Find the probability that the card number 18 is the second jack that you deal.

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You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.




I have tried this numerous different ways, but I cant seem to get it.



I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.






probability combinatorics







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up vote
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down vote

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You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.




I have tried this numerous different ways, but I cant seem to get it.



I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.






probability combinatorics







share|cite|improve this question









New contributor




Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
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up vote
3
down vote

favorite









up vote
3
down vote

favorite












You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.




I have tried this numerous different ways, but I cant seem to get it.



I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.






probability combinatorics







share|cite|improve this question









New contributor




Nathaniel Sexton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












You deal from a well-shuffled $52$-card deck, one card at a time. Find the probability that the card number 18 is the second jack that you deal. Include at least $4$ digits after the decimal point in your answer.




I have tried this numerous different ways, but I cant seem to get it.



I've tried
$$fracdbinom4816 cdot dfrac452dbinom5216 cdot dfrac335$$
and I don't know why this doesn't work.






probability combinatorics




probability combinatorics






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edited 1 hour ago









N. F. Taussig

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  • Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
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  • Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
    – N. F. Taussig
    1 hour ago















Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
– N. F. Taussig
1 hour ago




Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site.
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1 hour ago










3 Answers
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If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
$$fracdbinom41dbinom4816dbinom5217$$
The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
$$fracdbinom41dbinom4816dbinom5217 cdot frac335$$






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    up vote
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    There are $17$ ways to choose when the first Jack is drawn.



    $$17tag1$$



    Given this, there are $4$ ways to choose which Jack that is.



    $$4 tag2$$



    There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...



    $$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$



    Finally there are three choices for which Jack is drawn second:



    $$3 tag4$$



    The number of successful draws is the product of values $(1)$ through $(4)$ above.



    The total number of possible draws of $18$ cards in order is:



    $$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$



    The final probability is the number of successful possibilities, divided by the total number of possibilities:



    $$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$






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      The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$






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        3 Answers
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        3 Answers
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        If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
        $$fracdbinom41dbinom4816dbinom5217$$
        The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
        $$fracdbinom41dbinom4816dbinom5217 cdot frac335$$






        share|cite|improve this answer
























          up vote
          3
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          If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
          $$fracdbinom41dbinom4816dbinom5217$$
          The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
          $$fracdbinom41dbinom4816dbinom5217 cdot frac335$$






          share|cite|improve this answer






















            up vote
            3
            down vote










            up vote
            3
            down vote









            If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
            $$fracdbinom41dbinom4816dbinom5217$$
            The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
            $$fracdbinom41dbinom4816dbinom5217 cdot frac335$$






            share|cite|improve this answer












            If the $18$th card is the second Jack you deal, then there must be one Jack and 16 non-Jacks among the first $17$ cards, then a second Jack at the $18$th card. The probability of obtaining one of the four Jacks and $16$ of the $48$ non-Jacks in the first $17$ deals is
            $$fracdbinom41dbinom4816dbinom5217$$
            The probability of dealing one of the three remaining jacks from among the $52 - 17 = 35$ remaining cards is $3/35$. Hence, the probability that the second Jack you deal occurs on the $18$th deal is
            $$fracdbinom41dbinom4816dbinom5217 cdot frac335$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            N. F. Taussig

            40.1k93253




            40.1k93253




















                up vote
                2
                down vote













                There are $17$ ways to choose when the first Jack is drawn.



                $$17tag1$$



                Given this, there are $4$ ways to choose which Jack that is.



                $$4 tag2$$



                There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...



                $$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$



                Finally there are three choices for which Jack is drawn second:



                $$3 tag4$$



                The number of successful draws is the product of values $(1)$ through $(4)$ above.



                The total number of possible draws of $18$ cards in order is:



                $$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$



                The final probability is the number of successful possibilities, divided by the total number of possibilities:



                $$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$






                share|cite|improve this answer
























                  up vote
                  2
                  down vote













                  There are $17$ ways to choose when the first Jack is drawn.



                  $$17tag1$$



                  Given this, there are $4$ ways to choose which Jack that is.



                  $$4 tag2$$



                  There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...



                  $$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$



                  Finally there are three choices for which Jack is drawn second:



                  $$3 tag4$$



                  The number of successful draws is the product of values $(1)$ through $(4)$ above.



                  The total number of possible draws of $18$ cards in order is:



                  $$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$



                  The final probability is the number of successful possibilities, divided by the total number of possibilities:



                  $$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$






                  share|cite|improve this answer






















                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    There are $17$ ways to choose when the first Jack is drawn.



                    $$17tag1$$



                    Given this, there are $4$ ways to choose which Jack that is.



                    $$4 tag2$$



                    There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...



                    $$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$



                    Finally there are three choices for which Jack is drawn second:



                    $$3 tag4$$



                    The number of successful draws is the product of values $(1)$ through $(4)$ above.



                    The total number of possible draws of $18$ cards in order is:



                    $$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$



                    The final probability is the number of successful possibilities, divided by the total number of possibilities:



                    $$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$






                    share|cite|improve this answer












                    There are $17$ ways to choose when the first Jack is drawn.



                    $$17tag1$$



                    Given this, there are $4$ ways to choose which Jack that is.



                    $$4 tag2$$



                    There are $16$ non-Jack cards drawn before the second Jack, and there are $48$ choices for the first of these cards, $47$ for the second, and so on ...



                    $$48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33 tag3$$



                    Finally there are three choices for which Jack is drawn second:



                    $$3 tag4$$



                    The number of successful draws is the product of values $(1)$ through $(4)$ above.



                    The total number of possible draws of $18$ cards in order is:



                    $$52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$



                    The final probability is the number of successful possibilities, divided by the total number of possibilities:



                    $$displaystylefrac17 ,cdot, 4 ,cdot,left(,48,cdot, 47, cdot, 46,, cdots ,, 34 ,cdot ,33,right), cdot , 3 52 , cdot , 51 ,cdot , 50 ,,cdots,,36 ,cdot, 35$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Zubin Mukerjee

                    14.5k32456




                    14.5k32456




















                        up vote
                        0
                        down vote













                        The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$






                          share|cite|improve this answer






















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                            0
                            down vote










                            up vote
                            0
                            down vote









                            The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$






                            share|cite|improve this answer












                            The first $17$ cards can be a single jack ($1$ from $4$) plus any $16$ from $48$. The $18$th card can be one of $3$ jacks from the $35$ remaining cards. So I get: $$4cdot fracbinom4816binom5217cdot frac335 = 0.03523$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









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