A Slice of Bad Luck

[bob012345]

There are nine people at a party. Dessert is a variety of cheesecake slices on two platters. Each platter contains 12 slices of 6 varieties. The two platters are identical. What is the probability that someone will not get what they want?

 

[FactChecker]

Have you stated the problem completely?
1) Are the varieties equally likely to be wanted by people? If not, I don't think this can be answered.
If everyone wants the same variety, there are not enough of that variety, so the answer is 1.
2) Are there two of each variety on each platter?
3) What is the significance of dividing them into two platters? Do the people take one from each?

 

[pbuk]

Are there two of each variety on each platter?

... how many slices does each person want (24 slices and 9 people, if they want 3 each the probability is again 1)?
... many other missing constraints

@bob012345 you need to think more about what constraints are needed to give this problem a meaningful solution.

 

[bob012345]

... how many slices does each person want (24 slices and 9 people, if they want 3 each the probability is again 1)?
... many other missing constraints

@bob012345 you need to think more about what constraints are needed to give this problem a meaningful solution.

This is a real life problem. The party is tomorrow. In my OP I stated that there are 12 slices of 6 varieties on each platter. That is two slices of each kind per platter. There are two platters because each platter only holds 12 slices and I wanted more for the party. That’s just how they are sold.

In order to approach such a problem one has to make some assumptions. I think it’s reasonable to assume each variety is equally desired by people.

 

[Mark44]

how many slices does each person want (24 slices and 9 people, if they want 3 each the probability is again 1)?

Except that 3 * 9 = 27, but there are only 24 slices. So not everyone would get 3 slices.

 

[Hornbein]

I think it’s reasonable to assume each variety is equally desired by people.

Then everyone gets something they like.

 

[FactChecker]

In order to approach such a problem one has to make some assumptions. I think it’s reasonable to assume each variety is equally desired by people.

That's fine. But if you ask a physics forum for a numerical answer, you should state those assumptions.
In the real world, you give them the 2 platters and tell anyone who complains to shut up. ;-)

At least that is what I would do, but I have very few friends. ;-)

 

[bob012345]

I believe one can think of this problem like the birthday problem (in a group of $n$ people, what are the odds that at least $m$ share the same birthday where $m<n$). In this case the ‘birthday’ is the preferred variety of cheesecake and there are only 6 of them.

 

[anuttarasammyak]

You have four slices of one kind. Cut one slice to equal three pieces. Now you have 12 pieces > 9 people.

Another approach
3 people have 2 cakes. 6 people have 3 cakes.
Let the three, say a b c, choose their favorite by two turns, ordered abc cba.
Then let the six, say d e f g h i, choose by two turns ordered defghi ihgfed. The six draw lot for the final turn order.
Someone gourmet but less appetite may want to belong to the three.

 

[kuruman]

Experimental solution

1. Call each person and ask them what kind of cheesecake and how many pieces they want.
2. Order appropriately.
3. The probability that someone will not get what they want is zero.

 

[bob012345]

Experimental solution

1. Call each person and ask them what kind of cheesecake and how many pieces they want.
2. Order appropriately.
3. The probability that someone will not get what they want is zero.

I could buy just one platter and call each person offering them a choice among the available varieties. The first person can choose among the six…assorted possibilities happen…for the ninth person there are four slices left so they still have a choice. The last person has a choice which could be from two to four but doesn't know the first person had a choice of six. Everyone is satisfied and I only had to buy one platter!

 

[bob012345]

Then everyone gets something they like.

Not necessarily. This is referring to the quote “I think it’s reasonable to assume each variety is equally desired by people”. That statement does not mean each of the nine likes each variety the same. It means among the population as a whole each variety is equally popular.

 

[bob012345]

Here is what I got.

Spoiler

Well it turns out that there is about a 5.37% chance that someone will be unhappy and not get what they want. If I buy only one platter of cheesecake, that raises to an 84.25% chance someone will be unhappy. This is exactly the birthday problem using 6 for the number of days of the year which represents preference and using 5 for the number of people with the same preference. That’s because there are only four slices of a given kind so with 5 people preferring that someone is unhappy. For a case of only one platter, that number is 3 since only two slices of a given kind are available.

The exact numbers are $\frac{22549}{419904} ≈0.0537$ and $\frac{6551}{7776} ≈0.8425$

[\Spoiler]

 

[bob012345]

It turned out that only 7 of the 9 people came and I bought only one cheesecake platter, which had twelve slices, two of each kind. That computed to a 54% chance at least someone would be unhappy. There were at least 5 slices left over so maybe that’s why? Here is a good site with a birthday problem calculator;

https://www.dcode.fr/birthday-problem

 

[FactChecker]

There were at least 5 slices left over so maybe that’s why? Here is a good site with a birthday problem calculator;

This may say more about cheesecake than it does about the birthday problem. ;-)

 

[bob012345]

This may say more about cheesecake than it does about the birthday problem. ;-)

Someone brought really nice big cookies which might have skewed the experiment and also people are unpredictable as they don’t always obey the probabilities. But the cheesecake was good and I used the occasion to talk a little bit about math.

 

[sbrothy]

I hate cheesecake. Count me out.