Tag Archives: Probability

In like a Lion; Out like a Lamb?

This morning as I was walking to work through whipping wind and freezing cold temps, I thought to myself “March, you’re supposed to be going out like a lamb.”  Goodness knows it sure came in like a lion!  Then, I started thinking about Groundhog’s Day and how unreliable that good old Phil actually is!  That made me wonder . . . is there really any truth to this whole lion/lamb thing?

I spent the better part of my morning trying to track down an answer!

This is what I did:

I’m only concerned with how March comes in and out, and not what happens in the middle of the month, so I thought I’d look at the first 7 days of March and the last 7 days of March.  Then, I did a quick search and found that the average March temperature in Iowa for the past 150 years is 34.5 degrees Fahrenheit.

Given this information I decided I would define “Lion” to be a 7-day span in which 4 or more of the days had an average temperature that was less than the average monthly temperature.  Then, a “Lamb” was a 7-day span in which 4 or more of the days had an average temperature that was greater than or equal to the average monthly temperature.

I obtained daily average temperature data for Des Moines from The University of Dayton Average Daily Temperature Archive dating back to 1995.  Because the month of March isn’t over yet for 2014, I didn’t use the temperature data for any of the days in March 2014.  This is what I came up with:

Screen Shot 2014-03-26 at 12.05.03 PM

In the past 19 years March has followed the “In like a Lion; Out like a Lamb,” pattern in 13 different years (or 68% of the time).  It has followed an “In like a Lamb; Out like a Lamb” pattern 5 different years (or 26% of the time).  And once in the last 19 years March has come in like a Lion and gone out like a Lion . . . according to the average daily temperature in Des Moines.

This got me thinking about a few things:

1. This saying seems to be a little more accurate then the Groundhog shadow thing.

2. There weren’t any “In like a Lamb; Out like a Lion” years . . . I wonder how many times that has happened in the past 150 years (if at all)?

3. What do you think of my definition for Lion-like weather and Lamb-like weather?  Would you define it another way?  If so, how?

M&M’s Revisited (for the Last Time!)

If you haven’t been here before, then you don’t know that we’ve already talked about M&M’s twice (here and here) and you don’t know that we’ve talked a little bit about the colors of M&M’s in the bags.

Well, today I want to keep talking about the different colors of M&M’s in the bags.  Except today I want to talk about the percent of M&M’s that are red, orange, yellow, green, blue, brown.  Before we continue our M&M discussion, do you have a guess?   That is, what percent of the M&M’s manufactured are red, orange, yellow, green, blue, brown?

Hmmm . . . Let’s pretend that we don’t know (maybe you really don’t!).  I think a pretty educated guess would be that 16.67% of the M&M’s are red, 16.67% of them are orange, 16.67% yellow, etc., etc.  Can you live with that guess?

I’m going to use the data I collected in my last M&M post, except instead of individual bags I’m going to look at my entire sample of M&M’s.

Here’s the percentage breakdown of M&M’s:

Screen Shot 2013-11-26 at 9.52.31 AM

Let’s make a nice table, based on what I would expect to get, given my educated guess of 16.67% of each color and what I actually got:

Screen Shot 2013-11-26 at 9.52.41 AM

So, I wonder if the distribution of colors I got in my sample would be likely, if the colors of M&M’s really were distributed evenly at the manufacturer?

Luckily for us there’s a statistical test we can use to answer that exact question.  And, luckily for us its a pretty straightforward test to understand!  It’s called the Chi-Square Goodness of Fit Test.  The Chi-Square Goodness of Fit test compares the observed values (in our case my M&M colors) to the expected values (if our initial assumption was true).  In our case we would subtract the expected value from the observed value and square the difference.  Then, we would divide by the expected value.  We’d do this for each color of M&M and add up the results.  Don’t worry, I’ll do it (actually, I did it with the help of this website). . .

Based on the Chi-Square Goodness of Fit Test it’s fairly reasonable to assume that I could have gotten this distribution of M&M colors given the fact that M&M Mars makes 16.67% of each color of M&M’s.

Screen Shot 2013-11-26 at 10.10.29 AM

So, here’s my next question?  Do they?

(So here’s the thing, about 5 years ago the M&M Mars website used to answer this exact question, but in 2008 they stopped.  This person wrote to M&M’s and posted the response)

Use the distribution for Milk Chocolate M&Ms detailed by M&M Mars and run another Chi Square Goodness of Fit Test with my data (or your own, if you collected any).  How does this compare to the 16.67% guess?


This is the second post in a trilogy!  Read the first post here.  And when you’re done with this post, read the last post here.

Have you been counting chocolate?


Did you eat the chocolate you counted?

Even better!

If we pick up where we left off last week, you’ll recall the M and M debacle (the debacle being the fact that my children fought, not the fact that I fed them chocolate before 8am)!

You remember the fight, don’t you?  You know, the one where my daughter said it wasn’t fair that her brother got all the colors of the M&M’s in his bag and she didn’t?

Then, I asked you to try to figure out how likely that situation was to occur?  And then . . . I asked you to buy lots and lots of M&M’s?

Well, did you?  I did!

See . . .


The question that led to all of this M&M consumption was “How likely is it that a bag of fun sized M&M’s will contain exactly two colors of M&M’s?”

I asked you to start by looking at a sample of M&Ms.

My sample contained 17 bags of fun sized M&M’s and 299 candies.    When collecting the data to answer this question, I did two things.

1) Counted the total number of M&M’s in the bag.

2) Counted the number of each color of M&M in the bag.

My piles looked like this:

photo copy photo

And the data I collected looked like this:

Screen Shot 2013-10-25 at 1.33.28 PM

If you look closely, you’ll see that many times during the data collection phase I opened a bag of M&Ms that was missing one color.  In fact, this particular event occurred approximately 35% of the time.

Screen Shot 2013-10-25 at 1.36.03 PM

And when a color was missing, 33.3% of the time it was the blue or brown M&Ms that were missing and 16.67% of the time it was the orange or yellow M&Ms that were missing.

In my study, two or more colors were missing 0% of the time.

Considering this data, it seems very, very unlikely that my daughter would have opened a bag of M&M’s containing only two colors of M&M’s.  In fact, in my data collection this occurred 0% of the time.

So, here’s my next question; since the instance my daughter described happened 0% of the time during my study does it mean it will never happen?  How do you know?

Could I use this data to make an educated guess that my daughter my have just been picking a fight with my son, and that her bag of M&M’s did, in fact, contain more than just two colors of M&M’s?

To be continued . . .

M and M’s

This actually ended up becoming a 3 part series . . . all devoted to M&M’s!.  This is the first post in the series, but you can read the other two here and here.

This morning a box of fun-sized M&M packages got delivered to our house (thanks to my mother-in-law aka Grandma).  Although this seems like a thoughtful gesture, its a major problem.  Why?  You might ask?  Easy.  I have absolutely no will power when it comes to M&M’s, so much so, that a package of M&M’s may have been opened at breakfast this morning.  And a package may have also been opened for my 7 year-old.  And my 5 year-old.  And my 2 year-old.  (Don’t be impressed with my husband, the only reason he didn’t open a package is because be was already at work!)

Chocolate in the morning.  Everyone should be in a good mood, right?  Even better is chocolate covered in brightly-colored candy coating.  Unless, of course, that candy coating happens to lay the groundwork for a fight between two pint-sized humans.

“Mom! Jack’s package of M&M’s has all the colors and mine only has two!  That’s not fair!  Tell him that’s not fair!’

Hmmm, I’m not sure that “not fair” is the phrase that should be used to describe this particular incident.  How about “Mom!  That’s not likely!”

I know, I know “not likely” is not nearly as moving as “not fair,” but this is a blog about mathematics after all.  Fairness is not something that I can speak to in this context, but likely, now that’s something we can talk about!

Here’s what I want you to do.  And you have to promise to not cheat.  (Promise).  Buy a few bags of M&M’s this weekend.  As many as you see fit and go ahead and try to determine how “likely” it is that someone will get all the colors of the M&M’s in one fun sized bag.  Also, see if you can’t figure out if M&M Mars makes equal numbers of red, orange, yellow, green, blue, and brown M&Ms.

Remember, you already promised you wouldn’t cheat.  So I’m trusting you not to just Google this as soon as you’re done here.  Besides, you’d get to eat the project when you’re done.  If you just Google it, there won’t be anything good to eat.

I’m going to answer this question too.  And I’ll show you how I went about answering it . . . next week.  In the meantime, send me your methods.  Let’s see what we can come up with!

And the Winner is . . .

So, did you see that last week some lucky person in South Carolina won the $400 million Powerball Jackpot? (Technically, it was $399.8 million, but $400 million is close enough!)

It seems like every time there’s a big jackpot won in the lottery you read statements like “you’re more likely to be struck by lightening,” or “you’re more likely to marry a prince,” or in the case of the CNN story about this particular Powerball winner “you’re more likely to get struck by lightening and bitten by a shark.” (Talk about a bad day!)  They also go an to say that the chances of winning a Powerball Jackpot are 1 in 175223510.  Don’t you wonder how people come up with all of these statistics?

Let’s take a look at how Powerball is played . . .

According to the Powerball website, lottery numbers are drawn from two drums.  The first drum contains 59 white balls and the second drum contains 35 red balls (These red balls are all potential Powerballs).  The jackpot is won by matching all five white balls in any order and the red Powerball.

In order to calculate the odds of winning, we need to figure out the odds of matching all five white balls, in any order, and the red ball.

Let’s start with the white balls:

I like to think about situations like the one described above by picturing an empty (in this case) lottery ticket.  Like this:

Screen Shot 2013-09-24 at 8.56.35 AM

The first ball that pops up could be any of the 59 balls, the second ball could be any of the 58 balls, the remaining blank spots on the ticket will be filled by drawing from the final 57, 56, and 55 balls respectively.

That looks like this:

Screen Shot 2013-09-24 at 8.58.23 AM

Now, the order doesn’t matter in the way I arrange the balls, remember?  That means if the winning white balls are 1, 2, 3, 4, 5 and my ticket is 2, 3, 4, 1, 5; I’m on my way to winning the Powerball!  So now we need to figure out how many different ways the 5 white ball numbers can be arranged.

The first white number could be any of the 5 numbers drawn from the drum, so I have 5 choices for the number in the first spot.  I only have 4 remaining numbers for the second spot, 3 for the 3rd spot, etc . . .

Because I can rearrange the numbers and still have a winning ticket, the possible number combinations I need to win has just been decreased!  Now, we can calculate the number of ways to get a winning combination from the white balls in the Powerball drawing:

The total number of combinations is :

Screen Shot 2013-09-24 at 9.05.05 AM


And now for the red Powerball!  The process we used above is going to be the same for red Powerball, except instead of having to match 5 numbers you only have to match 1 and because there’s only one number to be matched it doesn’t really make sense to talk about whether or not the order matters . . . there’s only one number.

Since you know a method to use and you know the answer, I’m pretty confident you are going to be able to figure how CNN could report that a person has a 1 in 175223510 chance in winning.

Good Luck!

P.S. I used a few different techniques from Discrete Mathematics or Counting Theory in this post that I didn’t explicitly name.  First, as in the case of the white lotto balls I was choosing 5 balls from a collection of 59 balls.  The order in which I arranged these 5 balls was not important.  This situation describes a combination or a binomial coefficient.  There is a formula associated with these types of situations.  We used this formula, although I did lots of canceling to make the numbers used look less overwhelming.  You could rewrite the situation that I described above as a combination using the formula provided from the combination link and see if you can get it to look like the one I used.

Also, we calculated the factorial of 5, denoted 5!.  If you attempt the challenge I’ve given you above, you’ll want to make sure you know what a factorial is.