# Water Balloon Fight

#### veronica1999

##### Member
7 people are having a water ballon fight. At the same time, each of the 7 people throws a water balloon at one of the other 6 people, chosen at random. What is the probability that there are 2 people who throw the balloon at each other?

I am having a lot of trouble on this one.

First, I calculated the total number of outcomes and that would be 6x6x6x6x6x6x6

I calculated the number of pairs i could make 7C2 =21

besides the pair there will be 5 people who can throw to any of the 6
so the outcome is 21x6x6x6x6x6
From here, I am sure I have to correct the overcounts and I am still trying...
Can I get some hints please?

the answer is 7847/15552
I tried working backwards from the answer and i still couldn't get it.

Thanks.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi veronica1999!

This is not an easy one.
The only way that I see is systematically counting all possibilities.
This is quite a bit of work, but still doable by hand.

Just now, I wrote a computer program that simply verified all possibilities and I got indeed your answer.

To do it manually, you could count the number of ways that no one throws a water ball at each other.
To find the actual number of ways that (at least) 2 people throw at each other, you'd have to subtract that number from the total of $6^7$.

Let's call the first person A.
Then A can throw at 6 people.
Let call his target B.

Then B can throw at 5 people.
Let's call his target C.
Then C can either throw at A, completing a triangle, which we'll count separately.
Or C can throw at one of the 4 people left, whom we'll call D.

At this point we have:
A closed triangle (ABC) for 6.5.1 ways.
A partial square ABCD for 6.5.4 ways.

If we continue from the close triangle (ABC), the next person can either throw at one of A, B, C, or he can throw at one of the 3 remaining people, which we'll call E.
We will have to count these 2 possibilities separately.

And so on...

#### Opalg

##### MHB Oldtimer
Staff member
This is not an easy one.
I thought so too, until I saw that veronica1999 was actually well on the way to a solution.

7 people are having a water ballon fight. At the same time, each of the 7 people throws a water balloon at one of the other 6 people, chosen at random. What is the probability that there are 2 people who throw the balloon at each other?

First, I calculated the total number of outcomes and that would be 6x6x6x6x6x6x6

I calculated the number of pairs i could make 7C2 =21

besides the pair there will be 5 people who can throw to any of the 6
so the outcome is 21x6x6x6x6x6
From here, I am sure I have to correct the overcounts and I am still trying...
Can I get some hints please?

the answer is 7847/15552
I tried working backwards from the answer and i still couldn't get it.
You have started in exactly the right way. There are $6^7$ possible outcomes, and there are $21\times6^5$ ways of getting a pair. But, as you realised, that includes some double counting. You can eliminate that by using the inclusion-exclusion method. First, you need to subtract the number of ways of getting two pairs who target each other. There are $35\times3$ ways of choosing two pairs out of seven, leaving three leftover who can throw at any of the $6$. That gives $105\times6^3$ outcomes. Finally, you need to add the number of ways of getting three pairs who target each other. There are $7\times5\times3$ ways of choosing three pairs out of seven, leaving one leftover who can throw at any of the $6$. That gives $105\times6$ outcomes.

Thus the number of outcomes in which at least one pair soak each other is $21\times6^5 - 105\times6^3 + 105\times6 = 163\,296 - 22\,680 + 630 = 141\,246$. Divide that by $6^7 = 279\,936$ and do some cancellation to get the answer.

#### chisigma

##### Well-known member
May be it is comfortable to proceed step by step as follows. If n is the number of players, then the possible plays are $n\ (n-1)$. Let's start...

a) n=2. We call the players A and B and the 2 possible plays are...

AB BA

... and the only 'good pair' is AB + BA, so that P=1...

b) n=3. We call the players A,B and C and the possible 6 plays are...

AB BA CA
AC BC CB

We have 3 'good pairs' AB+BA, AC+CA, BC+CB, so that is $\displaystyle P= \frac{1}{2}$...

c) n=4. We call the players A,B,C and D and the possible 12 plays are...

AB BA CA DA
AC BC CB DB
AD BD CD DC

We have 6 'good pairs' AB+BA, AC+CA, AD+DA,BC+CB,BD+DB,CD+DC, so that is $\displaystyle P=\frac{1}{2}$...

d) n=5. We call the players A,B,C,D and E and the possible 20 plays are...

AB BA CA DA EA
AC BC CB DB EB
AD BD CD DC EC
AE BE CE DE ED

We have10 'good pairs' AB+BA, AC+CA, AD+DA,AE+EA,BC+CB,BD+DB,BE+EB,CD+DC,CE+EC,DE+ED so that is $\displaystyle P=\frac{1}{2}$...

It is easy to see that in general we have $n\ (n-1)$ 'possible plays' and $1 + 2 + 3 + ...+\ n-1 = \frac{n\ (n-1)}{2}$ 'good pairs', so that, with the only exception n=2, in any case we have $P=\frac{1}{2}$...

Kind regards

$\chi$ $\sigma$

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#### veronica1999

##### Member
I thought so too, until I saw that veronica1999 was actually well on the way to a solution.

You have started in exactly the right way. There are $6^7$ possible outcomes, and there are $21\times6^5$ ways of getting a pair. But, as you realised, that includes some double counting. You can eliminate that by using the inclusion-exclusion method. First, you need to subtract the number of ways of getting two pairs who target each other. There are $35\times3$ ways of choosing two pairs out of seven, leaving three leftover who can throw at any of the $6$. That gives $105\times6^3$ outcomes. Finally, you need to add the number of ways of getting three pairs who target each other. There are $7\times5\times3$ ways of choosing three pairs out of seven, leaving one leftover who can throw at any of the $6$. That gives $105\times6$ outcomes.

Thus the number of outcomes in which at least one pair soak each other is $21\times6^5 - 105\times6^3 + 105\times6 = 163\,296 - 22\,680 + 630 = 141\,246$. Divide that by $6^7 = 279\,936$ and do some cancellation to get the answer.

Thanks!!!!
Now I think I understand how to correct the overcounting.
I am still amazed how you did it so easily.

First, I made the problem simpler and then applied what you did above.
Let's say there are 4 people A,B,C,D
There are a total of 3x3x3x3 outcomes.
And 4C2 x3x3 to make a pair.
In this case, it is easy to see that I overcounted when there are 2 pairs.
I subtract 6 to correct the double counting.

48/81

I tried again with 6 people

A B C D E F
I would have a total of 5x5x5x5x5x5 outcomes.
To make a pair it would be 6C2x5x5x5x5.
To correct for the double counts of 2 pairs
I would have to subtract (6C2X4C2)/2 x5x5.
Then I realized while i was subtracting all the 2 pairs
I subtracted all the 3 pairs so I have to add back 6C2X4C2/2.

Now finally for 7 people

A B C D E F G

There would be a total of 6x6x6x6x6x6x6 outcomes
And to make a pair, 7C2x6x6x6x6x6
Then subtract 7C2x5C2/2 x6x6x6
and then add (7C2X5C2X3C2)/6 x3

It took me a long time to figure out that I had to divide by 2 and 6 for the 2 pairs and 3pairs.
I hope my understanding is correct.

Once again, thanks!!!!!