Do we have any math genius's in the house?

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  • Do we have any math genius's in the house?

    Math is not my strong suit. Hopefully someone here can help.

    If you had a group of 100 blue pills, 10% were active, 90% were sugar substitutes

    You also had 100 yellow pills, 20% were active, 80% were sugar


    Statistically would consuming one yellow pill be the same as consuming 2 blue in terms of getting a positive drug reaction?

    To me it looks the same, 20% either way but 20 is 10 more than i have fingers so I could be wrong
    I heard the starting gun


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  • #2
    Re: Do we have any math genius's in the house?

    No. From what I read, I'm assuming you mean that of the 100 blue pills, 10 are active (and the 90 are sugar), and that of the 100 yellow pills, 20 are active (and the other 80 are sugar) - but you aren't sure which is which.

    If that's the case, you're confusing probability with actuality.

    A yellow pill may be twice as likely to be active than the blue pill, but that doesn't mean that 2 blue pills is equivalent to 1 yellow pill.

    Think of a coin toss. Theoretically, there's a 1-in-2 chance of getting heads (or tails). But if you were to flip a coin twice and get heads twice, the third time there is still a 50% chance (and you may still get tails). The 4th time, it's still 50%. But if you flip a coin enough times, the results will eventually even out to around 50/50 (but not necessarily exactly 50/50). And that "sample size" has to be large enough to even out the randomness of each individual "coin flip." Depending on the circumstances, it can be as low as 30 coin tosses, but maybe 100, or even 1,000 or more (I won't get into the math), depending on the nature of the study.

    In your case, 2 blue pills isn't the same as 1 yellow pill. But, depending on the circumstance, maybe 2,000 blue pills will yield close to the same as 1,000 yellow pills.

    Think of the 2:1 blue:yellow pills as just *one* coin flip. So just because a sample of 100 yellow pills are twice as likely to be active doesn't mean that *1* yellow pill is the same as *2* blue pills.

    Now, if you're assuming *every single blue pill* is 10/90 and *every single yellow pill* is 20/80, then your math would be correct.

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    • #3
      Re: Do we have any math genius's in the house?

      A good way to look at these things is to think of it as a tree with all the various outcomes as each of the branches.

      To establish numbers on this you'd need to be more specific about your drug taking activity. Specifically, the timing of your pill consumption, i.e. are you going to take both blue pills at the same time? Let's assume you are.

      So when you take two blue pills there are four possible branches...
      -- Pill 1 & Pill 2 are both sugar. Odds of that happening are 0.9 multiplied by 0.9, which is 0.81.
      -- Pill 1 Sugar, Pill 2 Drug. That has odds of 0.9 multiplied by 0.1, which is 0.09.
      -- Pill 1 Drug, Pill 2 Sugar. That has odds of 0.1 multiplied by 0.9, which is also 0.09.
      -- Pill 1 & Pill 2 are Drug. Odds for that are 0.1 multiplied by 0.1, which is 0.01.

      A good check for these kind of numbers is to add them up and make sure the total is 1, since probabilities always total 1.

      For the yellow pill it's simpler...
      -- Sugar is 0.8.
      -- Drug is 0.2.

      So if you take two blue pills you have an 81% chance of getting only sugar, so different odds than the yellow choice. But you also have a 1% chance of getting a double dose.

      If you tell us that taking a pill is fatally poisonous, then you'd take the two blue pills instead of one yellow, since it would be 81% versus 80%.

      Alternatively, if it's a lifesaving drug, you'd take the yellow, 20% versus 19%. If a double dose will kill you anyway, then you're even more likely to want the yellow, 20% versus 18%.

      I hope that helps, and I hope it's right...

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      • #4
        Re: Do we have any math genius's in the house?

        Howie has it right as far as he goes. But if you want to compare the probability between getting a single pill dose by selecting two blue pills versus the probability by selecting a single yellow pill, you have to add the two blue pill one dose possibilities together.
        Originally posted by Howie428 View Post
        So when you take two blue pills there are four possible branches...
        -- Pill 1 & Pill 2 are both sugar. Odds of that happening are 0.9 multiplied by 0.9, which is 0.81.
        -- Pill 1 Sugar, Pill 2 Drug. That has odds of 0.9 multiplied by 0.1, which is 0.09.
        -- Pill 1 Drug, Pill 2 Sugar. That has odds of 0.1 multiplied by 0.9, which is also 0.09.
        -- Pill 1 & Pill 2 are Drug. Odds for that are 0.1 multiplied by 0.1, which is 0.01.

        A good check for these kind of numbers is to add them up and make sure the total is 1, since probabilities always total 1.

        For the yellow pill it’s simpler...
        -- Sugar is 0.8.
        -- Drug is 0.2.
        Thus you'd add 0.9 + 0.9 = 0.18 which is closer to the 0.2 for the yellow pills. If you want results for any amount of drug then add in the last scenario as well to get 0.19.

        Actually if you want to get really precise the probabilities for the blue scenarios they are actually affected by the removal of a pill so to be exact:

        --first pill sugar, second pill drug - .9 multiplied by 10/99 because there is one less sugar pill.
        -- first pill drug, second pill sugar - .1 multiplied by 90/99 because there is one less drug pill.

        But really, those differences are small so just ignore them.

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        • #5
          Re: Do we have any math genius's in the house?

          Or spelling genius's. Either one would be helpful.

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          • #6
            Re: Do we have any math genius's in the house?

            Originally posted by Paul Striver View Post
            Or spelling genius's. Either one would be helpful.
            I could probably benefit from all sort of geni-asses, but their hourly rate is too high
            I heard the starting gun


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            • #7
              Re: Do we have any math genius's in the house?

              Originally posted by tinlizzie View Post
              Howie has it right as far as he goes. But if you want to compare the probability between getting a single pill dose by selecting two blue pills versus the probability by selecting a single yellow pill, you have to add the two blue pill one dose possibilities together.


              Thus you'd add 0.9 + 0.9 = 0.18 which is closer to the 0.2 for the yellow pills. If you want results for any amount of drug then add in the last scenario as well to get 0.19.

              Actually if you want to get really precise the probabilities for the blue scenarios they are actually affected by the removal of a pill so to be exact:

              --first pill sugar, second pill drug - .9 multiplied by 10/99 because there is one less sugar pill.
              -- first pill drug, second pill sugar - .1 multiplied by 90/99 because there is one less drug pill.

              But really, those differences are small so just ignore them.
              0.9 + 0.9 = 1.8 (not 0.18). Perhaps you're attempting to employ the "multiplication rule" which attempts to estimate the probability of two independent events occurring coincidentally. For example, if a couple has a 0.50 chance of having a child with freckles, and a 0.25 chance of producing a child who can roll their tongue, then the probability of that couple producing a freckled child who can roll their tongue = 0.50 x 0.25, or 0.125 (or 12.5%).

              As was stated, the probability of an event occurring is unrelated to the result it produces, if and when the event occurs - unless, of course, the event significantly changes the frequency (occurrence) of alternative outcomes. For example, if a guy swallows a yellow pill and dies, then it is very unlikely that he will swallow a blue (or any other colored) pill in the future.
              Last edited by bioprofessor; 03-11-2015, 01:50 AM.

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              • #8
                Re: Do we have any math genius's in the house?

                Originally posted by Southern_land View Post
                Math is not my strong suit. Hopefully someone here can help.

                If you had a group of 100 blue pills, 10% were active, 90% were sugar substitutes

                You also had 100 yellow pills, 20% were active, 80% were sugar


                Statistically would consuming one yellow pill be the same as consuming 2 blue in terms of getting a positive drug reaction?

                To me it looks the same, 20% either way but 20 is 10 more than i have fingers so I could be wrong
                All you've stated here is, that if you randomly sampled a "population" of blue pills, you'd have a 10% chance of selecting an active pill. Likewise, the probability of selecting an active pill from the yellow pile is 20%. Again, the probability of an event occurring, has little to do with the result it produces.

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                • #9
                  Re: Do we have any math genius's in the house?

                  There's a very simple way to calculate the probability of getting a favorable result ("F"), which is to subtract the probability of getting an unfavorable result ("U") from 1:

                  F = 1 - U

                  So for one yellow pill, this is trivial: U = 0.8, so F = 1 - 0.8 = 0.2

                  But for two blue pills, the chance of failure is U = 0.9^2 (i.e., the individual blue pill failure rate raised to the number of pills) = 0.9*0.9 = 0.81, so F = 1 - 0.81 = 0.19

                  Note that, no matter how many pills you take, U will always be greater than zero (however slightly)-- which is why there will always be at least an infinitesimal chance that the blue (or yellow) pills won't work favorably.

                  So you may want to consider taking the red pill...

                  EDIT: Above incorrectly assumes infinite supply of blue and yellow pills.
                  Last edited by AP Oz; 03-13-2015, 11:58 PM.

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                  • #10
                    Re: Do we have any math genius's in the house?

                    Thanks for your help. Now I need to amalgamate that into two sentences so the prof can explain it simply to his blond assistant
                    I heard the starting gun


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                    • #11
                      Re: Do we have any math genius's in the house?

                      It's "math geniuses." But that's all I know.

                      I think this is a put-on, right, you're just making stuff up? Because trying to read this makes my head hurt. I'm going to have to take ALL the yellow and blue pills now to feel better . . .

                      I think this thread should be made a stickie and titled "Nonsense Verse."

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                      • #12
                        Re: Do we have any math genius's in the house?

                        Originally posted by bioprofessor View Post
                        All you've stated here is, that if you randomly sampled a "population" of blue pills, you'd have a 10% chance of selecting an active pill. Likewise, the probability of selecting an active pill from the yellow pile is 20%. Again, the probability of an event occurring, has little to do with the result it produces.
                        LOL. That's what happens when I come in, skim a thread and try to do math in five minutes after 20 years of not trying. So now I get to feel like an idiot.

                        Also, AP Oz did what I was trying to do (or half remembered needed to be done). Think maybe I should just stick to writing now.

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                        • #13
                          Re: Do we have any math genius's in the house?

                          Are you kidding? You've given me 3-4 lines of dialog? Arguing over the exact mathematics of this. Thanks again!
                          I heard the starting gun


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                          • #14
                            Re: Do we have any math genius's in the house?

                            The second blue pill is from a pool of 99, not 100...unless you regurgitate the first one...but then it wouldn't be as blue.

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                            • #15
                              Re: Do we have any math genius's in the house?

                              Originally posted by anu View Post
                              The second blue pill is from a pool of 99, not 100...unless you regurgitate the first one...but then it wouldn't be as blue.
                              Ah, right-- I incorrectly assumed an infinite supply of pills.

                              So the probability of two blue pills yielding an unfavorable result in the OP's original example is actually U = (90/100)*(89/99) = 0.809

                              And the chances of a favorable result are F = 1 - U = 0.191

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