Posted by: caitlincrump | October 10, 2008

Who likes cake?

Well, I saw this problem and I thought it was very interesting. So, I thought I would include it in my posting this week.  The problem is as follows:

It’s time for dessert at a party, and everyone wants dessert, (duh).  A total of 25 people at a party want cake, 16 people want vanilla ice cream, and 10 want apple pie.  Additionally, 6 people want both cake with ice cream, 9 want apple pie with ice cream, and 4 people want all three desserts at once.  How many people are at the party?

I thought the answer to this question would be fairly intuitive, but if you just think about the question, it can be really confusing!  So, I did some research and found an EASY way to figure it out.

First of all, they are a little tricky with the actual question. Since everyone at the party wants desert, then what you are really trying to find out is how many people want at least one dessert.  One way we can look at this question that will make it easy to solve is the following: if we pick one person, what is the probability that they want at least one dessert?  Since everyone wants a dessert – cake, ice cream, or pie – then we know our answer is going to be 1/N, where N is the number of people who want a dessert.

A mathematical proposition I found was the following: P(E1 U E2 U… En) = P(E1) + P(E2) +… P(En) – P(E1 E2) – P(E2 E3)… – P(En-1 En) + P(E1 E2 E3)

NOTE:  In my problem any elements in a row of a probability, such as P(E1 E2), represents the intersection of those elements.  Thus, P(E1 E2) is the probability that they like both E1 and E2.  I’ve never been taught to use the intersection symbol and it just drives me crazy… Sorry for the confusion.

I wrote this proposition making it consist of only three elements, since we have three probabilities we are concerned with here, the probability of liking each respective dessert.  However, you could expand upon this proposition to encompass more elements.  You only need to alter between addition and subtraction when you increase the number of intersections in your probability (which is done in the example).

Anyway, back to the problem.  So, I used  this proposition, and suddenly the problem became very, very easy.

We know our probability is 1/N, but with the proposition we can rewrite this as

1/P(C U I U P) = 1/[P(C) + P(I) + P(P) – P(C I) – P(I P) + P(C I P)]

where C = wanting cake, I = wanting ice cream, and P = wanting pie.

Now, we have these probabilities, because they are the number of people given in the problem!  So…

1/N = 1/[25 + 16 + 10 – 6 – 9 + 4] = 1/40

This means that if we pick a random person from the party, there is a 1/40 chance that they want at least one dessert.  We know no one can pass up dessert at a birthday party, so everyone wants at least one dessert!  This means there are 40 people at the party!  Woo-hoo!

I hope this was understandable… I just thought it was a really cool way to solve the problem, because it looks very complicated, but it is quite simple.  If you have any questions, just comment away.  I’ll respond.  That is, if I know the answer or help you find it.

Responses

1. I know itʼs a little bit late, but I just recently found this blog. I have some issues with the solution to this problem, though.

caitlincrump, October 10, 2008:
> if we pick one person, what is the probability
> that they want at least one dessert? Since
> everyone wants a dessert – cake, ice cream, or
> pie – then we know our answer is going to be
> 1/N, where N is the number of people who want a dessert.

In my opinion the probability, that a randomly picked person wants dessert, is one, as it is premised, as everyone at the party wants a dessert. I donʼt understand your reasoning here.

> A mathematical proposition I found was the
> following:
> P(E1 U E2 U… En) = P(E1) + P(E2) +… P(En)
> – P(E1 E2) – P(E2 E3)… – P(En-1 En)
> + P(E1 E2 E3)

I donʼt know how you arrived at this proposition or found it, but isnʼt it faulty? Isnʼt it more like:
P(E₁ ∪ E₂ ∪ E₃) = P(E₁) + P(E₂) + P(E₃) − P(E₁ ∩ E₂) − P(E₁ ∩ E₃) − P(E₂ ∩ E₃) + P(E₁ ∩ E₂ ∩ E₃)
That would be a major flaw in your argumentation, as you forget to include P(E₁ ∩ E₃), which in this case equals P(E₁ ∩ E₂ ∩ E₃).

> We know our probability is 1/N, but with the
> proposition we can rewrite this as
>
> 1/P(C U I U P) = 1/[P(C) + P(I) + P(P)
> – P(C I) – P(I P) + P(C I P)]
>
> where C = wanting cake, I = wanting ice cream,
> and P = wanting pie.
>
> Now, we have these probabilities, because they
> are the number of people given in the problem!
> So…
>
> 1/N = 1/[25 + 16 + 10 – 6 – 9 + 4] = 1/40

How can it be, anyway, that the probabilities you used exceed 1? I found no viable way of interpreting this.

> This means that if we pick a random person from
> the party, there is a 1/40 chance that they
> want at least one dessert. We know no one can
> pass up dessert at a birthday party, so
> everyone wants at least one dessert! This means
> there are 40 people at the party! Woo-hoo!

Once more, I think the chance of someone to want at least one dessert is 100%. If it were only 1/40 and the party hosted 40 people, then the chance, that there was 1 among 40 picked people, who wants at least one dessert, equaled 1 resp. 100%.

I used a different means of reasoning and came to a different conclusion (skip to END_OF_SOLUTION if uninterested). I hope this sketch is of argumentative aid (and I hope it renders well with non-monospaced fonts):

(C)ake: 25(21)(19)
|
|
CV: 6(2) —————
| |
| |
(V)an: 16(12)(10)(5) CVA 4
| |
| |
VA: 9(5) —————
|
|
(A)pple: 10(6)(1)

There shall be 5 sets, which represent the groups of people at the party:
C, V, A ((C)ake, (V)anilla ice cream, (A)pple pies),
CV ((C)ake AND (V)anilla ice cream),
VA (analogous to CV),
CVA (analogous to CV and VA) and
E (everyone).
The following holds:
#(C) = 25, #(V) = 16, #(A) = 10
#(CV) = 6, #(VA) = 9, #(CVA) = 4
Because #(CVA) = 4, #(E) ≥ 4. But obviously the following statements are also true:
CVA ⊂ C, CVA ⊂ V, CVA ⊂ A
CVA ⊂ CV, CVA ⊂ VA
That however means, that:
#(CV \ CVA) = #(CV) − #(CVA) = 6 − 4 = 2
#(VA \ CVA) = #(VA) − #(CVA) = 9 − 4 = 5
Now obviously (CV \ CVA) and (VA \ CVA) do not share any people anymore, meaning their intersection would be empty:
((C ∩ V) \ (C ∩ V ∩ A)) ∩ ((V ∩ A) \ (C ∩ V ∩ A)) = ((C ∩ V) \ A) ∩ ((V ∩ A) \ C) = ∅
So now, that we know, that
CVwCVA ≔ CV \ CVA and
VAwCVA ≔ VA \ CVA
contain people who are actually distinct from CVA and from each other, we can safely assume, that
#(E) ≥ #(CVA) + #(CVwCVA) + #(VAwCVA)
= 4 + 2 + 5
Now we only need to ascertain how many people there are in the sets C, V and A that we havenʼt counted so far. Because we know, that
CVA ⊂ C, CVA ⊂ V, CVA ⊂ A
we can say, that
#(C \ CVA) = 25 − 4 = 21
#(V \ CVA) = 16 − 4 = 12
#(A \ CVA) = 10 − 4 = 6
and because we know, that neither CVA, CVwCVA nor VAwCVA share any people (the intersection of each pair of these sets is empty) and that
CV ⊂ C, CV ⊂ V, VA ⊂ V, VA ⊂ A
we can exclude further:
#((C \ CVA) \ CVwCVA) = 25 − 4 − 2 = 19
#(((V \ CVA) \ CVwCVA) \ VAwCVA) = 16 − 4 − 2 − 5 = 5
#((A \ CVA) \ VAwCVA) = 10 − 4 − 5 = 1
Thus we obtain:
E ≥ #(CVA) + #(CVwCVA) + #(VAwCVA) + #(C \ CVwCVA) + #(V \ CVwCVA \ VAwCVA) + #(A VAwCVA)
= 4 + 2 + 5 + 19 + 5 + 1 = 36
I canʼt possibly think of any people I missed. Thus E = 36.

END_OF_SOLUTION

This clearly does not equal your solution to the problem. However, taking into account that it seems you forgot to subtract P(E₁ ∩ E₃) (which would, superficially following your argumentation (once more, I donʼt understand it), equal 4), your “corrected” result actually equaled mine.

As you (assumingly) understand your line of argumentation, I would be delighted to learn where the differences in our deductions come from and if thereʼs either an error in your or in my argumentation… or maybe Iʼm just taking it too seriously.

Regards, Shiranai

2. Well, seems like the sketch didnʼt render well after all. Nevertheless: There should be lines connecting C-CV, CV-V, V-VA, VA-A, CV-CVA and VA-CVA.

Regards, Shiranai

3. Well, I also got 36, because it’s small enough to do the naive way: essentially what Shiranai did (mathematicians use the word ‘naive’ without the disparaging connotation: it means the most direct and elementary approach).

But the author’s solution is more exciting, a more unusual approach and one that generalises to larger examples where it would save a lot of tedious work. A little problem: yes, the formula has an error and forgets the term P(E1 E3), which, if added in, corrects the mistake and gives 36. (The way the question is stated is confusing here: if 4 people want all three, then clearly 4 people want cake and apple pie. But the question doesn’t mention that.)

I’m also a little confused why the solutions writes 1/N, but the formula used is essentially a formula for the sizes of sets (since probability is largely set theory), so you can forget the probability if you like and just talk about cardinalities of sets. A formula of this type (in various guises) is often called ‘inclusion-exclusion’ and it’s useful all over the place.