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schmidtdancer

  • one year ago

Back to the cards! In poker, a flush is when all five cards are the same suit. Find the probability of being dealt a flush (when being dealt five cards). Start by just considering clubs.

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  1. schmidtdancer
    • one year ago
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    a) What is the probability that the first card dealt is a club? b) What is the probability that the second card dealt is a club given that the first one was a club? c) What is the probability that the third card dealt is a club given that the first two were clubs? d) What is the probability that the fourth card dealt is a club given that the first three were clubs? e) What is the probability that the fifth card dealt is a club given that the first four were clubs? f) The probability of being dealt all five clubs is the product of the above probabilities. Why is this true and what is this probability? g) You have now found the probability of being dealt a flush in clubs. This is the same as the probability of being dealt a flush in diamonds, hearts, or spades. Then, what is the proability of being dealt a flush?

  2. hba
    • one year ago
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    How many cards in a deck ?

  3. schmidtdancer
    • one year ago
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    Umm... im not sure

  4. schmidtdancer
    • one year ago
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    @mathmate

  5. schmidtdancer
    • one year ago
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    @Hero @radar

  6. schmidtdancer
    • one year ago
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    @marsss

  7. EulerGroupie
    • one year ago
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    There are 52 cards in a standard deck. Divide by 4 to get the number of cards of each suit.

  8. radar
    • one year ago
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    That would mean there are 13 club cards out of 52 cards. Wouldn't that mean the probability of drawing a club is 1 out of 4?

  9. mathmate
    • one year ago
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    Yes, drawing the first club out of the full deck of 52 cards has a probability of 1 out of 4, or 0.25. What about the probabiliity of the second card: - how many clubs are left (assuming first one is a club) ? - how many cards are left in the deck?

  10. radar
    • one year ago
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    Now on the 2nd draw there would be 13 club cards out of 51 3rd draw 12 out of 50 4th draw 11 out of 49 5th draw 10 out of 48

  11. mathmate
    • one year ago
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    ...huh?

  12. radar
    • one year ago
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    Wouldn't that be the chances for each draw assuming he draws into a 5 card flush. Now take the product of all those chances to get the probability. 1/4 * 13/51 * 6/25 * 11/49 * 5/24

  13. radar
    • one year ago
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    I would guesstimate a little less than 1400 to 1

  14. radar
    • one year ago
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    What does google say?

  15. schmidtdancer
    • one year ago
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    hey I'm here now

  16. schmidtdancer
    • one year ago
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    so a i got 1/4. what is b?

  17. radar
    • one year ago
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    Chance according to Google is 0.003940. or about 254 to 1 so it looks like I probably erred.

  18. schmidtdancer
    • one year ago
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    so b is 254/1?

  19. schmidtdancer
    • one year ago
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    or .004?

  20. radar
    • one year ago
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    I am afraid I have been away from those kind of problems too long. Hopfully, @mathmate will provide further assistance.

  21. schmidtdancer
    • one year ago
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    Do u know anyone online right now that can give me assistance now?

  22. schmidtdancer
    • one year ago
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    @Agent_Sniffles

  23. schmidtdancer
    • one year ago
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    @zepdrix

  24. radar
    • one year ago
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    The Google solution involved a flush in any suit not just clubs.

  25. schmidtdancer
    • one year ago
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    huh?

  26. radar
    • one year ago
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    The probability of a flush can occur in hearts, diamonds, spades or clubs, just as long as all cards are the same suit!

  27. schmidtdancer
    • one year ago
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    So what would my answers be?

  28. radar
    • one year ago
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    You requested the probability of a flush in clubs only.

  29. schmidtdancer
    • one year ago
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    Yea. thats what the question said

  30. schmidtdancer
    • one year ago
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    how would I start answering B?

  31. radar
    • one year ago
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    I answered that, if the first one was a club, you now have a deck of 51 cards of which 12 are clubs; 12/51 or 0.235294

  32. schmidtdancer
    • one year ago
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    Ok so b would then be 12/51?

  33. radar
    • one year ago
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    Here is mathmate. hopefully shed some light on this.

  34. schmidtdancer
    • one year ago
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    I have (a) 1/4 (b) 12/51 now c?

  35. radar
    • one year ago
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    Ask yourself how many cards are now in the deck, how many are clubs and figure out the probability using the method you have been taught.

  36. mathmate
    • one year ago
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    Sorry for being away. @radar sorry, I was just questioning in case there was a typo. For (b) After the first card, there are 13-1=12 clubs left out of 51. So the probability is 12/51. Or, using conditional probabilities: P(1)=13/52 P(1&2)=13/52*12/51 P(2|1)=P(2&1)/P(1)=(13/52*12/51) / (13/52) = 12/51, same as before.

  37. schmidtdancer
    • one year ago
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    Okay I got that for (b) too!:) Im not sure how to find c now

  38. radar
    • one year ago
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    Sorry but I have to now run, you are in good hands.

  39. schmidtdancer
    • one year ago
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    Would (c) be 11/51 then?

  40. mathmate
    • one year ago
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    @schmidtdancer The questions are made in such a way to guide you to the final answer. I suggest that after a and b have been explained and answered, it would be advantageous for you to continue the logic and post your suggested responses for verification. What do you think?

  41. schmidtdancer
    • one year ago
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    ? I just need clarification on how to find C. can u guide me through the steps....and ill figure it out by myself then u can check?

  42. schmidtdancer
    • one year ago
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    @mathmate

  43. schmidtdancer
    • one year ago
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    @precal

  44. mathmate
    • one year ago
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    Sure!

  45. schmidtdancer
    • one year ago
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    Thanks! how do i begin c?

  46. schmidtdancer
    • one year ago
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    c) What is the probability that the third card dealt is a club given that the first two were clubs?

  47. mathmate
    • one year ago
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    For the third club in a row, how many clubs are left? and how many cards are left in the deck?

  48. schmidtdancer
    • one year ago
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    since there are 14 clubs in a set, then we would have 11 left right?

  49. schmidtdancer
    • one year ago
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    because b is 12/51? and were losing another so would c be 11/51?

  50. mathmate
    • one year ago
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    I'll make it clear: After the first two clubs are drawn and before we draw the third card, how many clubs remain in the deck, and how many cards total remain?

  51. schmidtdancer
    • one year ago
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    there are 14 clubs total in a deck... and 52 cards total in a deck... so, after two are drawn, then we have 12 clubs and 50 cards?

  52. schmidtdancer
    • one year ago
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    is that for b or c?

  53. schmidtdancer
    • one year ago
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    @mathmate ????

  54. schmidtdancer
    • one year ago
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    @countonme123

  55. mathmate
    • one year ago
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    Each deck has 52 cards, divided by 4 suits gives 13 cards per suit to start with (not 14). (a) before drawing any card, we have 13 clubs and 52 cards. (b) given the first card drawn was a club, before drawing the second card, we have 12 clubs and 51 cards. (c) given the first 2 cards drawn were clubs, before drawing the third card, we have how many clubs and how many cards in the deck?

  56. schmidtdancer
    • one year ago
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    10/49?

  57. schmidtdancer
    • one year ago
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    @mathmate

  58. mathmate
    • one year ago
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    We have only drawn 2 clubs out of 13, how many left?

  59. schmidtdancer
    • one year ago
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    11

  60. mathmate
    • one year ago
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    Sorry, the OS seems to be very selective in response. I cannot get to your question unless I go by your profile. When I go by "mathematics", it never responds for the past two days.

  61. schmidtdancer
    • one year ago
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    Its ok. but is 11 right?

  62. mathmate
    • one year ago
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    (c) Also, drawn two cards (clubs) out of 52, how many left?

  63. schmidtdancer
    • one year ago
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    50

  64. schmidtdancer
    • one year ago
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    so 11/52 would be c?

  65. schmidtdancer
    • one year ago
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    I mean 11/50!!!!

  66. schmidtdancer
    • one year ago
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    right @mathmate

  67. mathmate
    • one year ago
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    Yes, that is correct for (c). Again, your response was not updated. I had to check through you profile every time I suspect a response. You're comfortable continuing?

  68. schmidtdancer
    • one year ago
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    Okay, and yes

  69. schmidtdancer
    • one year ago
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    How would I find (d) now @mathmate

  70. mathmate
    • one year ago
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    Each deck has 52 cards, divided by 4 suits gives 13 cards per suit to start with (not 14). (a) before drawing any card, we have 13 clubs and 52 cards. (b) given the first card drawn was a club, before drawing the second card, we have 12 clubs and 51 cards. (c) given the first 2 cards drawn were clubs, before drawing the third card, we have how many clubs and how many cards in the deck? (d) given the first 3 cards drawn were clubs, before drawing the fourth card, we have how many clubs and how many cards in the deck?

  71. schmidtdancer
    • one year ago
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    so (d) is: 10/49?

  72. schmidtdancer
    • one year ago
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    This is what I have.... is it correct @mathmate ? (a) 1/4 (b) 12/51 (c) 11/50 (d) 10/49 (e) 9/48

  73. mathmate
    • one year ago
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    Looks good so far. Keep it up, you're almost there!

  74. schmidtdancer
    • one year ago
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    Im a little confused on f and g!

  75. mathmate
    • one year ago
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    Have you done conditional probability before?

  76. schmidtdancer
    • one year ago
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    Kinda

  77. schmidtdancer
    • one year ago
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    When it says product, do I just multiply a-e?

  78. schmidtdancer
    • one year ago
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    ?@mathmate

  79. mathmate
    • one year ago
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    Yes, the numerical part is just the product of the 5 probabilities. Give me a minute for the explanation part of (f). Once you have the numerical probability of clubs obtained in (f), how would you propose to find (g)?

  80. schmidtdancer
    • one year ago
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    so f is then: 11880/23990400??

  81. schmidtdancer
    • one year ago
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    @mathmate

  82. schmidtdancer
    • one year ago
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    Then would g be the decimal value of f?

  83. mathmate
    • one year ago
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    It's almost correct, but you need to simplify it to the simplest form. Can you do that?

  84. schmidtdancer
    • one year ago
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    I was gonna ask u lol, I'm not sure how to simplify it... im trying

  85. mathmate
    • one year ago
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    No. Think of (g) is for the case where there are 4 suits instead of just clubs. Imagine buying raffle tickets. What are the changes of winning if you bought 4 instead of one?

  86. schmidtdancer
    • one year ago
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    I have 495/999600 so far

  87. schmidtdancer
    • one year ago
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    for f..

  88. schmidtdancer
    • one year ago
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    33/66640

  89. schmidtdancer
    • one year ago
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    Is this f?? 33/66640

  90. mathmate
    • one year ago
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    Factors that you can cancel are like 36...

  91. schmidtdancer
    • one year ago
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    is this right @mathmate 33/66640

  92. mathmate
    • one year ago
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    That is correct.

  93. schmidtdancer
    • one year ago
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    Ok thank you! now, g?

  94. mathmate
    • one year ago
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    Now try (g)

  95. schmidtdancer
    • one year ago
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    hmm?

  96. schmidtdancer
    • one year ago
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    g) You have now found the probability of being dealt a flush in clubs. This is the same as the probability of being dealt a flush in diamonds, hearts, or spades. Then, what is the proability of being dealt a flush?

  97. schmidtdancer
    • one year ago
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    im not sure what its asking for an answer....@mathmate

  98. mathmate
    • one year ago
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    Means a flush of any of the 4 suits. Imagine buying raffle tickets. What are the changes of winning if you bought 4 instead of one?

  99. mathmate
    • one year ago
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    *chances

  100. schmidtdancer
    • one year ago
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    4 times as much?@mathmate

  101. schmidtdancer
    • one year ago
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    ?

  102. mathmate
    • one year ago
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    Right! That's. I hope you are better prepared for the next question.

  103. schmidtdancer
    • one year ago
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    But what would g be??

  104. schmidtdancer
    • one year ago
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    Is the answer: 4 times as much? @mathmate

  105. mathmate
    • one year ago
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    4 times what you got for (f).

  106. schmidtdancer
    • one year ago
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    so is it: .002?

  107. schmidtdancer
    • one year ago
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    132/266560 in fraction form <-- is this the answer @mathmate

  108. mathmate
    • one year ago
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    I would rather put it as 4*(33/66640) =33/16660. In probabilities, small numbers like 0.002 is better represented by fractions.

  109. schmidtdancer
    • one year ago
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    thx! so g @mathmate is 33/16660?

  110. schmidtdancer
    • one year ago
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    but thats lower than f.

  111. mathmate
    • one year ago
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    Yes. I am sorry it is probably painful for you as much as for me because the system does not respond (does not update). So I don't really know when you put in a response.

  112. schmidtdancer
    • one year ago
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    its ok but @mathmate how is g 33/16660? when f is higher than that value?

  113. mathmate
    • one year ago
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    When the denominator is 4 times smaller, it means that the fraction is 4 times bigger. For example, 1/4 is smaller than 1/1.

  114. schmidtdancer
    • one year ago
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    Oh oak!! thanks for all your help

  115. mathmate
    • one year ago
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    You're welcome! Good luck with your homework/exam! :)

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