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Math Help - What are the odds to tie in a single-deck game of 'casino war' four consecutive times

  1. #1
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    What are the odds to tie in a single-deck game of 'casino war' four consecutive times

    Hi everybody,

    I played poker the other night and we used casino war in order to determine the dealer for the first hand. I tied with another player four times in a row and this seemed most unlikely. Let's quantify that!

    For clearity: We used a shuffled poker deck (52 cards: 13 faces a 4 suits) and two players alternately picked cards from the pile (without putting the cards back afterwards). Cards with the same face, e.g. two Kings, tie regardless of the suit. What is the probability of four ties in a row?

    I've come up with an answer but since it's combinatorics I am not sure if I'm correct. In my opinion, three cases have to be distinguished (The cards Jack, Queen, King, and Aces are abbreviated by their first letter.):

    (a) Each two of the four pairs are different, e.g. JJQQKKAA: The probability of this happening is
    P_1 = \frac{4}{52}  \frac{3}{51}  \frac{4}{50}  \frac{3}{49}  \frac{4}{48}  \frac{3}{47}  \frac{4}{46}  \frac{3}{45}.
    We are not restricted to these specific cards but can pick four of the 13 faces: So the binomial coefficient \binom{13}{4} arises as a factor. Another factor of \frac{4!}{1!1!1!1!} = 4! is in the game, since the sequence of pairs can be permutated (JJQQKKAA, JJQQAAKK, etc). So eventually, the probability of picking four different pairs is 4! \binom{13}{4} P_1 .

    (b) There are only two different kinds of cards, e.g. JJJJAAAA: The probability of this happening is (analog to (a)) P_2 = 4!\cdot 4!\cdot 44! / 52!. Again, lifting the restriction of having Jacks and Aces gives rise to a factor of \binom{13}{2}. This time, \frac{4!}{2!2!} = 6 permutations(*) are possible, thus there is a factor of 6. This leaves us with the probability 6 \binom{13}{2} P_2.
    (*) Those are: {JJJJAAAA, JJAAAAJJ, AAAAJJJJ, AAJJAAJJ, JJAAJJAA, AAJJJJAA}

    (c) Exactly one pair is picked twice, e.g. JJJJQQKK: The probability for this is P_3 =  4! \cdot 4\cdot3 \cdot 4\cdot3\cdot 44! / 52!. Just as before, an extra factor of \frac{4!}{2!1!1!} \cdot\binom{13}{3} is mandatory. This results in 12\cdot \binom{13}{3}\cdot P_3. The factor 12 is due to the possible permutations {JJJJQQKK, JJJJKKQQ, JJQQKKJJ, JJQQJJKK, JJKKQQJJ, JJKKJJQQ, QQJJJJKK, QQJJKKJJ, QQKKJJJJ, KKJJJJQQ, KKQQJJJJ, KKJJQQJJ}.

    The sum of these probabilities gives the probability to tie four times in a row in a game of casino war, which is
    4! \binom{13}{4} P_1 + 6 \binom{13}{2} P_2 + 12 \binom{13}{3} P_3.

    Am I correct or does anybody disagree? I tried different approaches and this one seems /for me/ to be the least confusing one.

    Thanks in advance for your help!
    needsomemathhelp
    Last edited by needsomemathhelp; October 5th 2010 at 07:11 AM. Reason: Factors arev now achieved systematically.
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  2. #2
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    I don't know if my answer is correct but I thought it'd be fun to take a shot at the problem. I didn't use combinations but rather just did:

    First tie:

    1*(\frac{3}{51})

    Second tie:

    (\frac{2}{50}*\frac{1}{49})+(\frac{48}{50}*\frac{3  }{49})

    Third tie:

    (\frac{2}{48}*\frac{1}{47})+(\frac{2}{48}*\frac{1}  {47})+(\frac{46}{48}*\frac{3}{47})

    Fourth tie:

    (\frac{2}{46}*\frac{1}{45})+(\frac{2}{46}*\frac{1}  {45})+(\frac{2}{46}*\frac{1}{45})+(\frac{44}{46}*\  frac{3}{45})

    The just multiply each of the answers from tie 1 through 4.

    The answer is \approx 1.50292*10^{-5}=0.00150292\%
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  3. #3
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    A new angle...

    Hi everyone,

    First of all I'd like to thank downthesun01 for his answer. It certainly gives me a new angle to the problem and I tried it this way myself. I've come up with a few extra terms, though.

    But there's no doubt about the first tie. The probability for that is
     1 \cdot \frac{3}{51}.

    And the probability for the second tie is (in concurrence with downthesun01's answer)
     1 \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} + 1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot\frac{3}{49} .
    The first term is the probability to pick four cards of the same kind (xxxx), whereas two different pairs (xxyy) are described by the second one.

    Continuing the scheme up to the fourth tie results in the following probabilites (combinations of pairs in square brackets):

    (b) [xxxxyyyy]  1 \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} \cdot 1 \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (c) [xxxxyyzz]  1 \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} \cdot 1 \cdot \frac{3}{47} \cdot \frac{44}{46} \cdot \frac{3}{45}

    (b) [xxyyxxyy]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48} \cdot \frac{1}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (c) [xxyyxxzz]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48} \cdot \frac{1}{47} \cdot \frac{44}{46} \cdot \frac{3}{45}

    (b) [xxyyyyxx]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48} \cdot \frac{1}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (c) [xxyyyyzz]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48} \cdot \frac{1}{47} \cdot \frac{44}{46} \cdot \frac{3}{45}

    (c) [xxyyzzxx]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{44}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (c) [xxyyzzyy]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{44}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (c) [xxyyzzzz]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{44}{48} \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    (a) [xxyyzzaa]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{44}{48} \cdot \frac{3}{47} \cdot \frac{40}{46} \cdot \frac{3}{45}

    Here, identical products are given the same letter in parenthesis.

    Product (a) arises only once and is identical with the product derived in my first approach (counting possibilities), i.e.  1\cdot (a) =  4!\binom{13}{4} P_1.

    Product (b) arises thrice and  3\cdot (b)  = 6\binom{13}{2}P_2, so this works out as well.


    Product (c) arises six times BUT  6\dcot (c)  \neq 12 \binom{13}{3}P_3. But  2\dcot (c)  = 12 \binom{13}{3}P_3 is correct.

    Why does this not work out? If product (c) arised only twice, everything would be fine. I reckon the new approach (calculating probabilities) counts some combinations thrice, but I cannot see why. Although I do understand why I don't have to worry about (yyyyxxxx) when I've already taken care of (xxxxyyyy).


    Sorry for my elongated answer, but in my opinion writing everything down explicitly makes it much easier to comprehend each step. Although I admit it looks a bit deterrent.


    I really appreciate your help on this one!

    Thanks in advance,
    needsomemathhelp
    Last edited by needsomemathhelp; October 6th 2010 at 02:59 AM.
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  4. #4
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    I tried to doing this using the hypergeometric distribution formula but I may have confused myself somehow.

    First tie:
    \frac{{{4}\choose{2}}{{48}\choose{0}}}{{{52}\choos  e{2}}}

    Second tie:
    \frac{{{2}\choose{2}}{{4}\choose{0}}{{44}\choose{0  }}}{{{50}\choose{2}}}+\frac{{{2}\choose{0}}{{4}\ch  oose{2}}{{44}\choose{0}}}{{{50}\choose{2}}}

    Third tie:
    \frac{{{2}\choose{2}}{{2}\choose{0}}{{4}\choose{0}  }{{40}\choose{0}}}{{{48}\choose{2}}}+\frac{{{2}\ch  oose{0}}{{2}\choose{2}}{{4}\choose{0}}{{40}\choose  {0}}}{{{48}\choose{2}}}+\frac{{{2}\choose{0}}{{2}\  choose{0}}{{4}\choose{2}}{{40}\choose{0}}}{{{48}\c  hoose{2}}}

    Fourth tie:
    \frac{{{2}\choose{2}}{{2}\choose{0}}{{2}\choose{0}  }{{4}\choose{0}}{{36}\choose{0}}}{{{46}\choose{2}}  }+\frac{{{2}\choose{0}}{{2}\choose{2}}{{2}\choose{  0}}{{4}\choose{0}}{{36}\choose{0}}}{{{46}\choose{2  }}}+\frac{{{2}\choose{0}}{{2}\choose{0}}{{2}\choos  e{2}}{{4}\choose{0}}{{36}\choose{0}}}{{{46}\choose  {2}}}+\frac{{{2}\choose{0}}{{2}\choose{0}}{{2}\cho  ose{0}}{{4}\choose{2}}{{36}\choose{0}}}{{{46}\choo  se{2}}}

    When you solve everything, the final answer actually reduces down to \frac{1}{4447625}

    That just seems like a really small probability... But after looking at a stats textbook I have, I believe that this is the correct answer. The book gives an example with playing cards and you can see the tiny probability associated with a particular poker hand.

    In a poker hand consisting of five cards, find the probability of holding 2 aces and 3 jacks

    Number of ways to choose 2 aces:
    {{4}\choose{2}}

    Number of ways to choose 3 jacks:
    {{4}\choose{3}}

    Number of ways to choose 5 cards:
    {{52}\choose{5}}

    So,
    \frac{{{4}\choose{2}}{{4}\choose{3}}}{{{52}\choose  {5}}}=\frac{24}{2598960}=\frac{1}{108290}
    Last edited by downthesun01; October 5th 2010 at 02:33 PM.
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  5. #5
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    I find this thread interesting and it is good to see so many different approaches. Unfortunately I don't have time right now to look over everything carefully. (Although in general I saw some good reasoning, obviously there must be mistakes somewhere since there are different answers obtained.) May I suggest starting with some smaller cases like three ties in a row to ensure that your multiple methods agree? Then once you can get the methods to agree, when you get the same answer ~three different ways you will have pretty good confidence that the answer is correct. Another good tool if you have the resources is computer simulations, which would tend to work a bit better if the probability isn't too terribly small. And you can consider decks that have for example 4*7 cards altogether to make the numbers a little more manageable, then generalize.
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  6. #6
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    Hi everyone,

    thanks again your effort. The hypergeometric series introduces another point of view and I gave it a try. Unfortunately I am basically stuck with the same problem.

    First of all, the hypergeometric series reproduces all of the probabilities from my previous post. Yet, the way downthesun01 used the series does not give the probabilities to tie, as can easily be seen at the first tie. Instead,  \frac{\binom{4}{2} \binom{48}{0}}{\binom{52}{2}} is the probability to tie with a specific pair, e.g. aces [AA]. Just multiply this by the number of ways to choose a specific card, i.e. \binom{13}{1} = 13, and you've got the probability for the first tie, that is [xx].

    Going on in this fashion reconfirms the probabilites already calculated in the "calculating probabilities" approach.

    Unfortunately, although the hypergeometric series is used to calculate the probabilities for [xxyyxxyy] etc., the tree scheme
    Code:
                         ---[xxyyxx]---(...)
            ---[xxyy]--- ---[xxyyyy]---(...)
           |             ---[xxyyzz]---(...)    
           |
    [xx]--- 
           |
           |                           ---[xxxxyyyy]
            ---[xxxx]-------[xxxxyy]---
                                       ---[xxxxyyzz]
    remains unchanged. Hence, Prob([xxyyxxyy]) = Prob([xxyyyyxx]) = Prob([xxxxyyyy]) occurs again three times and the prefactors in the "calculating probabilities" approach remain unaltered.

    Is this tree scheme correct, i.e. do I really have to account for product (c) six times? Or is my initial "counting combinations" approach correct and there's an error I just don't see with "calculating probabilities" (with or without employing the hypergeometric series).

    Unfortunately, both approaches appear rather convicing to me.

    Any ideas on what went wrong?

    Thanks again for helping,
    needsomemathhelp


    @undefined: Your suggestion to try a reduced deck of cards is surely very promising in order to get a more detailed view on the combinatorics. And you're right: A smaller deck renders a numerical treatment of the problem possible. It ought to be easy to implement this problem numerically and thereby decide whether the "calculating probabilities" approach is correct or "counting combination" does the job (or both fail). Yet, to put it in the words of Eugene Wigner: 'It is nice to know that the computer understands the problem. But I would like to understand it too.'
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    This post has been removed.
    Last edited by needsomemathhelp; October 6th 2010 at 11:33 AM. Reason: Misleading content.
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  8. #8
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    So I'm just looking through this thread from the first post onward, so hopefully I don't repeat anything that's already been said.

    Initially I could find no error in post #1, but while working through the problem some more I believe I uncovered one error. See below for details.

    Regarding post #2

    Quote Originally Posted by downthesun01 View Post
    I don't know if my answer is correct but I thought it'd be fun to take a shot at the problem. I didn't use combinations but rather just did:

    First tie:

    1*(\frac{3}{51})

    Second tie:

    (\frac{2}{50}*\frac{1}{49})+(\frac{48}{50}*\frac{3  }{49})

    Third tie:

    (\frac{2}{48}*\frac{1}{47})+(\frac{2}{48}*\frac{1}  {47})+(\frac{46}{48}*\frac{3}{47})
    The first and second tie are all right, but note that the probability of getting uuuu is different from the probability of getting uuvv and thus we cannot just lump them together, so this method falls apart on the third tie.

    I find it fairly easy to make mistakes with these problems, but I will attempt to get the methods from posts #1 and #2 to match for three ties.

    First, the method from post #1 reduced to three ties in a row. We have two cases

    (a) uuvvww

    gives 4*3*4*3*4*3*46!/52! * C(13,3) * 3!

    (b) uuuuvv

    gives 4!*4*3*46!/52! * C(13,2) * 2 * 3!/2!/1!

    Note the 2 in bold; this is there because uuuuvv is different from vvvvuu. I believe this consideration is lacking from post #1, part (c), where there should be an additional factor of 3 to account for the fact that there are 3 ways to choose the card value for which all four are drawn. (Perhaps the sentence is confusing; consider that JJJJQQKK is different from QQQQJJKK.)

    This comes out to 207/978775.

    Then the method from post #2

    Probability of getting first two ties is (3/51) ((2/50 * 1/49) + (48/50 * 3/49)).

    Let a = 2/50 * 1/49
    Let b = 48/50 * 3/49
    Let q = a / (a+b)
    Let r = b / (a+b)

    Then probability of getting three ties is

    (3/51) (a+b) (q * (48/48 * 3/47) + r * (2/48 * 1/47 + 2/48 * 1/47 + 44/48 * 3/47))

    This also comes out to 207/978775. So I'm pretty confident the answer is right.
    Last edited by undefined; October 6th 2010 at 08:02 AM.
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  9. #9
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    Quote Originally Posted by undefined View Post
    Note the 2 in bold; this is there because uuuuvv is different from vvvvuu.
    Can you please explain why you think there's a difference?

    Quote Originally Posted by undefined View Post
    I believe this consideration is lacking from post #1, part (c), where there should be an additional factor of 3 to account for the fact that there are 3 ways to choose the card value for which all four are drawn. (Perhaps the sentence is confusing; consider that JJJJQQKK is different from QQQQJJKK.)
    Though I'm afraid it is confusing, I think I've basically the same problem again: I doubt that JJJJQQKK is different from QQQQJJKK, because only the letters are switched. But even if the positions of the pairs were altered, since everything's made of products, everything commutes.

    The good news is that you came up with the same answer in two different ways. Unfortunately I don't understand it. Yet.
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  10. #10
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    Quote Originally Posted by needsomemathhelp View Post
    Can you please explain why you think there's a difference?
    No problem. (See below.)

    Quote Originally Posted by needsomemathhelp View Post
    Though I'm afraid it is confusing, I think I've basically the same problem again: I doubt that JJJJQQKK is different from QQQQJJKK, because only the letters are switched. But even if the positions of the pairs were altered, since everything's made of products, everything commutes.
    Hmm I'm sure this isn't as concise as it could be, but hopefully it will work for you.

    I use the phrase "card value" to mean one of {2,3,...10,J,Q,K,A}. Perhaps there is a more standard term. Anyway I mean that in a deck of cards there are exactly four suits and thirteen card values.

    For post #1, part (c), we wish to count all possible ways to get four ties in a row such that exactly three different card values appear. Just counting individual cards, this means there is one card value that appears four times and two card values that appear twice each. I will call an arrangement that satisfies these conditions "valid" in the context of the rest of this post.

    So think enumeratively (that is, listing all possible valid arrangements). We first select the three cards values, for example {J,Q,K}. Then we must count all possible valid arrangements using these three card values.

    The list you provided in post #1

    {JJJJQQKK, JJJJKKQQ, JJQQKKJJ, JJQQJJKK, JJKKQQJJ, JJKKJJQQ, QQJJJJKK, QQJJKKJJ, QQKKJJJJ, KKJJJJQQ, KKQQJJJJ, KKJJQQJJ}

    is only 1/3 of the possible arrangements. You fully miss all those arrangements that have Q appearing four times, and those arrangements that have K appearing four times.

    Note that the probability you gave

    P_3 =  4! \cdot 4\cdot3 \cdot 4\cdot3\cdot 44! / 52!

    is the probability of some particular valid arrangement. You must count all valid arrangements to get a correct result.

    If it's still not clear, think about how we could write the probability of (c) as a massive sum

    P(JJJJQQKK) + P(JJJJKKQQ) + P(JJQQKKJJ) + ...

    You see why we need to count JJJJQQKK and QQQQJJKK separately?
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  11. #11
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    Now seems a perfect time to respond to post #3.

    Quote Originally Posted by needsomemathhelp View Post
    (b) [xxxxyyyy]  1 \cdot \frac{3}{51} \cdot \frac{2}{50} \cdot \frac{1}{49} \cdot 1 \cdot \frac{3}{47} \cdot \frac{2}{46} \cdot \frac{1}{45}

    ...

    (a) [xxyyzzaa]  1 \cdot \frac{3}{51} \cdot \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{44}{48} \cdot \frac{3}{47} \cdot \frac{40}{46} \cdot \frac{3}{45}

    Here, identical products are given the same letter in parenthesis.
    What you have done here is modify the method of post #2 so that it now gives the right answer. It's cleaner than my modification in post #8.

    Quote Originally Posted by needsomemathhelp View Post
    Product (c) arises six times BUT  6\dcot (c)  \neq 12 \binom{13}{3}P_3. But  2\dcot (c)  = 12 \binom{13}{3}P_3 is correct.
    This is precisely the issue that I brought up in post #8 and explained in post #10. You are missing a factor of 3 in your original formulation.
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    Thanks for your quick reply!

    I think we're getting closer now. The point is that when introducing the factor C(13, 3), permutations of the three card values chosen are not considered! I thought I'd fixed that by the factor 4!/2!1!1!, which generates the list of permutations you quoted. But since permutations are not considered when choosing 3 card values out of 13, I am still missing QQQQJJKK when I start with JJJJQQKK. Unfortunately this is not fixed by replacing C(13, 3) with 3!C(13, 3), since some of the permutations already occur in the list. Only the permutations, where the card values of the double pair (JJJJ in JJJJQQKK) is replaced, have to be added. Since J is already accounted for, this leaves us with 2 further possibilities, K and Q. So I was indeed missing 2/3 of the arrangements. This can be fixed by an additional factor of 3, which fixes the differences of the results achieved by apporach one (1st post) and two (3rd post).

    Did I get that right?
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    Quote Originally Posted by needsomemathhelp View Post
    Thanks for your quick reply!

    I think we're getting closer now. The point is that when introducing the factor C(13, 3), permutations of the three card values chosen are not considered! I thought I'd fixed that by the factor 4!/2!1!1!, which generates the list of permutations you quoted. But since permutations are not considered when choosing 3 card values out of 13, I am still missing QQQQJJKK when I start with JJJJQQKK. Unfortunately this is not fixed by replacing C(13, 3) with 3!C(13, 3), since some of the permutations already occur in the list. Only the permutations, where the card values of the double pair (JJJJ in JJJJQQKK) is replaced, have to be added. Since J is already accounted for, this leaves us with 2 further possibilities, K and Q. So I was indeed missing 2/3 of the arrangements. This can be fixed by an additional factor of 3, which fixes the differences of the results achieved by apporach one (1st post) and two (3rd post).

    Did I get that right?
    Yes, that's right; good job!
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    Solution

    So the probability to tie four consecutive times is

    4!\binom{13}{4} P_1 + 6\cdot \binom{13}{2}P_2 + 3\cdot 12 \cdot \binom{13}{3}P_3 = \frac{1453}{5\cdot17\cdot23\cdot25\cdot47\cdot49} \approx 1.29088\cdot10^{-5}.

    Thank you very much for the discussions!
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  15. #15
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    Quote Originally Posted by needsomemathhelp View Post
    So the probability to tie four consecutive times is

    4!\binom{13}{4} P_1 + 6\cdot \binom{13}{2}P_2 + 3\cdot 12 \cdot \binom{13}{3}P_3 = \frac{1453}{5\cdot17\cdot23\cdot25\cdot47\cdot49} \approx 1.29088\cdot10^{-5}.

    Thank you very much for the discussions!
    You're welcome!

    I get the same answer you got.
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