A game involving binomial probabilities

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Consider the following tables:



enter image description here



Given a number of independent trials $n=2,3,4ldots$ each of these tables represents the list of all the possible $binomk+n-1n=binomn+2n=6,10,15ldots$ ways (rows) in which a certain random process may occur (downward arrows) in correspondence of $k=3$ distinguishable and equiprobable states (here defined as three colors: red, green and blue).



In particular, I underline the following events:



  • $L$: "To get at least one red state and at least one blue state". This event occurs in $binomn2=1,3,6ldots$ different cases (half-red-half-blue dots on the left of each table).


  • $E$: "To get always the green state", which occurs only once for each $n$ (green dot on the right of each table).


The problem is to




Find a physical system such that the ratios
$$p_L=fracbinomn2binomn+2n
text and
p_E=frac1binomn+2n
$$
represent the probabilities of two events, with the above characteristics, related to that system.




The key should be that the order of the "trials" does not affect the occurrence of these events (I have the suspect that there should be some game which is described by such probabilities, but I don't know which one).



The context and the motivation of this problem can be found in this post, together with many useful and interesting comments.



Thanks for your suggestions!







share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    Consider the following tables:



    enter image description here



    Given a number of independent trials $n=2,3,4ldots$ each of these tables represents the list of all the possible $binomk+n-1n=binomn+2n=6,10,15ldots$ ways (rows) in which a certain random process may occur (downward arrows) in correspondence of $k=3$ distinguishable and equiprobable states (here defined as three colors: red, green and blue).



    In particular, I underline the following events:



    • $L$: "To get at least one red state and at least one blue state". This event occurs in $binomn2=1,3,6ldots$ different cases (half-red-half-blue dots on the left of each table).


    • $E$: "To get always the green state", which occurs only once for each $n$ (green dot on the right of each table).


    The problem is to




    Find a physical system such that the ratios
    $$p_L=fracbinomn2binomn+2n
    text and
    p_E=frac1binomn+2n
    $$
    represent the probabilities of two events, with the above characteristics, related to that system.




    The key should be that the order of the "trials" does not affect the occurrence of these events (I have the suspect that there should be some game which is described by such probabilities, but I don't know which one).



    The context and the motivation of this problem can be found in this post, together with many useful and interesting comments.



    Thanks for your suggestions!







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Consider the following tables:



      enter image description here



      Given a number of independent trials $n=2,3,4ldots$ each of these tables represents the list of all the possible $binomk+n-1n=binomn+2n=6,10,15ldots$ ways (rows) in which a certain random process may occur (downward arrows) in correspondence of $k=3$ distinguishable and equiprobable states (here defined as three colors: red, green and blue).



      In particular, I underline the following events:



      • $L$: "To get at least one red state and at least one blue state". This event occurs in $binomn2=1,3,6ldots$ different cases (half-red-half-blue dots on the left of each table).


      • $E$: "To get always the green state", which occurs only once for each $n$ (green dot on the right of each table).


      The problem is to




      Find a physical system such that the ratios
      $$p_L=fracbinomn2binomn+2n
      text and
      p_E=frac1binomn+2n
      $$
      represent the probabilities of two events, with the above characteristics, related to that system.




      The key should be that the order of the "trials" does not affect the occurrence of these events (I have the suspect that there should be some game which is described by such probabilities, but I don't know which one).



      The context and the motivation of this problem can be found in this post, together with many useful and interesting comments.



      Thanks for your suggestions!







      share|cite|improve this question













      Consider the following tables:



      enter image description here



      Given a number of independent trials $n=2,3,4ldots$ each of these tables represents the list of all the possible $binomk+n-1n=binomn+2n=6,10,15ldots$ ways (rows) in which a certain random process may occur (downward arrows) in correspondence of $k=3$ distinguishable and equiprobable states (here defined as three colors: red, green and blue).



      In particular, I underline the following events:



      • $L$: "To get at least one red state and at least one blue state". This event occurs in $binomn2=1,3,6ldots$ different cases (half-red-half-blue dots on the left of each table).


      • $E$: "To get always the green state", which occurs only once for each $n$ (green dot on the right of each table).


      The problem is to




      Find a physical system such that the ratios
      $$p_L=fracbinomn2binomn+2n
      text and
      p_E=frac1binomn+2n
      $$
      represent the probabilities of two events, with the above characteristics, related to that system.




      The key should be that the order of the "trials" does not affect the occurrence of these events (I have the suspect that there should be some game which is described by such probabilities, but I don't know which one).



      The context and the motivation of this problem can be found in this post, together with many useful and interesting comments.



      Thanks for your suggestions!









      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 8 at 6:36
























      asked Aug 7 at 20:41









      Andrea Prunotto

      953520




      953520




















          1 Answer
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          up vote
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          down vote



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          Does this game work?



          Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.

          Let $L$ be the event that neither $1$ nor $n+2$ is drawn.

          Let $E$ be the event that both $1$ and $n+2$ are drawn.




          Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.



          Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.




          Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively
          $$
          fracr+1r+b+g+3,qquad
          fracg+1r+b+g+3,qquad
          fracb+1r+b+g+3.$$



          This also produces the same distribution.






          share|cite|improve this answer























          • Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
            – Andrea Prunotto
            Aug 7 at 23:18











          • Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
            – Mike Earnest
            Aug 7 at 23:23






          • 1




            $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
            – Mike Earnest
            Aug 7 at 23:33







          • 1




            @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
            – Mike Earnest
            Aug 11 at 17:07






          • 1




            @AndreaPrunotto See edits
            – Mike Earnest
            Aug 13 at 17:03










          Your Answer




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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Does this game work?



          Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.

          Let $L$ be the event that neither $1$ nor $n+2$ is drawn.

          Let $E$ be the event that both $1$ and $n+2$ are drawn.




          Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.



          Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.




          Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively
          $$
          fracr+1r+b+g+3,qquad
          fracg+1r+b+g+3,qquad
          fracb+1r+b+g+3.$$



          This also produces the same distribution.






          share|cite|improve this answer























          • Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
            – Andrea Prunotto
            Aug 7 at 23:18











          • Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
            – Mike Earnest
            Aug 7 at 23:23






          • 1




            $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
            – Mike Earnest
            Aug 7 at 23:33







          • 1




            @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
            – Mike Earnest
            Aug 11 at 17:07






          • 1




            @AndreaPrunotto See edits
            – Mike Earnest
            Aug 13 at 17:03














          up vote
          1
          down vote



          accepted










          Does this game work?



          Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.

          Let $L$ be the event that neither $1$ nor $n+2$ is drawn.

          Let $E$ be the event that both $1$ and $n+2$ are drawn.




          Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.



          Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.




          Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively
          $$
          fracr+1r+b+g+3,qquad
          fracg+1r+b+g+3,qquad
          fracb+1r+b+g+3.$$



          This also produces the same distribution.






          share|cite|improve this answer























          • Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
            – Andrea Prunotto
            Aug 7 at 23:18











          • Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
            – Mike Earnest
            Aug 7 at 23:23






          • 1




            $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
            – Mike Earnest
            Aug 7 at 23:33







          • 1




            @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
            – Mike Earnest
            Aug 11 at 17:07






          • 1




            @AndreaPrunotto See edits
            – Mike Earnest
            Aug 13 at 17:03












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Does this game work?



          Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.

          Let $L$ be the event that neither $1$ nor $n+2$ is drawn.

          Let $E$ be the event that both $1$ and $n+2$ are drawn.




          Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.



          Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.




          Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively
          $$
          fracr+1r+b+g+3,qquad
          fracg+1r+b+g+3,qquad
          fracb+1r+b+g+3.$$



          This also produces the same distribution.






          share|cite|improve this answer















          Does this game work?



          Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.

          Let $L$ be the event that neither $1$ nor $n+2$ is drawn.

          Let $E$ be the event that both $1$ and $n+2$ are drawn.




          Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.



          Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.




          Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively
          $$
          fracr+1r+b+g+3,qquad
          fracg+1r+b+g+3,qquad
          fracb+1r+b+g+3.$$



          This also produces the same distribution.







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 13 at 17:02


























          answered Aug 7 at 21:02









          Mike Earnest

          15.7k11645




          15.7k11645











          • Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
            – Andrea Prunotto
            Aug 7 at 23:18











          • Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
            – Mike Earnest
            Aug 7 at 23:23






          • 1




            $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
            – Mike Earnest
            Aug 7 at 23:33







          • 1




            @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
            – Mike Earnest
            Aug 11 at 17:07






          • 1




            @AndreaPrunotto See edits
            – Mike Earnest
            Aug 13 at 17:03
















          • Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
            – Andrea Prunotto
            Aug 7 at 23:18











          • Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
            – Mike Earnest
            Aug 7 at 23:23






          • 1




            $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
            – Mike Earnest
            Aug 7 at 23:33







          • 1




            @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
            – Mike Earnest
            Aug 11 at 17:07






          • 1




            @AndreaPrunotto See edits
            – Mike Earnest
            Aug 13 at 17:03















          Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
          – Andrea Prunotto
          Aug 7 at 23:18





          Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables.
          – Andrea Prunotto
          Aug 7 at 23:18













          Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
          – Mike Earnest
          Aug 7 at 23:23




          Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely.
          – Mike Earnest
          Aug 7 at 23:23




          1




          1




          $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
          – Mike Earnest
          Aug 7 at 23:33





          $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers.
          – Mike Earnest
          Aug 7 at 23:33





          1




          1




          @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
          – Mike Earnest
          Aug 11 at 17:07




          @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit
          – Mike Earnest
          Aug 11 at 17:07




          1




          1




          @AndreaPrunotto See edits
          – Mike Earnest
          Aug 13 at 17:03




          @AndreaPrunotto See edits
          – Mike Earnest
          Aug 13 at 17:03












           

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