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A Slice of Bad Luck
Дата публикации: 13-10-2026 15:51:44
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- Level: High School
- Thread starter Thread starter bob012345
- Start date Start date Oct 8, 2026
Discussion
... many other missing constraints
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 the real world, you give them the 2 platters and tell anyone who complains to shut up. ;-)
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... 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.
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... 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.
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Hornbein
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Then everyone gets something they like.I think it’s reasonable to assume each variety is equally desired by people.
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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. ;-)
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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.
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Experimental solution
- Call each person and ask them what kind of cheesecake and how many pieces they want.
- Order appropriately.
- The probability that someone will not get what they want is zero.
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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!Experimental solution
- Call each person and ask them what kind of cheesecake and how many pieces they want.
- Order appropriately.
- The probability that someone will not get what they want is zero.
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bob012345
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Here is what I got.
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]
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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;
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