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This math problem has got my whole office debating each other holy **** (hard)(Srs)
This math problem has got my whole office debating each other holy **** (hard)(Srs)
09-09-2016, 08:35 AM
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#61
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Originally Posted By DoItForCuddles⏩
compsci pleb reporting
95% chance Sarah doesn't give you a bj
95% chance Mindy doesn't give you a bj
Assumin independent 0.95^2= 0.9025 chance you get no bj
Giving 9.75% chance of bj
Where did I **** up
95% chance Sarah doesn't give you a bj
95% chance Mindy doesn't give you a bj
Assumin independent 0.95^2= 0.9025 chance you get no bj
Giving 9.75% chance of bj
Where did I **** up
Originally Posted By TrettinR⏩
reps for both of youThis
Edit: .25% chance you get 2 blowjobs!
Edit: .25% chance you get 2 blowjobs!
09-09-2016, 08:36 AM
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#62
Originally Posted By Bothang⏩
that would be 10 out of 200 times dumassAssuming Sarah and Mindy are independant sloots
5 out of 100 times Sarah gives you a bj
5 out of 100 times Mindy gives you a bj
10 out of 100 times you get BJ
= 10%
5 out of 100 times Sarah gives you a bj
5 out of 100 times Mindy gives you a bj
10 out of 100 times you get BJ
= 10%
09-09-2016, 08:39 AM
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#63
the reason you don't just add them to be 10% is because -
If there were 20 chicks, all with 5% chance of blowing you...that does not mean there is 100% chance of a bj (it should tho misc, it should.)...there has to me some room for error there.
If there were 20 chicks, all with 5% chance of blowing you...that does not mean there is 100% chance of a bj (it should tho misc, it should.)...there has to me some room for error there.
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09-09-2016, 08:45 AM
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#64
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LOL @ potatoes still providing wrong answers when it was correctly solved a dozen or so posts in.
09-09-2016, 08:51 AM
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#65
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5%....
"ERMUHGAWD Crew"
"ERMUHGAW LOL...."
"Oh you, ERMUHGAWD!"
09-09-2016, 08:51 AM
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#66
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Originally Posted By kcuviello⏩
Well you could, but it would get uglythe reason you don't just add them to be 10% is because -
If there were 20 chicks, all with 5% chance of blowing you...that does not mean there is 100% chance of a bj (it should tho misc, it should.)...there has to me some room for error there.
If there were 20 chicks, all with 5% chance of blowing you...that does not mean there is 100% chance of a bj (it should tho misc, it should.)...there has to me some room for error there.
Let's say Mindy, Sarah, and Betty all have 5% chances of blowing you... Chances you get blown today are:
Sarah + Mindy + Betty - Sarah and Mindy - Sarah and Betty - Sarah Mindy and Betty
Point is, with two people, addition/subtraction is the way to go. As you add more girls, then you can use those autistic formulas
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09-09-2016, 08:51 AM
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#67
09-09-2016, 09:01 AM
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#68
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Originally Posted By zilaniz⏩
The odds aren't the same if you buy multiple tickets, assuming they're not the exact same numbers... there's just a minuscule change in the probability. i.e. going from 1 chance in 292,201,338 (0.0000003422%) to 2 chances in 292,201,338 (0.0000006844%). Or if you buy 500 tickets (0.000171114%). By your logic though, if I bought every single possible number combination, I still would only have a 1 in 292,000,000 chance of winning, which obviously isn't true since there is only a finite set of numbers and combinations.Still think its 5%, just like the lottery, no matter how many tickets you buy, the odds are still the same, just playing multiple times.
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09-09-2016, 09:07 AM
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#69
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Law man to the rescue: why it's not 5%
The big circle represents your universe: either you will get blown or you won't
The gray represents the chance you don't get blown
The red, the chance you get blown by Mindy. The blue, the chance you get blown by Sarah. The purple, the chance you get blown by both.
Let's say it's 9AM, and you're at the office. Mindy isn't in yet, and Sarah isn't in yet. The chance you get a blow job today (either Sarah or Mindy will come in and blow you) is 9.75%.
Let's say now it's noon, and Mindy called, saying she won't be coming in today. Now the chance you get a blow job today (Sarah will come in and blow you) is 5%.
But if you have no prior information, either Sarah or Mindy can polish your knob, so your chance of a blowjob is greater than 5%.
The big circle represents your universe: either you will get blown or you won't
The gray represents the chance you don't get blown
The red, the chance you get blown by Mindy. The blue, the chance you get blown by Sarah. The purple, the chance you get blown by both.
Let's say it's 9AM, and you're at the office. Mindy isn't in yet, and Sarah isn't in yet. The chance you get a blow job today (either Sarah or Mindy will come in and blow you) is 9.75%.
Let's say now it's noon, and Mindy called, saying she won't be coming in today. Now the chance you get a blow job today (Sarah will come in and blow you) is 5%.
But if you have no prior information, either Sarah or Mindy can polish your knob, so your chance of a blowjob is greater than 5%.
Bish Don't Kill my Vibe
09-09-2016, 09:10 AM
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#70
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Two pages in and still no pics
Order Will Be Restored
09-09-2016, 09:12 AM
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#71
Originally Posted By zilaniz⏩
Actually your odds get miniscule-ly better by playing more tickets assuming you don't pick the same numbers, as one combination has been excluded (your first non winning ticket).Still think its 5%, just like the lottery, no matter how many tickets you buy, the odds are still the same, just playing multiple times.
But this example in the OP is two separate events, and they should be summed (not multiplied). So you have the first event with a probability of 1 in 20, and a second event with a probability of 1 in 20.
1/20 + 1/20 = 2/40 = 5%
For you guys making the mistake of multiplying, here's why that is wrong. If we wanted the % chance of getting consecutive bj's, we'd multiply as the events are dependent on each other. The chance of getting a bj from the first girl, and also the second would be 1/20 * 1/20
09-09-2016, 09:14 AM
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#72
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Just do a weighted average.
(5% chance of bj from one slut+50% weight)+(5% chance of bj from other slut+50% weight)
=5%
(5% chance of bj from one slut+50% weight)+(5% chance of bj from other slut+50% weight)
=5%
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09-09-2016, 09:16 AM
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#73
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I wouldn't be able to tell you the exact answer like the other guy, but common sense tells you 25% chance is ridiculous, and that it has to be above 5%. If you have a 5% chance of getting a blowie from one sloot, and also a 5% chance with another, your chances go up the more you add sloots. If you have 10 girls that want to zuk your dik with a 5% chance each, obviously your chances of getting a single BJ is higher than 5%. That said it's not going to be quite 10% either, since the probability doesn't simply double.
09-09-2016, 09:17 AM
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#74
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Originally Posted By rootcon⏩
Incorrect. Why would you assume you can not collect another blow job just because you got one already?Actually your odds get miniscule-ly better by playing more tickets assuming you don't pick the same numbers, as one combination has been excluded (your first non winning ticket).
But this example in the OP is two separate events, and they should be summed (not multiplied). So you have the first event with a probability of 1 in 20, and a second event with a probability of 1 in 20.
1/20 + 1/20 = 2/40 = 5%
For you guys making the mistake of multiplying, here's why that is wrong. If we wanted the % chance of getting consecutive bj's, we'd multiply as the events are dependent on each other. The chance of getting a bj from the first girl, and also the second would be 1/20 * 1/20
But this example in the OP is two separate events, and they should be summed (not multiplied). So you have the first event with a probability of 1 in 20, and a second event with a probability of 1 in 20.
1/20 + 1/20 = 2/40 = 5%
For you guys making the mistake of multiplying, here's why that is wrong. If we wanted the % chance of getting consecutive bj's, we'd multiply as the events are dependent on each other. The chance of getting a bj from the first girl, and also the second would be 1/20 * 1/20
What I find amazing is all of the people calling others potatoes and most of them had the wrong answer.
09-09-2016, 09:28 AM
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#75
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Originally Posted By bartosh⏩
Exactly. If you're saying that each of two girls has a 50% probability of blowing you, then you're saying that there's 0% of somebody not blowing you, so you do have 100% chance.Lol at anybody saying 10%
Brb let's say both girls have a 50% chance. Brb that means there's a 100% chance according to your logic
Brb let's say both girls have a 50% chance. Brb that means there's a 100% chance according to your logic
09-09-2016, 09:33 AM
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#76
09-09-2016, 09:38 AM
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#77
Originally Posted By BigPoppaPumpin⏩
Of course the autistic stemcells have to go and make an easy problem look complicated. It takes a law man to simplify it for the people.
The chances you get a blowjob today are:
5% (sarah) + 5% (mindy) - .25% (both) = 9.75%
You subtract .25% because if Sarah blows you, whether or not Mindy also blows you, you got your bj for the day. And likewise, if Mindy blows you, whether or not Sarah also blows you, you got your bj for the day.
The chances you get a blowjob today are:
5% (sarah) + 5% (mindy) - .25% (both) = 9.75%
You subtract .25% because if Sarah blows you, whether or not Mindy also blows you, you got your bj for the day. And likewise, if Mindy blows you, whether or not Sarah also blows you, you got your bj for the day.
Originally Posted By TrettinR⏩
I was always bad at probability, so why can you subtract the probability of getting a bj twice in the first quote, but add the probability of 2 bj's in the second quote?The way doitforcuddles laid it out is the simplest. But it can also be shown by adding all the possibilities of getting at least 1 blowjob
(Girl1 BJ) 5%* (Girl 2 NoBJ) 95% = 4.75%
(Girl1 No BJ) 95%* (Girl 2 BJ) 5% = 4.75%
(Girl1 BJ) 5%* (Girl 2 BJ) 5% = .25%
4.75% + 4.75% + 0.25% = 9.75%
(Girl1 BJ) 5%* (Girl 2 NoBJ) 95% = 4.75%
(Girl1 No BJ) 95%* (Girl 2 BJ) 5% = 4.75%
(Girl1 BJ) 5%* (Girl 2 BJ) 5% = .25%
4.75% + 4.75% + 0.25% = 9.75%
09-09-2016, 09:38 AM
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#78
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Originally Posted By rootcon⏩
And people keep posting the wrong answer, even after it's been calculated at least three times...Actually your odds get miniscule-ly better by playing more tickets assuming you don't pick the same numbers, as one combination has been excluded (your first non winning ticket).
But this example in the OP is two separate events, and they should be summed (not multiplied). So you have the first event with a probability of 1 in 20, and a second event with a probability of 1 in 20.
1/20 + 1/20 = 2/40 = 5%
For you guys making the mistake of multiplying, here's why that is wrong. If we wanted the % chance of getting consecutive bj's, we'd multiply as the events are dependent on each other. The chance of getting a bj from the first girl, and also the second would be 1/20 * 1/20
But this example in the OP is two separate events, and they should be summed (not multiplied). So you have the first event with a probability of 1 in 20, and a second event with a probability of 1 in 20.
1/20 + 1/20 = 2/40 = 5%
For you guys making the mistake of multiplying, here's why that is wrong. If we wanted the % chance of getting consecutive bj's, we'd multiply as the events are dependent on each other. The chance of getting a bj from the first girl, and also the second would be 1/20 * 1/20
Best thread: http://forum.obnoxiousbrutes.com/showthread.php?t=168274783
***Black Crew*** (emeritus)
09-09-2016, 09:40 AM
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#79
09-09-2016, 09:41 AM
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#80
Originally Posted By TitanFall⏩
Chance can't be determined with a fixed variable.I'll put it in misc terms:
Sarah has a 5% chance to give you a blowjob today.
Mindy has a 5% chance to give you a blowjob today.
What's the % chance you get a blowjob today?
One camp says 10%
Another says 5%
Another says 25%
What's correct?
Sarah has a 5% chance to give you a blowjob today.
Mindy has a 5% chance to give you a blowjob today.
What's the % chance you get a blowjob today?
One camp says 10%
Another says 5%
Another says 25%
What's correct?
May as well ask what is 5% of randomness.
in misc terms;
chance = ?
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I'm on a diet.
09-09-2016, 09:42 AM
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#81
09-09-2016, 09:43 AM
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#82
09-09-2016, 10:04 AM
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#83
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its 5% you idiots
theyre both independent entities
theyre both independent entities
09-09-2016, 10:07 AM
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#84
Expected value 1(opposite) + expected value 2(opposite) - opposite
.05(.95)+.05(.95) - (.025)= .0975. Fixed my game theory equation
.05(.95)+.05(.95) - (.025)= .0975. Fixed my game theory equation
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09-09-2016, 10:08 AM
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#85
Originally Posted By ImpressiveGainz⏩
Bruh you need to go back to week one of a probability course where they cover dependent and independent events, there's a big difference.
09-09-2016, 10:12 AM
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#86
09-09-2016, 10:12 AM
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#87
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Originally Posted By NoHerro⏩
Method 1: Sarah BJ + Mindy BJ - Sarah and Mindy BJI was always bad at probability, so why can you subtract the probability of getting a bj twice in the first quote, but add the probability of 2 bj's in the second quote?
In this universe there are 4 possible outcomes: Sarah BJ, Mindy BJ, Sarah and Mindy BJ, and no BJ. The question asks what are the chances you get at least one BJ for the day, that is, either Sarah or Mindy gives you a blowjob. If both give you a blowjob, that doesn't affect the equation because all you need is one BJ. So you gotta subtract the chances that both give you a BJ. So the equation is, 5% (sarah bj) + 5% (mindy bj) - .25% (sarah and mindy bj).
Method 2: Sarah BJ and no Mindy BJ + Mindy BJ and no Sarah BJ + Sarah BJ and Mindy BJ
Again, the question asks what are the chances you get at least one BJ for the day, that is, either Sarah or Mindy gives you a bj. Here you're adding all the possible ways for you to get a blowjob for the day. Sarah could give you a bjwith Mindy not giving you a BJ; Mindy could give you a bjwith Sarah not giving you a bj; and both sarah and mindy could give you a bj. Here the both is added at the end because the first two terms are limited: they only include the chances one of the girls give you a bj. The first two terms weren't so limited in the first method.
Graphically:
Bish Don't Kill my Vibe
09-09-2016, 10:13 AM
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#88
09-09-2016, 10:13 AM
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#89
09-09-2016, 10:15 AM
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#90
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0% because girls are GROCE.
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