Probability after $n$ execution with two different probability distributions












1












$begingroup$


Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.



I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.



    I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.



      I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?










      share|cite|improve this question









      $endgroup$




      Lets say there's a statement and you want to know of, whether it is correct or not. Given an algorithm which outputs "probably true" or "probably false". In case the statement is correct the probability of "probably true" is $frac{9}{10}$ and in case the statement is incorrect the probability of "probably false" is $frac{6}{10}$.



      I see that executing the algorithm many times will in-/decrease the probability of the statement being correct depending on the outputs. But what is the exact probability after $n$ executions?







      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 '18 at 21:34









      upeupe

      123




      123






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.



          If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.



          This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014960%2fprobability-after-n-execution-with-two-different-probability-distributions%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.



            If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.



            This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.



              If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.



              This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.



                If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.



                This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.






                share|cite|improve this answer









                $endgroup$



                This is more easily done as an odds calculation. The odds of the statement being correct based solely on a single output of "probably true" are $9:4$ (since the algorithm outputs "probably true" with probability $frac{9}{10}$ for a correct statement and $frac{4}{10}$ for an incorrect statement). The odds based on a single output of "probably false" are $1:6$ for the same reason.



                If different runs of the algorithm are independent, then we can just multiply these together. For example, if the algorithm says "probably true" $3$ times and "probably false" twice, then the combined odds are $$(9:4)^3 cdot (1:6)^2 = 9^3 : 4^3cdot6^2 = 729: 2304 = 81:256$$ so the probability of the statement being correct is $frac{81}{81+256}$.



                This calculation assumes that initially the odds are $1:1$ and the statement is equally likely to be true or false without further information. If you have other information about the odds, you can just use that instead by multiplying that in, as well.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 21:40









                Misha LavrovMisha Lavrov

                45.1k656107




                45.1k656107






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014960%2fprobability-after-n-execution-with-two-different-probability-distributions%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    How to change which sound is reproduced for terminal bell?

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents