interesting brainteaser

[boytoy]

Today the pouch contains a random quantity of marbles, and we’re going to withdraw a handful. But first, consider:

• If the bag contains an even number of marbles, then we are equally likely to withdraw an even or an odd number. For instance, if it contains 4 marbles, then we are equally likely to withdraw 2 or 4 as 1 or 3.
• But if the pouch contains an odd number of marbles, then we’re more likely to withdraw an odd number, as there’s one more way of choosing an odd number than an even number. For example, if the pouch contains 5 marbles then we’re more likely to draw 1, 3, or 5 than 2 or 4.

This is troubling. Without even opening the pouch we seem to have decided that, on balance, we’re more likely to withdraw an odd number of marbles than an even. Indeed, this seems to mean that handfuls in general are more commonly odd than even. How can this be?


taken from futility closet

 

[euroazn]

"Handfuls in general" are basically withdrawals from a set of infinite marbles, so the ratio is still 1:1.

Btw, with the huge volume of brainteaser threads, I seriously think it's time to make a subforum for this...

 

[pruse]

"But if the pouch contains an odd number of marbles, then we’re more likely to withdraw an odd number, as there’s one more way of choosing an odd number than an even number. For example, if the pouch contains 5 marbles then we’re more likely to draw 1, 3, or 5 than 2 or 4."

This is wrong. Whether (n) is even or odd, we always have

(\binom{n}{1}+\binom{n}{3}+\binom{n}{5}\cdots = \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots)

 

[Joy Pathak]

Btw, with the huge volume of brainteaser threads, I seriously think it's time to make a subforum for this...
It's on my to-do list so you will see them soon-ish.

Yes. That would be very usefull. It would be nice to have the brainteasers tagged well for searches.

 

[euroazn]

Nice one peterruse! I'm such a klutz
I fail at even simple binomial distributions

 

[koupparis]

"But if the pouch contains an odd number of marbles, then we’re more likely to withdraw an odd number, as there’s one more way of choosing an odd number than an even number. For example, if the pouch contains 5 marbles then we’re more likely to draw 1, 3, or 5 than 2 or 4."

This is wrong. Whether (n) is even or odd, we always have

(\binom{n}{1}+\binom{n}{3}+\binom{n}{5}\cdots = \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots)

That is true if you use the same number of terms on each side. Also we're not given the option of drawing 0 marbles.

If we were given that option, then for an even number, say 4, we could draw 0, 2 or 4 as even sets but only 1 or 3 for odd sets. Thus the question is reversed, but remains the same.

Also I don't think "handfuls in general" should refer to drawing from infinite sets. It should refer to the aggregate characteristics of all drawings from all sets (even, odd and infinite).

This seems more like a philosophical question, than a brainteaser ...:-k

 

[pruse]

That is true if you use the same number of terms on each side.

No, that identity is always true. For example,

(\binom{2}{1}=\binom{2}{0}+\binom{2}{2})

even though the numbers of terms differ.

If 0 marbles are allowed, then the claim in the original post is wrong, and both odd and even are equally likely.

If 0 marbles are not allowed, then the probability of removing an odd number of marbles from a random pouch is indeed greater than 0.5, because from the identity we see that there are always more ways to remove an odd number than an even number, whatever the number of marbles in the pouch.

 

[euroazn]

Oh, true - I don't think 0 marbles are allowed (from the phrasing of the original Q.) =S

 

[koupparis]

No, that identity is always true. For example,

(\binom{2}{1}=\binom{2}{0}+\binom{2}{2})

even though the numbers of terms differ.

If 0 marbles are allowed, then the claim in the original post is wrong, and both odd and even are equally likely.

If 0 marbles are not allowed, then the probability of removing an odd number of marbles from a random pouch is indeed greater than 0.5, because from the identity we see that there are always more ways to remove an odd number than an even number, whatever the number of marbles in the pouch.

Yes I was pairing up the combinations and not looking at their sum :-\"

I don't think the marbles are distinct though, so counting permutations doesn't make sense.

Dunno ... bad question.

Actually come to think of it.

For even sets:
Picking an even number of objects creates a new even set and leaves behind an even set. Thus even sets increase by 1.
Picking an odd number of objects creates a new odd set and leaves behind an odd set. Even sets decrease by one, and odd sets increase by 2.

For odd sets:
Picking an even number of objects creates a new even set and leaves behind an odd set. Thus even sets increase by 1.
Picking an odd number of objects creates a new odd set and leaves behind an even set. Thus even sets increase by 1.

So it seems that the odd sets will generate more even sets, but the even sets generate more odd sets ... Does this balance out?

Turns out it does. On average evens create 1 odd set. And on average odds create 1 even set. Hence if we start with an equal proportion, on average we'll end up with an equal proportion.

However, if this process continues indefinitely, we'll end up with a bunch of singletons, and hence more odd sets.

Like I said, bad question. Done thinking about it, edited this post 3 times.

 

[euroazn]

Yes I was pairing up the combinations and not looking at their sum :-\"

I don't think the marbles are distinct though, so counting permutations doesn't make sense.

Dunno ... bad question.

Actually come to think of it.

For even sets:
Picking an even number of objects creates a new even set and leaves behind an even set. Thus even sets increase by 1.
Picking an odd number of objects creates a new odd set and leaves behind an odd set. Even sets decrease by one, and odd sets increase by 2.

For odd sets:
Picking an even number of objects creates a new even set and leaves behind an odd set. Thus even sets increase by 1.
Picking an odd number of objects creates a new odd set and leaves behind an even set. Thus even sets increase by 1.

So it seems that the odd sets will generate more even sets, but the even sets generate more odd sets ... Does this balance out?

Turns out it does. On average evens create 1 odd set. And on average odds create 1 even set. Hence if we start with an equal proportion, on average we'll end up with an equal proportion.

However, if this process continues indefinitely, we'll end up with a bunch of singletons, and hence more odd sets.

Like I said, bad question. Done thinking about it, edited this post 3 times.

..or maybe... it's a GREAT question

 

[pruse]

Yes I was pairing up the combinations and not looking at their sum :-\"

I don't think the marbles are distinct though, so counting permutations doesn't make sense.

Dunno ... bad question.

I was looking at those sums because to me the question seemed to imply that all "handfuls" of marbles are equally likely, small and large ones alike. (How big is your hand??) Of course, that's an unreasonable assumption. To make the question more realistic, we need a reasonable distribution on the number of marbles drawn, given the number of marbles in the pouch.

 

[koupparis]

..or maybe... it's a GREAT question

A great question for debate possibly

 

[Andy Nguyen]

Btw, with the huge volume of brainteaser threads, I seriously think it's time to make a subforum for this...

Done
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