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Here's a braintease I made up (Math Question)
Its not as easy as it sounds...
If you're shooting 20% from a certain spot on the field. If you take 8 shots from that spot what are the odds of you making 2 goals (exactly 2 goals). Think about this one real good. |
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one in three or 30% |
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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. |
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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%" |
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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" |
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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 |
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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! Quote:
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 |
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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?
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Have a good night :) |
1 in 8 odds?
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Of course it is. |
1 in 8 odds????????
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Hmmmm 5% or 10% ? :)
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The question is... lacking information. But I'd guess either 4% or 100% by what is given.
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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% |
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the odds of getting exactly 2 is 2 times success and 6 times failure thats .8^6 * .2^2 |
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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.
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too early for this :helpme
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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
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its 5 percent
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noone has got it right yet :)
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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% |
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its 8!/2! *(.2^2)*(.8^6) |
16% is that right
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