What is the probability that the nth card becomes the top card after shuffling a certain way?Probability of drawing the top two cardsshuffle a deck of cards and cut it into three piles ,what probability that (at least) a court card will turn up on top of one of the piles?What is the probability that the first ace in a deck is the 30th card?The probability that a card is not chosen after a number of drawsProbability of drawing 3 cards in a 16 card deck where one of those 3 cards a duplicate in the deck?Probability of drawing certain card types in an opening handProbability that no five-card hands have each card with the same rank?What is the probability that at least 10 cards go by before the first jack?Probability of picking a card randomly vs always picking the first card?What is the probability that only one of the cards will have the matching suit?

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What is the probability that the nth card becomes the top card after shuffling a certain way?


Probability of drawing the top two cardsshuffle a deck of cards and cut it into three piles ,what probability that (at least) a court card will turn up on top of one of the piles?What is the probability that the first ace in a deck is the 30th card?The probability that a card is not chosen after a number of drawsProbability of drawing 3 cards in a 16 card deck where one of those 3 cards a duplicate in the deck?Probability of drawing certain card types in an opening handProbability that no five-card hands have each card with the same rank?What is the probability that at least 10 cards go by before the first jack?Probability of picking a card randomly vs always picking the first card?What is the probability that only one of the cards will have the matching suit?













3












$begingroup$


The following problem I can only seem to solve by simulation.



Suppose we take a deck and just label the cards from 1-52 in order, with 1 being the card on top. Now suppose we cut the deck at approximately the middle and complete the cut.



We could assume that there's an equal probability that we cut at each of 3 cards near the exact middle; that is, we either cut at exactly the middle (26 cards in hand), or we cut up to 29 cards or as few as 23 cards, all with equal probability.



Then we could ask, what's the probability that the $n$th card is now on top? The answer is simply $0$ for most of the cards, and $frac17$ that cards 24, 25, 26, 27, 28, 29, or 30 are on top.



But suppose we perform this cut twice, what then? I think the simplest answer unfortunately is just to sum up all the ways you can make each outcome and total the probability. For example, obviously card #1 is most likely to return back on top after cutting twice. This can happen if you cut exactly in the middle twice, if you're short one and then long one, if you're short two and then long two, etc. In total, there is a $frac749$ chance card 1 is on top, a $frac649$ chance that card 2 is on top, etc.



I'm having trouble finding a general pattern here. If you have an odd number of cuts, the most likely cards are somewhere near the middle of the range 1-52; an even number of cuts and the most likely cards are near the edges. But how do I describe this mathematically?










share|cite|improve this question









$endgroup$
















    3












    $begingroup$


    The following problem I can only seem to solve by simulation.



    Suppose we take a deck and just label the cards from 1-52 in order, with 1 being the card on top. Now suppose we cut the deck at approximately the middle and complete the cut.



    We could assume that there's an equal probability that we cut at each of 3 cards near the exact middle; that is, we either cut at exactly the middle (26 cards in hand), or we cut up to 29 cards or as few as 23 cards, all with equal probability.



    Then we could ask, what's the probability that the $n$th card is now on top? The answer is simply $0$ for most of the cards, and $frac17$ that cards 24, 25, 26, 27, 28, 29, or 30 are on top.



    But suppose we perform this cut twice, what then? I think the simplest answer unfortunately is just to sum up all the ways you can make each outcome and total the probability. For example, obviously card #1 is most likely to return back on top after cutting twice. This can happen if you cut exactly in the middle twice, if you're short one and then long one, if you're short two and then long two, etc. In total, there is a $frac749$ chance card 1 is on top, a $frac649$ chance that card 2 is on top, etc.



    I'm having trouble finding a general pattern here. If you have an odd number of cuts, the most likely cards are somewhere near the middle of the range 1-52; an even number of cuts and the most likely cards are near the edges. But how do I describe this mathematically?










    share|cite|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      The following problem I can only seem to solve by simulation.



      Suppose we take a deck and just label the cards from 1-52 in order, with 1 being the card on top. Now suppose we cut the deck at approximately the middle and complete the cut.



      We could assume that there's an equal probability that we cut at each of 3 cards near the exact middle; that is, we either cut at exactly the middle (26 cards in hand), or we cut up to 29 cards or as few as 23 cards, all with equal probability.



      Then we could ask, what's the probability that the $n$th card is now on top? The answer is simply $0$ for most of the cards, and $frac17$ that cards 24, 25, 26, 27, 28, 29, or 30 are on top.



      But suppose we perform this cut twice, what then? I think the simplest answer unfortunately is just to sum up all the ways you can make each outcome and total the probability. For example, obviously card #1 is most likely to return back on top after cutting twice. This can happen if you cut exactly in the middle twice, if you're short one and then long one, if you're short two and then long two, etc. In total, there is a $frac749$ chance card 1 is on top, a $frac649$ chance that card 2 is on top, etc.



      I'm having trouble finding a general pattern here. If you have an odd number of cuts, the most likely cards are somewhere near the middle of the range 1-52; an even number of cuts and the most likely cards are near the edges. But how do I describe this mathematically?










      share|cite|improve this question









      $endgroup$




      The following problem I can only seem to solve by simulation.



      Suppose we take a deck and just label the cards from 1-52 in order, with 1 being the card on top. Now suppose we cut the deck at approximately the middle and complete the cut.



      We could assume that there's an equal probability that we cut at each of 3 cards near the exact middle; that is, we either cut at exactly the middle (26 cards in hand), or we cut up to 29 cards or as few as 23 cards, all with equal probability.



      Then we could ask, what's the probability that the $n$th card is now on top? The answer is simply $0$ for most of the cards, and $frac17$ that cards 24, 25, 26, 27, 28, 29, or 30 are on top.



      But suppose we perform this cut twice, what then? I think the simplest answer unfortunately is just to sum up all the ways you can make each outcome and total the probability. For example, obviously card #1 is most likely to return back on top after cutting twice. This can happen if you cut exactly in the middle twice, if you're short one and then long one, if you're short two and then long two, etc. In total, there is a $frac749$ chance card 1 is on top, a $frac649$ chance that card 2 is on top, etc.



      I'm having trouble finding a general pattern here. If you have an odd number of cuts, the most likely cards are somewhere near the middle of the range 1-52; an even number of cuts and the most likely cards are near the edges. But how do I describe this mathematically?







      probability combinatorics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 5 hours ago









      HiddenBabelHiddenBabel

      1547




      1547




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          For the first iterations you get a convolution of discrete uniforms. Afterwards, there is a cyclic overlapping, so I don't think an analytic expression will be very simple.



          You can solve this numerically by modelling it as a Markov chain with 52 states (positions).



          Then, if $P$ is the transition matrix, the desired probabilities (after $n$ cuts) can be found in the first row of $P^n$.



          For example, in Octave/Matlab



          P = zeros(52,52);
          for i=1:52
          for k=23:29
          P(i,mod(i-1+k,52)+1) = 1/7;
          endfor
          endfor

          P(:,1) % probabilities after the first cut
          (P^2)(:,1) % probabilities after the second cut
          (P^3)(:,1) % probabilities after the third cut...


          enter image description here






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
            $endgroup$
            – HiddenBabel
            2 hours ago


















          2












          $begingroup$

          There are seven possible cuts, each corresponding to a permutation of the cards, so each determining a $52times 52$ permutation matrix. Let $P_1,P_2,dots,P_7$ be these matrices. Let $M=frac17(P_1+dots+P_7)$. Finally, let $x$ be the $52times 1$ column vector whose first coordinate is $1$ and whose other coordinates are zero. Then, the probability that card number $i$ is on top after $n$ cuts is just the $i^th$ coordinate of $M^nx$.






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            One possibility is to put algebraic structure on the cards. They may be the elements $0,1,2,3,dots, 51$ of the abelian group $Bbb Z/52$ of the integers taken modulo $52$. Then let us start we $S_0=0$.



            Completing a cut means adding to $S_0$ a random variable $X_1$ which is taking the values $26+kinBbb Z/52$ for $kin0,pm1,pm2,pm3$ with equal probability $1/7$, and any other number with probability zero.



            We can perform further cuts.



            Then we add further random variables $X_2,X_3,dots$ which have the same "shape" (repartition) as $X_1$. It is natural to write $X_k=26+Z_k$, so $Z_k$ takes values in $0,pm1,pm2,pm3$ with probability one.



            We have $S_0=0$, then




            • $S_1=S_0+X_1=26+Z_1$ is "near" the middle $26$,


            • $S_2=S_1+X_2=Z_1+Z_2$ is "near" the start $0$,


            • $S_3=S_2+X_3=26+Z_1+Z_2+Z_3$ is "near" the middle $26$,


            • $S_4=S_3+X_4=Z_1+Z_2+Z_3+Z_4$ is "near" the start $0$,

            and so on. I would start the repartition of the process $(S_n)$ using this language...






            share|cite|improve this answer









            $endgroup$












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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              For the first iterations you get a convolution of discrete uniforms. Afterwards, there is a cyclic overlapping, so I don't think an analytic expression will be very simple.



              You can solve this numerically by modelling it as a Markov chain with 52 states (positions).



              Then, if $P$ is the transition matrix, the desired probabilities (after $n$ cuts) can be found in the first row of $P^n$.



              For example, in Octave/Matlab



              P = zeros(52,52);
              for i=1:52
              for k=23:29
              P(i,mod(i-1+k,52)+1) = 1/7;
              endfor
              endfor

              P(:,1) % probabilities after the first cut
              (P^2)(:,1) % probabilities after the second cut
              (P^3)(:,1) % probabilities after the third cut...


              enter image description here






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
                $endgroup$
                – HiddenBabel
                2 hours ago















              2












              $begingroup$

              For the first iterations you get a convolution of discrete uniforms. Afterwards, there is a cyclic overlapping, so I don't think an analytic expression will be very simple.



              You can solve this numerically by modelling it as a Markov chain with 52 states (positions).



              Then, if $P$ is the transition matrix, the desired probabilities (after $n$ cuts) can be found in the first row of $P^n$.



              For example, in Octave/Matlab



              P = zeros(52,52);
              for i=1:52
              for k=23:29
              P(i,mod(i-1+k,52)+1) = 1/7;
              endfor
              endfor

              P(:,1) % probabilities after the first cut
              (P^2)(:,1) % probabilities after the second cut
              (P^3)(:,1) % probabilities after the third cut...


              enter image description here






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
                $endgroup$
                – HiddenBabel
                2 hours ago













              2












              2








              2





              $begingroup$

              For the first iterations you get a convolution of discrete uniforms. Afterwards, there is a cyclic overlapping, so I don't think an analytic expression will be very simple.



              You can solve this numerically by modelling it as a Markov chain with 52 states (positions).



              Then, if $P$ is the transition matrix, the desired probabilities (after $n$ cuts) can be found in the first row of $P^n$.



              For example, in Octave/Matlab



              P = zeros(52,52);
              for i=1:52
              for k=23:29
              P(i,mod(i-1+k,52)+1) = 1/7;
              endfor
              endfor

              P(:,1) % probabilities after the first cut
              (P^2)(:,1) % probabilities after the second cut
              (P^3)(:,1) % probabilities after the third cut...


              enter image description here






              share|cite|improve this answer









              $endgroup$



              For the first iterations you get a convolution of discrete uniforms. Afterwards, there is a cyclic overlapping, so I don't think an analytic expression will be very simple.



              You can solve this numerically by modelling it as a Markov chain with 52 states (positions).



              Then, if $P$ is the transition matrix, the desired probabilities (after $n$ cuts) can be found in the first row of $P^n$.



              For example, in Octave/Matlab



              P = zeros(52,52);
              for i=1:52
              for k=23:29
              P(i,mod(i-1+k,52)+1) = 1/7;
              endfor
              endfor

              P(:,1) % probabilities after the first cut
              (P^2)(:,1) % probabilities after the second cut
              (P^3)(:,1) % probabilities after the third cut...


              enter image description here







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 3 hours ago









              leonbloyleonbloy

              41.7k647108




              41.7k647108











              • $begingroup$
                I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
                $endgroup$
                – HiddenBabel
                2 hours ago
















              • $begingroup$
                I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
                $endgroup$
                – HiddenBabel
                2 hours ago















              $begingroup$
              I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
              $endgroup$
              – HiddenBabel
              2 hours ago




              $begingroup$
              I had discounted Markov chains because I thought you needed a 52!x52!-sized matrix! What are the other rows in this case?
              $endgroup$
              – HiddenBabel
              2 hours ago











              2












              $begingroup$

              There are seven possible cuts, each corresponding to a permutation of the cards, so each determining a $52times 52$ permutation matrix. Let $P_1,P_2,dots,P_7$ be these matrices. Let $M=frac17(P_1+dots+P_7)$. Finally, let $x$ be the $52times 1$ column vector whose first coordinate is $1$ and whose other coordinates are zero. Then, the probability that card number $i$ is on top after $n$ cuts is just the $i^th$ coordinate of $M^nx$.






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                There are seven possible cuts, each corresponding to a permutation of the cards, so each determining a $52times 52$ permutation matrix. Let $P_1,P_2,dots,P_7$ be these matrices. Let $M=frac17(P_1+dots+P_7)$. Finally, let $x$ be the $52times 1$ column vector whose first coordinate is $1$ and whose other coordinates are zero. Then, the probability that card number $i$ is on top after $n$ cuts is just the $i^th$ coordinate of $M^nx$.






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  There are seven possible cuts, each corresponding to a permutation of the cards, so each determining a $52times 52$ permutation matrix. Let $P_1,P_2,dots,P_7$ be these matrices. Let $M=frac17(P_1+dots+P_7)$. Finally, let $x$ be the $52times 1$ column vector whose first coordinate is $1$ and whose other coordinates are zero. Then, the probability that card number $i$ is on top after $n$ cuts is just the $i^th$ coordinate of $M^nx$.






                  share|cite|improve this answer









                  $endgroup$



                  There are seven possible cuts, each corresponding to a permutation of the cards, so each determining a $52times 52$ permutation matrix. Let $P_1,P_2,dots,P_7$ be these matrices. Let $M=frac17(P_1+dots+P_7)$. Finally, let $x$ be the $52times 1$ column vector whose first coordinate is $1$ and whose other coordinates are zero. Then, the probability that card number $i$ is on top after $n$ cuts is just the $i^th$ coordinate of $M^nx$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Mike EarnestMike Earnest

                  25.2k22151




                  25.2k22151





















                      1












                      $begingroup$

                      One possibility is to put algebraic structure on the cards. They may be the elements $0,1,2,3,dots, 51$ of the abelian group $Bbb Z/52$ of the integers taken modulo $52$. Then let us start we $S_0=0$.



                      Completing a cut means adding to $S_0$ a random variable $X_1$ which is taking the values $26+kinBbb Z/52$ for $kin0,pm1,pm2,pm3$ with equal probability $1/7$, and any other number with probability zero.



                      We can perform further cuts.



                      Then we add further random variables $X_2,X_3,dots$ which have the same "shape" (repartition) as $X_1$. It is natural to write $X_k=26+Z_k$, so $Z_k$ takes values in $0,pm1,pm2,pm3$ with probability one.



                      We have $S_0=0$, then




                      • $S_1=S_0+X_1=26+Z_1$ is "near" the middle $26$,


                      • $S_2=S_1+X_2=Z_1+Z_2$ is "near" the start $0$,


                      • $S_3=S_2+X_3=26+Z_1+Z_2+Z_3$ is "near" the middle $26$,


                      • $S_4=S_3+X_4=Z_1+Z_2+Z_3+Z_4$ is "near" the start $0$,

                      and so on. I would start the repartition of the process $(S_n)$ using this language...






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        One possibility is to put algebraic structure on the cards. They may be the elements $0,1,2,3,dots, 51$ of the abelian group $Bbb Z/52$ of the integers taken modulo $52$. Then let us start we $S_0=0$.



                        Completing a cut means adding to $S_0$ a random variable $X_1$ which is taking the values $26+kinBbb Z/52$ for $kin0,pm1,pm2,pm3$ with equal probability $1/7$, and any other number with probability zero.



                        We can perform further cuts.



                        Then we add further random variables $X_2,X_3,dots$ which have the same "shape" (repartition) as $X_1$. It is natural to write $X_k=26+Z_k$, so $Z_k$ takes values in $0,pm1,pm2,pm3$ with probability one.



                        We have $S_0=0$, then




                        • $S_1=S_0+X_1=26+Z_1$ is "near" the middle $26$,


                        • $S_2=S_1+X_2=Z_1+Z_2$ is "near" the start $0$,


                        • $S_3=S_2+X_3=26+Z_1+Z_2+Z_3$ is "near" the middle $26$,


                        • $S_4=S_3+X_4=Z_1+Z_2+Z_3+Z_4$ is "near" the start $0$,

                        and so on. I would start the repartition of the process $(S_n)$ using this language...






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          One possibility is to put algebraic structure on the cards. They may be the elements $0,1,2,3,dots, 51$ of the abelian group $Bbb Z/52$ of the integers taken modulo $52$. Then let us start we $S_0=0$.



                          Completing a cut means adding to $S_0$ a random variable $X_1$ which is taking the values $26+kinBbb Z/52$ for $kin0,pm1,pm2,pm3$ with equal probability $1/7$, and any other number with probability zero.



                          We can perform further cuts.



                          Then we add further random variables $X_2,X_3,dots$ which have the same "shape" (repartition) as $X_1$. It is natural to write $X_k=26+Z_k$, so $Z_k$ takes values in $0,pm1,pm2,pm3$ with probability one.



                          We have $S_0=0$, then




                          • $S_1=S_0+X_1=26+Z_1$ is "near" the middle $26$,


                          • $S_2=S_1+X_2=Z_1+Z_2$ is "near" the start $0$,


                          • $S_3=S_2+X_3=26+Z_1+Z_2+Z_3$ is "near" the middle $26$,


                          • $S_4=S_3+X_4=Z_1+Z_2+Z_3+Z_4$ is "near" the start $0$,

                          and so on. I would start the repartition of the process $(S_n)$ using this language...






                          share|cite|improve this answer









                          $endgroup$



                          One possibility is to put algebraic structure on the cards. They may be the elements $0,1,2,3,dots, 51$ of the abelian group $Bbb Z/52$ of the integers taken modulo $52$. Then let us start we $S_0=0$.



                          Completing a cut means adding to $S_0$ a random variable $X_1$ which is taking the values $26+kinBbb Z/52$ for $kin0,pm1,pm2,pm3$ with equal probability $1/7$, and any other number with probability zero.



                          We can perform further cuts.



                          Then we add further random variables $X_2,X_3,dots$ which have the same "shape" (repartition) as $X_1$. It is natural to write $X_k=26+Z_k$, so $Z_k$ takes values in $0,pm1,pm2,pm3$ with probability one.



                          We have $S_0=0$, then




                          • $S_1=S_0+X_1=26+Z_1$ is "near" the middle $26$,


                          • $S_2=S_1+X_2=Z_1+Z_2$ is "near" the start $0$,


                          • $S_3=S_2+X_3=26+Z_1+Z_2+Z_3$ is "near" the middle $26$,


                          • $S_4=S_3+X_4=Z_1+Z_2+Z_3+Z_4$ is "near" the start $0$,

                          and so on. I would start the repartition of the process $(S_n)$ using this language...







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 5 hours ago









                          dan_fuleadan_fulea

                          6,7781312




                          6,7781312



























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