Fair gambler's ruin problem intuitionProbability of Gambler's Ruin with Unequal Gain/LossAdaptive gambler's ruin problemGambler's Ruin with no set target for winGambler's ruin problem - unsure about the number of roundsEffect of Gambler's Ruin Bet Size on DurationGambler's ruin: verifying Markov propertyComparison of duration of two gambler's ruin gamesGambler's Ruin - Probability of Losing in t StepsGambler's Ruin: Win 2 dollars, Lose 1 dollarGambler's ruin Markov chain

How does a dynamic QR code work?

Mathematica command that allows it to read my intentions

One verb to replace 'be a member of' a club

If a warlock makes a Dancing Sword their pact weapon, is there a way to prevent it from disappearing if it's farther away for more than a minute?

Finding the reason behind the value of the integral.

How could indestructible materials be used in power generation?

Can compressed videos be decoded back to their uncompresed original format?

Sums of two squares in arithmetic progressions

Car headlights in a world without electricity

How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction

Forgetting the musical notes while performing in concert

How to stretch the corners of this image so that it looks like a perfect rectangle?

Is there a hemisphere-neutral way of specifying a season?

Why is the sentence "Das ist eine Nase" correct?

Placement of More Information/Help Icon button for Radio Buttons

Does int main() need a declaration on C++?

Can someone clarify Hamming's notion of important problems in relation to modern academia?

How many wives did king shaul have

Ambiguity in the definition of entropy

What is the opposite of "eschatology"?

How to show a landlord what we have in savings?

Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?

Why were 5.25" floppy drives cheaper than 8"?

Where would I need my direct neural interface to be implanted?



Fair gambler's ruin problem intuition


Probability of Gambler's Ruin with Unequal Gain/LossAdaptive gambler's ruin problemGambler's Ruin with no set target for winGambler's ruin problem - unsure about the number of roundsEffect of Gambler's Ruin Bet Size on DurationGambler's ruin: verifying Markov propertyComparison of duration of two gambler's ruin gamesGambler's Ruin - Probability of Losing in t StepsGambler's Ruin: Win 2 dollars, Lose 1 dollarGambler's ruin Markov chain













2












$begingroup$


In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_k-1 = p_k-1 - p_k-2 = ... = p_1 - p_0 = p_1$.



Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



Is there an intuitive reason why this is the case?










share|cite









$endgroup$
















    2












    $begingroup$


    In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



    In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_k-1 = p_k-1 - p_k-2 = ... = p_1 - p_0 = p_1$.



    Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



    Is there an intuitive reason why this is the case?










    share|cite









    $endgroup$














      2












      2








      2





      $begingroup$


      In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



      In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_k-1 = p_k-1 - p_k-2 = ... = p_1 - p_0 = p_1$.



      Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



      Is there an intuitive reason why this is the case?










      share|cite









      $endgroup$




      In a fair gambler's ruin problem, where the gambler starts with k dollars, wins $1 with probability 1/2 and loses $1 with probability 1/2, and stops when he/she reaches $n or $0.



      In the solution (from Dobrow's Introduction to Stochastic Processes with R), they let $p_k$ be defined as the probability of reaching $n with $k in one's inventory. Then they use the fact that $p_k - p_k-1 = p_k-1 - p_k-2 = ... = p_1 - p_0 = p_1$.



      Intuitively this means the probability of reaching $n with $k minus the probability of reaching $n with $k-1 is equivalent to the probability of reaching $n with only $1.



      Is there an intuitive reason why this is the case?







      probability






      share|cite













      share|cite











      share|cite




      share|cite










      asked 3 hours ago









      platypus17platypus17

      416




      416




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



          $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



          Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



          Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



          $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



          based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



          $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



          Having the summations only include the common terms on both sides gives



          $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



          Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_k-1$ term on the left to the RHS, eqrefeq4 becomes



          $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



          Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqrefeq2 to get that $p_i+1 - p_i = p_i - p_i-1$, like John Doe's answer states.






          share|cite|improve this answer











          $endgroup$




















            4












            $begingroup$

            The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_k-1+p_k+1)$$ Rearranging this gives $$2p_k=p_k-1+p_k+1\p_k-p_k-1=p_k+1-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






            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%2f3172677%2ffair-gamblers-ruin-problem-intuition%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



              $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



              Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



              Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



              $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



              based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



              $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



              Having the summations only include the common terms on both sides gives



              $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



              Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_k-1$ term on the left to the RHS, eqrefeq4 becomes



              $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



              Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqrefeq2 to get that $p_i+1 - p_i = p_i - p_i-1$, like John Doe's answer states.






              share|cite|improve this answer











              $endgroup$

















                2












                $begingroup$

                Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



                Having the summations only include the common terms on both sides gives



                $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_k-1$ term on the left to the RHS, eqrefeq4 becomes



                $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqrefeq2 to get that $p_i+1 - p_i = p_i - p_i-1$, like John Doe's answer states.






                share|cite|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                  $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                  Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                  Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                  $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                  based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                  $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



                  Having the summations only include the common terms on both sides gives



                  $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                  Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_k-1$ term on the left to the RHS, eqrefeq4 becomes



                  $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                  Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqrefeq2 to get that $p_i+1 - p_i = p_i - p_i-1$, like John Doe's answer states.






                  share|cite|improve this answer











                  $endgroup$



                  Regarding an "intuitive" reason for this relation, note that winning or losing a dollar has an equal chance and is independent of how much your currently have. Thus, the change in probability of winning or losing when starting off with $$1$ more is independent of what your starting value is. Note that if $q_k = 1 - p_k$ is the probability of losing when starting with $$k$, then plugging $p_k = 1 - q_k$ in gives that



                  $$q_k-1 - q_k = q_k-2 - q_k - 1 = ldots = q_1 - q_2 = q_0 - q_1 tag1labeleq1$$



                  Note you can reverse all the elements by multiplying by $-1$ to give the exact same relationship as with $p_k$.



                  Regarding how to get the relationship, this answer originally started with that, as the answer by John Doe states, the difference relation for reaching $n starting with $i is given by



                  $$p_i = frac12p_i - 1 + frac12p_i + 1 tag2labeleq2$$



                  based on the probabilities of either winning or losing the first time. Summing eqrefeq2 for $i$ from $1$ to $k - 1$ gives



                  $$sum_i=1^k-1 p_i = frac12sum_i=1^k-1 p_i - 1 + frac12sum_i=1^k-1 p_i + 1 tag3labeleq3$$



                  Having the summations only include the common terms on both sides gives



                  $$p_1 + sum_i=2^k - 2 p_i + p_k-1 = frac12p_0 + frac12p_1 + frac12sum_i=2^k - 2 p_i + frac12sum_i=2^k - 2 p_i + frac12p_k-1 + frac12p_k tag4labeleq4$$



                  Since the summation parts on both sides up to the same thing, they can be removed. Thus, after moving the $p_0$ and $p_1$ terms to the LHS and the $p_k-1$ term on the left to the RHS, eqrefeq4 becomes



                  $$frac12p_1 - frac12p_0 = frac12p_k - frac12p_k-1 tag5labeleq5$$



                  Multiplying both sides by $2$, then varying $k$ down, gives the relations you stated are used in the solution. However, it's generally simpler & easier to just manipulate eqrefeq2 to get that $p_i+1 - p_i = p_i - p_i-1$, like John Doe's answer states.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 1 hour ago

























                  answered 2 hours ago









                  John OmielanJohn Omielan

                  4,5362215




                  4,5362215





















                      4












                      $begingroup$

                      The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_k-1+p_k+1)$$ Rearranging this gives $$2p_k=p_k-1+p_k+1\p_k-p_k-1=p_k+1-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                      share|cite|improve this answer











                      $endgroup$

















                        4












                        $begingroup$

                        The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_k-1+p_k+1)$$ Rearranging this gives $$2p_k=p_k-1+p_k+1\p_k-p_k-1=p_k+1-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                        share|cite|improve this answer











                        $endgroup$















                          4












                          4








                          4





                          $begingroup$

                          The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_k-1+p_k+1)$$ Rearranging this gives $$2p_k=p_k-1+p_k+1\p_k-p_k-1=p_k+1-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.






                          share|cite|improve this answer











                          $endgroup$



                          The probability of reaching $n staring with $k can be split up by what possible first steps you can take - you either lose the first toss or win, each with probability 1/2. Then $$p_k=frac12(p_k-1+p_k+1)$$ Rearranging this gives $$2p_k=p_k-1+p_k+1\p_k-p_k-1=p_k+1-p_k$$ as required, and iterating it multiple times gets to $p_1-p_0$, and of course, $p_0=0$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 hours ago

























                          answered 3 hours ago









                          John DoeJohn Doe

                          11.5k11239




                          11.5k11239



























                              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%2f3172677%2ffair-gamblers-ruin-problem-intuition%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

                              Францішак Багушэвіч Змест Сям'я | Біяграфія | Творчасць | Мова Багушэвіча | Ацэнкі дзейнасці | Цікавыя факты | Спадчына | Выбраная бібліяграфія | Ушанаванне памяці | У філатэліі | Зноскі | Літаратура | Спасылкі | НавігацыяЛяхоўскі У. Рупіўся дзеля Бога і людзей: Жыццёвы шлях Лявона Вітан-Дубейкаўскага // Вольскі і Памідораў з песняй пра немца Адвакат, паэт, народны заступнік Ашмянскі веснікВ Минске появится площадь Богушевича и улица Сырокомли, Белорусская деловая газета, 19 июля 2001 г.Айцец беларускай нацыянальнай ідэі паўстаў у бронзе Сяргей Аляксандравіч Адашкевіч (1918, Мінск). 80-я гады. Бюст «Францішак Багушэвіч».Яўген Мікалаевіч Ціхановіч. «Партрэт Францішка Багушэвіча»Мікола Мікалаевіч Купава. «Партрэт зачынальніка новай беларускай літаратуры Францішка Багушэвіча»Уладзімір Іванавіч Мелехаў. На помніку «Змагарам за родную мову» Барэльеф «Францішак Багушэвіч»Памяць пра Багушэвіча на Віленшчыне Страчаная сталіца. Беларускія шыльды на вуліцах Вільні«Krynica». Ideologia i przywódcy białoruskiego katolicyzmuФранцішак БагушэвічТворы на knihi.comТворы Францішка Багушэвіча на bellib.byСодаль Уладзімір. Францішак Багушэвіч на Лідчыне;Луцкевіч Антон. Жыцьцё і творчасьць Фр. Багушэвіча ў успамінах ягоных сучасьнікаў // Запісы Беларускага Навуковага таварыства. Вільня, 1938. Сшытак 1. С. 16-34.Большая российская1188761710000 0000 5537 633Xn9209310021619551927869394п

                              Беларусь Змест Назва Гісторыя Геаграфія Сімволіка Дзяржаўны лад Палітычныя партыі Міжнароднае становішча і знешняя палітыка Адміністрацыйны падзел Насельніцтва Эканоміка Культура і грамадства Сацыяльная сфера Узброеныя сілы Заўвагі Літаратура Спасылкі НавігацыяHGЯOiТоп-2011 г. (па версіі ej.by)Топ-2013 г. (па версіі ej.by)Топ-2016 г. (па версіі ej.by)Топ-2017 г. (па версіі ej.by)Нацыянальны статыстычны камітэт Рэспублікі БеларусьШчыльнасць насельніцтва па краінахhttp://naviny.by/rubrics/society/2011/09/16/ic_articles_116_175144/А. Калечыц, У. Ксяндзоў. Спробы засялення краю неандэртальскім чалавекам.І ў Менску былі мамантыА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіГ. Штыхаў. Балты і славяне ў VI—VIII стст.М. Клімаў. Полацкае княства ў IX—XI стст.Г. Штыхаў, В. Ляўко. Палітычная гісторыя Полацкай зямліГ. Штыхаў. Дзяржаўны лад у землях-княствахГ. Штыхаў. Дзяржаўны лад у землях-княствахБеларускія землі ў складзе Вялікага Княства ЛітоўскагаЛюблінская унія 1569 г."The Early Stages of Independence"Zapomniane prawdy25 гадоў таму было аб'яўлена, што Язэп Пілсудскі — беларус (фота)Наша вадаДакументы ЧАЭС: Забруджванне тэрыторыі Беларусі « ЧАЭС Зона адчужэнняСведения о политических партиях, зарегистрированных в Республике Беларусь // Министерство юстиции Республики БеларусьСтатыстычны бюлетэнь „Полаўзроставая структура насельніцтва Рэспублікі Беларусь на 1 студзеня 2012 года і сярэднегадовая колькасць насельніцтва за 2011 год“Индекс человеческого развития Беларуси — не было бы нижеБеларусь занимает первое место в СНГ по индексу развития с учетом гендерного факцёраНацыянальны статыстычны камітэт Рэспублікі БеларусьКанстытуцыя РБ. Артыкул 17Трансфармацыйныя задачы БеларусіВыйсце з крызісу — далейшае рэфармаванне Беларускі рубель — сусветны лідар па дэвальвацыяхПра змену коштаў у кастрычніку 2011 г.Бядней за беларусаў у СНД толькі таджыкіСярэдні заробак у верасні дасягнуў 2,26 мільёна рублёўЭканомікаГаласуем за ТОП-100 беларускай прозыСучасныя беларускія мастакіАрхитектура Беларуси BELARUS.BYА. Каханоўскі. Культура Беларусі ўсярэдзіне XVII—XVIII ст.Анталогія беларускай народнай песні, гуказапісы спеваўБеларускія Музычныя IнструментыБеларускі рок, які мы страцілі. Топ-10 гуртоў«Мясцовы час» — нязгаслая легенда беларускай рок-музыкіСЯРГЕЙ БУДКІН. МЫ НЯ ЗНАЕМ СВАЁЙ МУЗЫКІМ. А. Каладзінскі. НАРОДНЫ ТЭАТРМагнацкія культурныя цэнтрыПублічная дыскусія «Беларуская новая пьеса: без беларускай мовы ці беларуская?»Беларускія драматургі па-ранейшаму лепш ставяцца за мяжой, чым на радзіме«Працэс незалежнага кіно пайшоў, і дзяржаву турбуе яго непадкантрольнасць»Беларускія філосафы ў пошуках прасторыВсе идём в библиотекуАрхіваванаАб Нацыянальнай праграме даследавання і выкарыстання касмічнай прасторы ў мірных мэтах на 2008—2012 гадыУ космас — разам.У суседнім з Барысаўскім раёне пабудуюць Камандна-вымяральны пунктСвяты і абрады беларусаў«Мірныя бульбашы з малой краіны» — 5 непраўдзівых стэрэатыпаў пра БеларусьМ. Раманюк. Беларускае народнае адзеннеУ Беларусі скарачаецца колькасць злачынстваўЛукашэнка незадаволены мінскімі ўладамі Крадзяжы складаюць у Мінску каля 70% злачынстваў Узровень злачыннасці ў Мінскай вобласці — адзін з самых высокіх у краіне Генпракуратура аналізуе стан са злачыннасцю ў Беларусі па каэфіцыенце злачыннасці У Беларусі стабілізавалася крымінагеннае становішча, лічыць генпракурорЗамежнікі сталі здзяйсняць у Беларусі больш злачынстваўМУС Беларусі турбуе рост рэцыдыўнай злачыннасціЯ з ЖЭСа. Дазволіце вас абкрасці! Рэйтынг усіх службаў і падраздзяленняў ГУУС Мінгарвыканкама вырасАб КДБ РБГісторыя Аператыўна-аналітычнага цэнтра РБГісторыя ДКФРТаможняagentura.ruБеларусьBelarus.by — Афіцыйны сайт Рэспублікі БеларусьСайт урада БеларусіRadzima.org — Збор архітэктурных помнікаў, гісторыя Беларусі«Глобус Беларуси»Гербы и флаги БеларусиАсаблівасці каменнага веку на БеларусіА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіУ. Ксяндзоў. Сярэдні каменны век (мезаліт). Засяленне краю плямёнамі паляўнічых, рыбакоў і збіральнікаўА. Калечыц, М. Чарняўскі. Плямёны на тэрыторыі Беларусі ў новым каменным веку (неаліце)А. Калечыц, У. Ксяндзоў, М. Чарняўскі. Гаспадарчыя заняткі ў каменным векуЭ. Зайкоўскі. Духоўная культура ў каменным векуАсаблівасці бронзавага веку на БеларусіФарміраванне супольнасцей ранняга перыяду бронзавага векуФотографии БеларусиРоля беларускіх зямель ва ўтварэнні і ўмацаванні ВКЛВ. Фадзеева. З гісторыі развіцця беларускай народнай вышыўкіDMOZGran catalanaБольшая российскаяBritannica (анлайн)Швейцарскі гістарычны15325917611952699xDA123282154079143-90000 0001 2171 2080n9112870100577502ge128882171858027501086026362074122714179пппппп

                              ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result