of giving stocks numbers, to give him/her a general method of calculating the probabilities. Kind of giving a fish vs teaching how to fish idea.

?

 

[Beth]

Yup, Beth is correct.

Wait a minute, your numbers still seem too high. Aren't there 45 ways to choose 2 out of 10? So .002 x 45 = .09

~~ Paul

Yes, you're right. I made an error in my combinations formula. Sorry.

 

[Beth]

I believe that is the correct answer to the first question, rather than the .1741 calculated by Beth. I also differ with Beth on the other three answers. I get, respectively, .0055 (not .0331); .0101 (not .1321); and .00002 (not .00004).

Yes, I made an error. Thanks for the correction. I now get the same as you except for question 3 - I now get .022 not .1321 or .0101.

 

[HappyCat]

Yes, you're right. I made an error in my combinations formula. Sorry.

I don't think that using (8/10)^28 * 45 works. It seems to be the intuitive answer to me as well, but then try the same method with only 9 events. Clearly there is not a 100% chance that 2 slots will be open, yet (8/10)^9 * 45 = 6.03979776. If this is not how you arrived at your answer, please post your math. I am interested to see how this problem is done.

 

[Beth]

I don't think that using (8/10)^28 * 45 works. It seems to be the intuitive answer to me as well, but then try the same method with only 9 events. Clearly there is not a 100% chance that 2 slots will be open, yet (8/10)^9 * 45 = 6.03979776. If this is not how you arrived at your answer, please post your math. I am interested to see how this problem is done.

Good point. I'll think about it some more.

Thanks.

 

[Ladewig]

I don't think that using (8/10)^28 * 45 works. It seems to be the intuitive answer to me as well, but then try the same method with only 9 events. Clearly there is not a 100% chance that 2 slots will be open, yet (8/10)^9 * 45 = 6.03979776. If this is not how you arrived at your answer, please post your math. I am interested to see how this problem is done.

I'm not sure of your notation so let me ask a question. Are you sure 45 is the right multiplier for this case. With nine events shouldn't one use 36 instead of 45?

ETA: oops, still too big. Nevermind.

ETA: After reading Paul's reply, I see I have misinterpreted your question completely. I thought you said nine slots instead of nine events

 

[Paul C. Anagnostopoulos]

I don't think that using (8/10)^28 * 45 works. It seems to be the intuitive answer to me as well, but then try the same method with only 9 events. Clearly there is not a 100% chance that 2 slots will be open, yet (8/10)^9 * 45 = 6.03979776. If this is not how you arrived at your answer, please post your math. I am interested to see how this problem is done.

With 9 slots you want to compute (7/9)^28 * 36, which is .0316.

~~ Paul

 

[Ladewig]

I don't think that using (8/10)^28 * 45 works. It seems to be the intuitive answer to me as well, but then try the same method with only 9 events. Clearly there is not a 100% chance that 2 slots will be open, yet (8/10)^9 * 45 = 6.03979776. If this is not how you arrived at your answer, please post your math. I am interested to see how this problem is done.

The problem with using 9 events is that 9<10 so the only way to have exact two empty slots is to have the first 8 events be placed in different spots and to have the last event placed in an already occupied slot.

Wouldn't the probability of the first 8 getting different slots be 10/10 x 9/10 x 8/10 x 7/10 x 6/10 x 5/10 x 4/10 x 3/10, and the probability of the last event hitting an already occupied slot be 8/10 for a product of .0145? Multiplying that times the 45 ways gives .652 which sounds like a reasonable number.

Raising (8/10) to a power produces too large of a number because it includes the possibility that all events fell into just 7 or just 6 or just 5 (etc.) categories.