Here's a braintease I made up (Math Question)

one in three or 30%

nope ......

That's not a good math question, at least not the way you phrased it because no matter how many times you shoot from the spot your odds of making it is always 20%(2-10).

Odds and percents are the same thing. So you are asking
"what is the percentage that you will make your percentage?"

But in your scenario the shooter does not take enough shots to make his actual percentage of exactly 20% (he needs 10 shots).

So the answer you are looking for is ZERO;
But that is not the real answer.

The real answer is that he would have to "better" his percentage to get exactly 2 baskets in 8 shots which means your real question is
"What are the odds the the shooter will better his percentage from that spot".

That cannot be deterimed in your scenario unless you provide the number of times this(bettering his percentage) was already attempted and achieved. The number of times he achieved it versus the number of times he tried it would be the odds that he would do it this time.


Nice try, but stay the fuck out of teaching math.

Not trying to teach anyone math..

and you're wrong.

Lets put it this way. If he took 8,000,000 shots. Then you broke it down into 1,000,000 sections of eight. What % of those sections would have exactly 2 goals. Its the same thing.

thats like saying
"every time I flip a coin there is a 50% chance it will be heads. But if i get heads 2 times in a row i bettered my odds. Therefore the odds of me getting heads twice is 0%"

Like I said...stay the fuck out of teaching math.


If you flip the coin and get heads then the odds that you will fip it again and get heads is the odds that you will get heads no matter what, which is 50%.

You are trying to add percentages of "turns" to try and determine total odds.
You can't do that.

See, reality and odds are not the same and you are trying to use reality as the odds.

The reality that the shooter could make 2 shots out of eight is really fucking high. But the math to figure the odds based on his actual percentage does not exist since the guy could hit 200 shots in a row and then miss 1000 times and still have the same shooting percentage.

Your 8,000,000 shot scenario is assuming that his hits and missess are evenally distributed and you can section them out and count the number of time he hit 2 shots out of eight. Which is total bullshit.


The way to make your question work is like this:

Everyday a guy takes 8 shots.
After a month his total percentage of made shots is 20%
(BTW the 20% means fucking nothing)
Then you would count how many days he made exactly 2 shots.
Then the number of days would be divided by one month to give you
the odds you are seeking.
Which would be the odds that he would go out today and make 2 of
eight shots.

Only under those circumstances would those odds have any meaning whatsoever.

Do you see how the fact that he had a total percentage of 20% does not show how many days he actually made excatly 2 shots?

Correction "200 shots in a row and then miss 1000"

Should have been "200 shots in a row and then miss 800"

So you're telling me there is no way to figure out the odds of you hitting heads twice? or tails twice?

Im not trying to be an asshole at all. Consider this a friendly debate.

If the player took 8,000,000 shots the hits and misses would not be evenly distributed. Thats the whole point of this question. At 8,000,000 tries sometimes he would make all 8(not often).. sometimes he would make 0.

But there is a way to calculate the probability.

i love this thread....


BY the way Im looking for interracial trades, hit me up

No, I'm saying you didn't provide the correct parameters to do such a thing in your brain teaser. Fliping a coin 100 times and getting heads 20% of the time does not give any indication whatsoever that heads turned up 2 times in a row at any time. Without that information there is no way to determine the odds.

If you had said that out of 100 tries two heads in a row came up 20 times.
Then the odds could be figure and they would be 20% that it woud happen now.

See, heads comming up twice in a row versus heads comming up period are two separate phenomenon and you are trying to take the properties of one phenomenon and determine the properties of another. CANT DO IT!






And he might make 200,000 shots in a row! Which would mean at NO time did he ever hit exactly 2 out of eight shots!!!!

He would have hit 8 out of 8 shots 25,000 times and then missed 8 out of 8 shots for the rest of the time.

How much would you bet on a guy like that making exactly 2 shots out of eight? ZERO?

I think the answer is 17...

ADG Webmaster

Yes but you know that the odds of it being heads is 50% because there are only 2 possibilities. Same thing here, 20% means there is 5 possibilities and 1 of them is a goal.






Yes he can make 200,000 shots in a row. There is even a way to figure out the odds of that happening. Probably something like 1 in a 1,000,000,000,000,000,000,000,000,000,000. But I think thats the part you don't get (or maybe ommiting to fit the criteria of your logic). That regardless of how you look at it the possibility is there.

The part you don't get is that I stated the odds can be figured out about 4 times already.

You just refuse to admit that you didn't provide the information needed to calculate those odds.

Your question is like " x * y = 12" but you don't tell what x or y is so the answer is (1,12);(2,6);(3,4);(12,1);(6,2);(4,3)...ect...(fra ctions and etc..)

If you really think about it you will realize that the only number you gave was 20% and that is not the number that relates to the odds you are seeking.

I got to go now but consider that I have a bachelors degree in computer science and mathematics.

1 in8 odds?

You're overcomplexing the whole thing (is that even a word).

Have a good night

1 in 8 odds?

Of course it is.

1 in 8 odds????????

Hmmmm 5% or 10% ?

The question is... lacking information. But I'd guess either 4% or 100% by what is given.

its not 1 out of 8

or 10% or 5%

and for those who want to say "well you didnt provide enough information"

I'll augment it for the stubborn ones.

Pretend its a 5 sided coin. 1 head, 2 arms, and 2 legs. (no matter how you spin it the odds of you getting heads is 20%, period)

you flip it 8 times, what are the odds of getting heads exactly 2 times.

easy
1.05%

(.2^2) * (.8^6)

You get the 8 shots you take and divide them by the automatic 20%[.2](heads) that is going to take place. Now, you want to see how many times that 20% (out of 8) is going to happen exactly 2 times, so you set it equal to 2. So you get:

8/.2=2
40=2
40=40
2/40= .05

.05%

wtf
the odds of getting exactly 2 is 2 times success and 6 times failure
thats .8^6 * .2^2

I just guessed, but the question might have been better put as:

Everytime you roll a 5 sided die you get X 20% of the time. If you flip it 8 times, what are the odds of getting X exactly 2 times.

Too obvious. 42.

too early for this

interracialtoons has no traffic.

The probablity of filliping a coin and getting 2 heads (HH) is:

(1/2)*(1/2)=1/4

Getting HT = (1/2)*(1/2)

Getting TH = (1/2)*(1/2)

Getting TT = HH = 1/4

So 50% of the time (1/4+1/4=1/2) you'll get the TH, HT combination and 25% of the time you'll get HH and 25% of the time you'll get TT.

Now just extending this to the problem from the initial post.

The probablity of getting 2 heads (HH) by rolling 2 times on this 5 sided dice is
(1/5)*(1/5)=(1/5)^2

The probablity of getting anything but heads (XXXXXX) and rolling 6 times is

(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)=(4/5)^6

Now the probablity of getting 2 out of 8 times of this 5 side dice is just
mutlitplying the above probabilities. (HHXXXXXX, where 'X' means anything but Heads)

=(1/5)^2 * (4/5)^6

I get the same answer as Myst.

(hopefully there aren't too many typos)

I think 3 shots in all off them

its 5 percent

noone has got it right yet

This is just a math question, if I wasnt so lazy I would grab my math book and look it up.

Because its exactly 2 goals your odds arent going to be that high.

It's a pretty simple question. There are 28 ways in which you can score exactly 2 goals. Each way has a probabilty of .0105

So the probabilty of scoring exactly 2 goals is .2936 or 29.36%

It doesn't matter how many times you flip a coin and it lands on heads every time you flip it there is a 50 50 shot of heads or tails

Always

pretty close you forgot to take the combinations into account

its 8!/2! *(.2^2)*(.8^6)

16% is that right

sorry typed it to fast

should have been 8!/6!/2!*.2^2*.8^6

or 4 *7= 28 times your number

1.6 now give the answere damn it

this is the right answer

Can I use a lifeline?

The answer to your orginal question is

Odds = 28/256 or 10.9%



The answer to the five sided coin is

Odds = 114688/390625 or 29.4% rounded



To show how I got the answer I will a simple example to prove the math

* A guy jumps out of a car, he will either land on his Head, Ass or feet.
What are the odds he will land on his Head exactly 2 times if he jumps
3 times.

The odds are calculated by determining all possibilties of his landings
over three jumps.

So he could land:
head, ass, feet (012) or
head, head, ass (001) or
ass, feet , head (120) etc.....

A total of 27 possiblities

But we are only interested about him landing on heads twice which means
these possible sets:

head,head,ass
head,head,feet
head,ass,head,
head,feet,head
ass,head,head
feet,head,head

A total of 6 relevent sets

No other sets satisifies our criteria.




Now run this simple perl script to prove my math




PERL cgi
------------

#!/usr/bin/perl
print "Content-type: text/html\n\n";


### copy right, 2007 interracialtoons.com


$head_ass_feet = "012"; ##### possible landings; head=0 ass=1 feet=2


@one = split(//, $head_ass_feet); #### array @one @two...etc are trys in the set of attempts
@two = @one;
@three = @one;

$total_permutes = 0; #### the total possible out comes of all trys in attempts
$heads2times = 0;

$a = 0;
foreach (@one) {
$b=0;
foreach(@two) {
$c = 0;
foreach (@three) {
$countheads = 0;

if ($one[$a] == 0) {$countheads++;}
if ($two[$b] == 0) {$countheads++;}
if ($three[$c] == 0) {$countheads++;}


print "$one[$a]$two[$b]$three[$c]<br>"; #### Dont print for 8 shots! Too many!

if ($countheads == 2) {$heads2times++;} #### head must occur EXACTLY 2 times

if (-e "killthisshit.txt") {print "ABORTED"; exit;}
$total_permutes++;

$c++;}
$b++;}

$a++;}


print "Total Permutations = $total_permutes<br><br>";
print "Heads occured exactly 2 times = $heads2times<br><br>";
print "Odds = $heads2times/$total_permutes<br><br>";



exit;



---END PERL---


RESULT ------

All The possible results of 3 trys

000
001
002
010
011
012
020
021
022
100
101
102
110
111
112
120
121
122
200
201
202
210
211
212
220
221
222
Total Permutations = 27

Heads occured exactly 2 times = 6

Odds = 6/27

Only one of the above permutations will happen in 3 trys

The odds reduce to 2/9 or 1 in 4.5 = 22% rounded





NOW run this program which has the exact same math but uses the
five sided coin in your example with sides = head, arm1, arm2, leg1, leg2



#!/usr/bin/perl
print "Content-type: text/html\n\n";





$head_arm1_arm2_leg1_leg2 = "01245"; ##### possible landings; head=0 arm1=1 arm2=2 ...


@one = split(//, $head_arm1_arm2_leg1_leg2); #### array @one @two...etc are trys in the set of attempts
@two = @one;
@three = @one;
@four = @one;
@five = @one;
@six = @one;
@seven = @one;
@eight = @one;

$total_permutes = 0; #### the total possible out comes of all trys in attempts
$heads2times = 0;

$a = 0;
foreach (@one) {
$b=0;
foreach(@two) {
$c = 0;
foreach (@three) {
$d = 0;
foreach (@four) {
$e = 0;
foreach (@five) {
$f = 0;
foreach (@six) {
$g = 0;
foreach (@seven) {
$h = 0;
foreach (@eight) {


$countheads = 0;

if ($one[$a] == 0) {$countheads++;}
if ($two[$b] == 0) {$countheads++;}
if ($three[$c] == 0) {$countheads++;}
if ($four[$d] == 0) {$countheads++;}
if ($five[$e] == 0) {$countheads++;}
if ($six[$f] == 0) {$countheads++;}
if ($seven[$g] == 0) {$countheads++;}
if ($eight[$h] == 0) {$countheads++;}

#####print "$one[$a]$two[$b]$three[$c]<br>"; #### Dont print for 8 shots! Too many!

if ($countheads == 2) {$heads2times++;} #### head must occur EXACTLY 2 times

if (-e "killthisshit.txt") {print "ABORTED"; exit;}
$total_permutes++;


$h++;}
$g++;}
$f++;}

$e++;}

$d++;}

$c++;}
$b++;}

$a++;}


print "Total Permutations = $total_permutes<br><br>";
print "Heads occured exactly 2 times = $heads2times<br><br>";
print "Odds = $heads2times/$total_permutes<br><br>";



exit;

---END PERL---






Now run the script for your orginal question


#!/usr/bin/perl
print "Content-type: text/html\n\n";





$hit_miss = "01"; ##### possible landings; hit=0 miss=1


@one = split(//, $hit_miss); #### array @one @two...etc are trys in the set of attempts
@two = @one;
@three = @one;
@four = @one;
@five = @one;
@six = @one;
@seven = @one;
@eight = @one;

$total_permutes = 0; #### the total possible out comes of all trys in attempts
$heads2times = 0;

$a = 0;
foreach (@one) {
$b=0;
foreach(@two) {
$c = 0;
foreach (@three) {
$d = 0;
foreach (@four) {
$e = 0;
foreach (@five) {
$f = 0;
foreach (@six) {
$g = 0;
foreach (@seven) {
$h = 0;
foreach (@eight) {


$countheads = 0;

if ($one[$a] == 0) {$countheads++;}
if ($two[$b] == 0) {$countheads++;}
if ($three[$c] == 0) {$countheads++;}
if ($four[$d] == 0) {$countheads++;}
if ($five[$e] == 0) {$countheads++;}
if ($six[$f] == 0) {$countheads++;}
if ($seven[$g] == 0) {$countheads++;}
if ($eight[$h] == 0) {$countheads++;}

#####print "$one[$a]$two[$b]$three[$c]<br>"; #### Dont print for 8 shots! Too many!

if ($countheads == 2) {$heads2times++;} #### head must occur EXACTLY 2 times

if (-e "killthisshit.txt") {print "ABORTED"; exit;}
$total_permutes++;


$h++;}
$g++;}
$f++;}

$e++;}

$d++;}

$c++;}
$b++;}

$a++;}


print "Total Permutations = $total_permutes<br><br>";
print "Heads occured exactly 2 times = $heads2times<br><br>";
print "Odds = $heads2times/$total_permutes<br><br>";



exit;


----END PERL----



That's the real math!

BUT Wait.

Lets make his shooting stick to 20%

meaning $hit_miss = "01111"; ##### possible landings for this 20% user is 1 in five hit=0 miss=1

The question here though is "is this valid" to respresent his shooting as such

Now run the last script with the new hit miss value.

The result is the same as the five sided coin.






Conclusion:

I can flip your five sided coin 1000 times a day for 30 days
and get a different heads percentage every day!
I can factor these together and get my total heads percentage for those
days and it could be 50%(or 20% shooting from that spot) or any number but when I do it on the 31st day
the math will still be the same as every other day reguarless of the
percentage I compiled previously.

Therefore the mathematical odds never change bases on my previous success rate of flipping
and getting heads.

Just because you can factor some numbers and get a result does not mean the
result is meaningfull.

So. The orignal question has no "meaningfull results".


BUT if the shooter had attempted 8 shots a day for 30 days and he got exactly 2 hits
on 10 days then that would yield a 30% chance he would do it today but that has nothing to do with his
overall percentage of 20%. You didn't provide the needed information to actually find
a meaningfull result.

so what are the odds of scoring 2 goals in a row?...

That is a an answer with intelligence but it is not the true way to find what you are looking for.

That's the odds that "the event" would occur not the odds that the paticular shooter would achieve "the event".

I figure, But the afro got me thinking and all giddy

take this shit to a math forum nick! lol

I do have interracial traffic to trade but not from that site.

That is a paysite and all the affiliate traffic goes to indexc.html and indexv.html, respresenting Ccbill and Verotel affiliates.

You should not take advise from people on this board as gospel, most don't have a clue about what others are doing.