Those "Guess How Many X's Are in the Jar" Games - Is There a Trick?

According to my calculations, you have an asshole the size of Manhattan.

Amazing that you found this. Where can I apply for work as an M&M imager?

Surely “packing efficiency” would only count towards a volume calculation, and not an array calculation.

I read this as, “Where can I apply for work as an M&M engineer.” No clue for either of them, though. :wink:

I read that link quite a time ago, it’s one of those things that just sticks in your head. I may have posted about it, if the search function was working I’d try to find the thread. There was an accompanying article that was a bit more scientific, but I can’t seem to find the link for that one anymore.


That’s what I thought when I read this.
A math approach may be to count the M&M around the bottom giving you the circumference. Square this then multiply this by the height and divide by 4pi. I don’t think this would have to be multiplied by 0.74.

**GuanoLad **was saying what JoeyP said, i.e.

And I think they are right. If you are counting how many M&Ms are present in a given direction, then that is how they are packed, so you don’t need to factor in a packing efficiency.

The packing efficiency figure describes how much of the total volume available can be taken up by M&M’s. If you assume 74%, then you would:

  1. Calculate volume of jar

  2. Multiply volume of jar by 0.74

  3. Divide the result by the volume of one M&M.

  4. Profit!!!

I won. :cool:

So what was your guess, what was the actual number, and how close were your nearest competitors? Did anyone go over? What did you win?

*** Ponder

The Dope triumphs!!!

Yeah, that makes sense. It’s probably be best to do each direction three times and average (or use the middle) result to account for randomness.

Congrats, Homie! I’ll take my fee in M&Ms, please! :wink:

Wait a minute!

There’s no x in jar.

My experience with M&Ms that were warmed in the packet (they don’t stay in my hand long enough to get warm there) is that the heat expansion of the chocolate ruptures the candy shell, which is then further displaced by the flow of the melted chocolate. I’m guessing that this happens because the shell does not expand as much with heat, and because the shell shape is weak when force is applied from inside.

Even if it didn’t rupture from the expansion of the chocolate, an M&M with a melted chocolate interior would have very little structural integrity. The slightest pressure and it will break and smear. Those shells just aren’t that strong without the solid chocolate to back them up.
Sorry for the hijack** HeyHomie**. Congratulations on your win. May your estimating skills ever increase.

Then how are we going to get the elephant in there? :smiley:

[spoiler]How do you fit an elephant into a Safeway bag?

Take the S out of Safe and the f out of way.

There’s no f in way![/spoiler]

For any uniform shape or volume, just fill up a known volume container with items in question (M&Ms, jelly bellys, etc) to the line and then fill the rest with water. Quickly separate the two mediums with a strainer and measure the water. Then solve for the packing efficiency. Repeat several times to get a more precise P.E. Then apply that P.E. to the container in question.

Are you implying that the Teeming Millions would try to catch someone on a technicality? Noooo, not us.

ETA, the questions was “How many X’s are in THE jar?”. The answer would still be zero, as the only letter contained in said jar was ‘m’.

I’m pretty sure I saw a bunch of W’s in there too.

Well, the OP did ask if it was a trick.

You can’t fool me that easy. There’s no x in the, either. And no m!

There are four lights!

Rats. Foiled again.

My trick would be to buy an identical jar and bags of the same sweets. Fill the jar and count how many you pull out until the jar is empty. Costly, time consuming and pathetic! Oh and it may not work - you could be one sweet out and someone else may snatch the prize! Worth a try though, eh? :smiley:

There were four jars: one Skittles, one M & M’s, one off-brand malted milk balls, and one a mix of candy corn and candy pumpkins.

My guess on the Skittles (which I won): 1,468. Actual number of Skittles: 1,489. I was within 21 Skittles (I’ll leave it to a Doper better at math to figure out how many percentage points I was away). The other guesses weren’t revealed. For correctly guessing how many Skittles were in a jar I won… a jar of Skittles.

I have no idea what my guesses were on the other products. I did hear that, on the M & M’s, there were two identical guesses, and the winner was determined by a flip of a coin.