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 choose their favorite by two turns.
Then let the six choose among 18 left by three turns.
Someone gourmet but less appetite may choose to be 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.