The draw for the fifth round of the FA Cup is about to be made. There are 16 teams, leading to eight matches. Your task is to pair the teams off, in an attempt to guess as many as possible of the actual matches in the real Cup draw. You are not asked which teams will be drawn at home, just which pairs will be selected.

What is the probability of at least four correct guesses?

This is a counting problem, I suppose you know your combinatorics (combinations and permutations), probability and set theory well. Have fun!

Hint: The numerator is 4705, denominator is 2027025.

Where are the first class students, you think say everything na cramming, come and show your self if you are truly good.

Are you willing to pay me 50k?

.

ositadima1:


No, if you solve it you gain my respect. I can't afford to pay you twice.

your reps means nothing to me bro, take care

Men, I am disappoint. Ok, does any one care for the answer?

NNAMDIII:
your reps means nothing to me bro, take care

I will give out the answer in 3 days time. I guess I am the only one on NL who can solve this question.

ositadima1:
The draw for the fifth round of the FA Cup is about to be made. There are 16 teams, leading to eight matches. Your task is to pair the teams off, in an attempt to guess as many as possible of the actual matches in the real Cup draw. You are not asked which teams will be drawn at home, just which pairs will be selected.

What is the probability of at least four correct guesses?

"You're not asked which pairs will be drawn at home..."

Could you explain this a bit?

Probability = Number of required outcomes / All possible outcomes.

Let's start with the denominator, it is much easier to deal with.

We have 16 teams and we want to pair off the teams, that is 8 pairs in all. To find the denominator we need to compute all possible pairs.

If we choose any team first we would have 15 choices to pair it with. After the first pair we are left with 14 teams. If we chose another team at random, we would have any of the 13 teams to pair it with. Continue like this for the remaining 6 pairs.

1st team to pair has 15 choices;
2nd team to pair has 13 choices;
3rd team to pair has 11 choices;
4th pair has 9 choices;
5th has 7;
6th has 5;
7th has 3;
8th team to pair has only 1 choice, the last team unpaired.

If you have n ways to do something and m ways to do another, then by logic we would have n x m ways to do both.

Therefore, we have 15 x 13 x 11 x 9 x 7 x 5 x 3x 1 ways in all.

This gives us 2027025, it is the total number of possible ways to pair these teams.

Darivie04:


"You're not asked which pairs will be drawn at home..."

Could you explain this a bit?

This phrase is there to simplify the question, it means that for any selection, that is any pair, you don't need to bother your self with which team is home or away. For example - Man U vs Arsenal counts the same as Arsenal vs Man U.

ositadima1:
Probability = Number of required outcomes / All possible outcomes.

Let's start with the denominator, it is much easier to deal with.

We have 16 teams and we want to pair off the teams, that is 8 pairs in all. To find the denominator we need to compute all possible pairs.

If we choose any team first we would have 15 choices to pair it with. After the first pair we are left with 14 teams. If we chose another team at random, we would have any of the 13 teams to pair it with. Continue like this for the remaining 6 pairs.

1st team to pair has 15 choices;
2nd team to pair has 13 choices;
3rd team to pair has 11 choices;
4th pair has 9 choices;
5th has 7;
6th has 5;
7th has 3;
8th team to pair has only 1 choice, the last team unpaired.

If you have n ways to do something and m ways to do another, then by logic we would have n x m ways to do both.

Therefore, we have 15 x 13 x 11 x 9 x 7 x 5 x 3x 1 ways in all.

This gives us 2027025, it is the total number of possible ways to pair these teams.

You said 3 days before you post solution na

I already had this part in my solution but I'm still trying to find the numerator.

Darivie04:

You said 3 days before you post solution na

I already had this part in my solution but I'm still trying to find the numerator.

Sorry, I will post d other part in 3 days time.

Hi,
Solve (√2 -1)^x + (√2+1)^x =34

ifada123:
Hi,
Solve (√2 -1)^x + (√2+1)^x =34

This thread is for a specific problem, their are other place where you can post ur math questions.




Easiest solution is to plot (√2 -1)^x + (√2+1)^x = y

and find x at point y = 34

ositadima1:


This thread is for a specific problem, their are other place where you can post ur math questions.




Easiest solution is to plot (√2 -1)^x + (√2+1)^x = y

and find x at point y = 34

I am aware of that graphical method,
I would like if u can solve numerically

ifada123:

I am aware of that graphical method,
I would like if u can solve numerically

Let's use algebra then

(√2 -1)^x + (√2 +1)^x = 34

if (√2 -1)^x = a
then (√2 +1)^x = 34 - a

We can now find log of both sides of d equation.

x[log(√2 -1)] = log(a)
x[log(√2 +1)] = log(34 -a)

therefore,

log(a) / log(√2 -1) = log(34 -a) / log(√2 +1)

Reorganizing;

{log(√2 +1) / log(√2 -1)} * log(a) = log(34 -a)

then

-1 * log(a) = log(34 -a)

then, antilog

1/a = 34 - a

a^2 - 34a +1


Can you find a and then plug it back to find x?

you might not get the accurate measurement of your IQ but you can know if you are an extraordinary person.....
see trait of extraordinary intelligent people here

https://explicitsuccess.com/traits-of-intelligent-people/

Darivie04:

You said 3 days before you post solution na

I already had this part in my solution but I'm still trying to find the numerator.

Some tips:

A) We have 8 pairs, we can visualize it as 8 boxes, each holding a pair of teams. If we are asked to guess at least 4 correct then we know that 4 of these boxes will contain 4 correct pairs. The other boxes can contain any pair, correct or not correct.

B) Assuming there is a set of correct pairs, all 8 pairs correct. If we where to populate the 4 correct boxes in our previous reasoning with this set, we would have to match any 4 with our 4 boxes. This gives us 8 combination 4 possible ways;

8! / (4! * 4!).

C) The remaining 4 boxes can be fitted with any pair, correct or not. That is, we now have 8 teams to be entered into 4 boxes pairwise. Same reasoning as in my earlier post, this can be done in 7 x 5 x 3 x 1 ways.

Put everything together gives to give us;

{8! / (4! x 4!)} x 7 x 5 x 3 x 1

Sweet right? But there is a big problem, we counted some pairs more than once. Here come the beauty of set theory to our rescue.


To be continued ...

Really sad, no one is interested, I have similar thought provoking problems I would love to post.

Is like this op is an engineer. See the way he dey chop maths

ositadima1:


Really sad, no one is interested, I have similar thought provoking problems I would love to post.

I would like to see your other problems.

The fact no one has come up with the solution doesn't mean no one has tried.

Even you self op struggled with the question when you first saw it

Let me expand a little here.

I mentioned an imaginary set with all 8 pair correct. Let's say I tag these pairs with alphabets; A, B, C, D, E, F, G, H. For example, Man U vs Arsenal pair could be labeled A.

Then the four boxes can be fitted like this:

| A | B | C | D | z | x | y | n |

this is just one iteration out of 8! / (4! x 4!) .

The other boxes will contain any pair that we haven't placed in d previous 4 boxes.


{8! / (4! x 4!)} x 7 x 5 x 3 x 1

This expression counts every possible combination except less than 4 boxes being correct. Yes 6 boxes correct is in there, 7 boxes correct is also in there, only 4 boxes correct is also accounted for. It is a big set of various 8 pair combinations.

Let's split it into subsets.

1) 4 correct pairs only, rest incorrect.

2) 5 correct pairs only

3) 6 correct pairs only

4) 7 correct pairs only

We have 4 subsets, 8 correct pairs only is not counted because is the same with 7 correct, last two teams most be correct if all others are.

For 5 correct boxes only (5PO) we have something like this;

| A | B | C | D | E | x | y | n |

This combination will be counted more than once. Which we don't want because order is not important. Eg

| A | E | C | D | B | x | y | n |
| E | B | C | D | A | x | y | n |
| A | B | E | D | C | x | y | n |

They are actually counted 5 combination 4 times more than required.

So this big set of 8 pairs;
{8! / (4! x 4!)} x 7 x 5 x 3 x 1

Can be represented as a set of subsets.

{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO}

Where PO stands for correct pairs only. Again I chose 8PO instead of 7PO, they are d same.

It is going to get a little more involved from now on. If you are following pls acknowledge, I don't want to type d whole thing for nothing.

Ok, let's proceed.

Why did I use d word set?

Let's assume I am given 3 letter: A, B and C. I want to form a set of two letter words, this set will contain the following 3 items.

AB
AC
BC

Where AB and BA is the same, the number of items in this set is 3. There is another way we can express this number, by using factorials.

Here, 3 combination 2, or 3! / (2! x 1!)

If we apply this same thinking to my problem, it starts to make sense.

{8! / (4! x 4!)} x 7 x 5 x 3 x 1

is the number of items in the set -- 8 team pairs with a minimum of 4 pairs being correct --

I know that 4 pairs are correct selections because I choose them from an imaginary set of correct pairs. I mean, assuming I cheated and found out all 8 correct pairs, I fit them in 4 boxes to make sure at the least 4 most be correct selections, there are 8! / (4! x 4!) ways this can be done.

How I wish I had a little percentage of your IQ in my brain for mathematics, Chai!


Although I love statistics.




Keep it up!!

Let's finish this motherfucker.

I have this set;

{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO}

But I need to eliminate repetitions and have something like this

{4PO, 5PO, 6PO, 8PO}

Where;

4PO stands for set of 8 team pairs with 4 correct only.

5PO stands for set of 8 team pairs with 5 correct only.

6PO stands for set of 8 team pairs with 6 correct only.

8PO stands for set of 8 team pairs with all 8 correct.
__________________________

This set;
{4PO, [5!/(4! x 1!)] x 5PO, [6!/(4! x 2!)] x 6PO, [8!/(4! x 4!)] x 8PO} = {4PO, 5 x 5PO, 15 x 6PO, 70 x 8PO}

has {8! / (4! x 4!)} x 7 x 5 x 3 x 1 = 7350 items.

I need to subtract 4 x 5PO to eliminate repetition and have only one 5PO.

Using
{8! / (5! x 3!)} x 5 x 3 x 1 = 840 ways

which is the set
{5PO, [6!/(5! x 1!)] x 6PO, [8!/(5! x 3!)] x 8PO} =

{5PO, 6 x 6PO, 56 x 8PO}

Multiplying with term [5!/(4! x 1!)] - 1 = 4

{4 x 5PO, 24 x 6PO, 4 x 56 x 8PO}


4 x 840 = 3360 ways

Need to subtract 3360 ways from 7350 = 3990

we have to add back these items 24 x 6PO and 4 x 56 x 8PO

For 24 x 6PO

24 x {8! / (6! x 2!)} x 3 x 1 = 2016 ways.

Which represents this set
24 x {6PO, 8!/(6! x 2!) x 8PO} = 24 x {6PO, 28 x 8PO}

Need to add 2016 ways to 3990 = 6006

Again, subtract these item 24 x 28 x 8PO

24 x 28 x 8!/(8! x 0!) = 672

Need to subtract 672 from 6006 = 5334

Add 4 x 56 x 8PO

4 x 56 x 8!/(8! x 0!) = 224

Need to add 224 to 5334 = 5558

Let's take stock, we are now at set {4PO, 5PO, 15 x 6PO, 70 x 8PO} with 5558 ways.

I need to subtract 14 x 6PO to eliminate repetition and have only one 6PO.

Using
{8! / (6! x 2!)} x 3 x 1 = 84 ways

which is the set
{6PO, [8!/(6! x 2!)] x 8PO} =

{6PO, 28 x 8PO}

Multiplying by 14

{14 x 6PO, 392 x 8PO}

= 14 x 84
= 1176

Need to subtract 1176 from 5558 = 4382

Add 392 x 1

Need to add 392 to 4382 = 4774


Let's take stock, we are now at set {4PO, 5PO, 6PO, 70 x 8PO} with 4774 ways.

I need to subtract 69 x 8PO to eliminate repetition and have only one 8PO.

Need to subtract 69 from 4774 = 4705

My guys, we have 4705 ways of arranging 8 team pairs with at least 4 correct choices. This can be represented as a set of subsets.

{4PO, 5PO, 6PO, 8PO}.

Now finally the probability of at least four correct guesses is

= 4705 / 2027025
= 0.0023211

Interestingly, I had another guy solve it this way;

https://math.stackexchange.com/questions/3142455/the-draw-for-the-fifth-round-of-the-fa-cup

That was when I was still battling with the problem.

The problem scattered my brain sha, at d time.

There is another elegant approach to solving d numerator of this problem that is less tasking.