Possibilites of dropping 4 coins in 4 cups: chance (expectation?) of empty cups

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RichardDL

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I made this up to remind/teach me some stats.

There are 4 coins and 4 cups to drop them in. They drop in randomly and equally. I'm interested in the chance (expectation?) of empty cups.

There are 4*4*4*4 = 256 possibilities of how they land. I've used a computer program to list them.

24 have 0 empty cups
144 have 1 empty cup
84 have 2 empty cups
4 have 3 empty cups
----
256

The expected number of empty cups for one go is (24*0 + 144*1 + 84*2 + 24*3) / 256 = 384/256 = 1.5

Does that mean the chance that any cup is empty is 1.5/4 = 0.375 ?

A different way (2):

For each coin the probability of landing in a particular cup is 0.25
A cup has 4 coins with that chance of landing in it.
What chance that it’s empty? (1 – 0.25) ^ 4 = 0.75^4 = 0.316

Another way (3):
Look at the list of results, how many have cup1 empty? 81 81/256 = 0.316
(It's the same for each cup).

I'm thinking that all three should agree. Seems to be the 0.375 is wrong, but why? Anything else wrong?
 
Well, I was hoping for some guidance. I think you are agreeing with the calcs up to "expected = 1.5 empty cups". From that is it possible to know the chance of one cup being empty?
 
Last edited:
RichardDL said:
I made this up to remind/teach me some stats.
There are 4 coins and 4 cups to drop them in. They drop in randomly and equally. I'm interested in the chance (expectation?) of empty cups.
There are 4*4*4*4 = 256 possibilities of how they land. I've used a computer program to list them.
24 have 0 empty cups
144 have 1 empty cup
84 have 2 empty cups
4 have 3 empty cups ---- 256
Click to expand...
Even though you made up this question it shows that there is a great deal that yon do not understand about counting theory
This question falls into the topic of occupancy: problems:
the number of ways that [imath]\large\bf {n\text{ objects}}[/imath] can occupy [imath]\large\bf {m\text{ cells}}[/imath]
Now here is what you did not tells us, Are the coins (the objects) distinguishable or not?
Are the cups (the cells) distinguishable or not?
The number of ways to place M distinguishable objects into N distinguishable cells is [imath]N^M[/imath].
That is also the number of functions from a set of M to a set of N. That seems to be the case here.
However, it is usually the case that coins are not distinguishable but cups(cells) are distinguishable.
Assuming in this case both cups and coins are distinguishable there are [imath]4^4[/imath] ways to put the coins in the cups.
However that counts all cases even empty cells. If you want certain to be empty or not then we must study the topic Stirling Numbers.
The number of ways of placing [imath]4[/imath] distinguishable coins into [imath]4[/imath] distinguishable cup none of which none is empty is simply [imath]4!=24[/imath] ways.
Now beyond there things are much more complicated. To have one and only one empty there are four way to pick the empty cup.
After that we need to know the number of ways to put [imath]4[/imath] distinguishable coins into [imath]3[/imath] distinguishable cups none of which is empty. That is the number of surjections from a set of four to a set of three.
[imath][/imath][imath][/imath][imath][/imath]
 
@pka. Thank you for that, sorry not to have been completely clear. The cups are indistinguishable, and so are the coins. N^M = 4^4 = 256 agreed. A cup may be empty or have all 4 coins. (I agree with your comments on distinguishable, but that's another situation.) Is my method of calculating the expected number of empty cups valid ...384/256 = 1.5 ?

Is there a way of going from the expected no. of empties to the chance of 1 cup being empty? It seems that 1.5/4=0.375 is not appropriate.
My calculations (2) and (3) agree, so presumably are correct?

I looked at Wikepedia on Stirling Numbers, I regret I've never found it easy to switch between that algebraic presentation and simple examples. Another page mentioned 'labelled' whereas I'm using indistinguishable.
 
Last edited:
RichardDL said:
I made this up to remind/teach me some stats.

There are 4 coins and 4 cups to drop them in. They drop in randomly and equally. I'm interested in the chance (expectation?) of empty cups.

There are 4*4*4*4 = 256 possibilities of how they land. I've used a computer program to list them.

24 have 0 empty cups
144 have 1 empty cup
84 have 2 empty cups
4 have 3 empty cups
----
256

Does that mean the chance that any cup is empty is 1.5/4 = 0.375 ?
Click to expand...
Shouldn't p(empty cup) = 24/256 using the numbers that your program gave you?
Can we see how you came up with 24?
 
@Steven G, Of the 256 possibilities 24 have no empty cups i.e. the 4 coins have fallen one in each cup. Results attached. The columns are:
- count 1..256
- number of cup (0..3) that coin 0 fell into
- number of cup (0..3) that coin 1 fell into
- number of cup (0..3) that coin 2 fell into
- number of cup (0..3) that coin 3 fell into

- count of empty cups
- cup 0 is empty 0..1 1=empty (totalled at end = 81)

1 0 0 0 0 3 0
2 0 0 0 1 2 0
3 0 0 0 2 2 0
4 0 0 0 3 2 0
5 0 0 1 0 2 0
...
256 3 3 3 3 3 1
81
 

Attachments

  • Results.txt
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@RichardDL, I think that you intend that the cups distinguishable and not the coins.
For if both are indistinguishable the there are just four ways to do it:
[imath]1111,~0112,~0013,\text{ or }0004[/imath] i.e. from zero to three empty cups.
If on the hand you have a first, second. third, & fourth cup then they are labeled & hence distinguishable.
In which case we can model with the string [imath]o\,o\,o\,o\,|\,|\,|[/imath] which can be rearranged in [imath]\dfrac{7!}{4!\cdot 3!}=35[/imath] ways.

[imath][/imath][imath][/imath]
 
@pka, yes, you're right, cups distinguishable by their position. I'm not familiar with that "o..." syntax, or how the 35 ways relates to my 256 results.
 
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