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Hello all!

I have a math problem which I struggle to solve. Any help is much appreciated so thank you in advance!

Let's assume we want to play lottery and 6 numbers are drawn.

I know that the odds of guessing 6 out of 6 in the exact order (order is important) = 1/1000000.

What are the odds of guessing:

5 of 6 = ?

4 of 6 = ?

3 of 6 = ?

2 of 6 = ?

1 of 6 = ?

Note: order is important.

Thank you!

So 6 of 6 is 1/1000000 = 0.0001% (min 000000 -> max 999999)

5 of 6 = 1/100000 = 0.001% (00000->99999)

4 of 6 = 1/10000 = 0.01% (0000->9999)

3 of 6 = 1/1000 = 0.1% (000->999)

2 of 6 = 1/100 = 1% (00->99)

1 of 6 = 1/10 = 10% (0->9 | 1/6 numbers are correct)

Maybe I'm wrong, I'm not too good at math.

Thanks for your answer!

I'm afraid you are wrong since 5 of 6 = 6 possibilities in 1000000. But I don't know the rest...

This task is really very interesting

I know the answer to this problem but I don't want to write it, let others think. We passed this task in the 7th grade

Good idea!

Maybe you can share it on PM? If you want of course.

Thanks anyway!

Hi, it's simple
1 of 6 = 1/6
2 of 6 = 1/6*1/6 = 1/36
3 of 6 = 1/6*1/6*1/6 = 1/216
.....
 

Got it, thank you!

n = number of possible drawn numbers,
k = number of drawn numbers in winning combination
x = matching drawn numbers
C(k,x) = k!/(x!*(k-x)!)

considering the numbers are drawn from separate drums:
P(x) = (1/n)^x * (1-1/n)^(k-x) * C(k,x)

k=6,
if n=10

P(6) = 1 / 10^6
P(5) = 9 * 6 / 10^6
p(4) = 9^2 * 15 / 10^6
p(3) = 9^3 * 20 / 10^6
p(2) = 9^4 * 15 / 10^6
p(1) = 9^5 * 6 / 10^6
p(0) = 9^6 / 10^6

Match    Combinations     Odds
Match 6   1         1 in 1,000,000
Match 5    54         1 in 18,518.52
Match 4    1,215      1 in 823.05
Match 3    14,580      1 in 68.59
Match 2    98,415      1 in 10.16
Match 1    354,294      1 in 2.82
Match 0    531,441      1 in 1.88

fulse
match 0  1 in 1.88 )) 

n = number of possible drawn numbers
k = number of drawn numbers in winning combination

if numbers can be repeated

1 of 10  1/10                   1 in 10
2 of 10  1/10*1/10           1 in 100 
3 of 10  1/10*1/10*1/10  1 in 1000
...

if the numbers cannot be repeated

1 of 10  1/10                     1 in 10
2 of 10  1/10*1/9               1 in 90
3 of 10  1/10*1/9*1/8        1 in 720
4 of 10  1/10*1/9*1/8*1/7  1 in 5040
...

Thank you for the answers guys!

@Patentico I think you are wrong  because matching 5 in 6 there are only 6 combination.

@Ankol99 there are only 6 numbers, that can be repeated so applying your formula would be:

1 of 6 =  1/6               = 1 in 6 = 16.666666666%
2 of 6 =  1/6*1/6         = 1 in 36 = 2.777777777%
3 of 6 =  1/6*1/6*1/6   = 1 in 216 = 0.462962962%
.........

This is the solution I am looking for, thanks again.

You can check different combinations on this site:
 https://www.lotterypost.com/odds  (https://www.lotterypost.com/odds)

You forget to take into account the probability of failure: 9 in 10 chance
If you want x out of 6 successes then there must be 6-x failures
plus there are C(6,x) ways to arrange these combinations of success & failures


P(x) = (1/10)^x * (9/10)^(6-x) * C(6,x)

There are 9 different cobinations of getting the 6th number wrong also,
so there are 6*9 = 54 different combinations!
 

In exact order?   That means there can only be one right number each time.   Each guess has one less chance (unless you can reuse numbers, which you did not specify)

Let's say you have 10 possible numbers
First ball you have a 1/10 chance
Second ball you have a 1/9 chance
3 = 1/8
4 = 1/7
5 = 1/6
6 = 1/5

That is 1 in 10*9*8*7*6*5 or 1 in 152,100 for just ten balls.

This puzzle is very difficult, I will not be able to solve it

Yes, sorry I forgot to specify that the numbers are reused and yes, you must guess them in the exact order.

For example:
- you pick a ticket with these numbers:    5 6 5 8 8 1.
- the numbers drawn by the lottery are    1 5 7 8 2 1.

You guessed 2 numbers, the 4th and the 6th.

I want to know what are the odd for this to happen. (also for hitting 3,4,5 and 6 numbers).

Thanks!