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Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo On Vacation
    Strange Egg
    21 Feb '13 19:42 / 1 edit
    If you roll 5 (six sided) dice, is it just as likely you roll 3 of a kind as you do a pair or am I under thinking this a bit?
  2. Subscriber joe shmo On Vacation
    Strange Egg
    21 Feb '13 22:24
    Never mind, That is definitely not correct.
  3. Subscriber joe shmo On Vacation
    Strange Egg
    22 Feb '13 00:54
    I think I figured it out...

    the number of all possible cases is 6^5

    The number of favorable cases is

    C(5,3)=10, (the number of ways you can select 3 of 5 dice, distinctly)

    C(5,3)*6=60, (each one of the ten ways above has 6 choices. e.g.1-6)

    C(5,3)*6*5*5 = 1500, (each one of the 60 choices has a remaining 5 choices each for the two remaining selections)

    P(3 of a kind) = 1500/6^5.

    Before, when I thought P(1 pair)=P(3 of a kind) was because I failed to factor in the remaining two numbers(thus leaving out the factor of 125 and 25 respectively). And since C(5,3)=C(5,2) thought they were equal...in case anyone was wondering what all this fuss was about.
  4. Standard member talzamir
    Art, not a Toil
    22 Feb '13 11:32
    The chances of a three of a kind and a two of a kind are equal, but only on dice with just two different sides, such as coins.

    Going through all the possibilities I got

    five-of-a-kind: 6
    four-of-a-kind: 150
    full house: 450
    three-of-a-kind: 1200
    pair: 5400
    two pairs: 1800
    straight: 240
    no combination, gap straight: 480

    ..which is not right as it should add up to 7,776 but doesn't. Probably a permutation where there should be combination somewhere, or vice versa.
  5. Subscriber joe shmo On Vacation
    Strange Egg
    22 Feb '13 15:41 / 1 edit
    Originally posted by talzamir
    The chances of a three of a kind and a two of a kind are equal, but only on dice with just two different sides, such as coins.

    Going through all the possibilities I got

    five-of-a-kind: 6
    four-of-a-kind: 150
    full house: 450
    three-of-a-kind: 1200
    pair: 5400
    two pairs: 1800
    straight: 240
    no combination, gap straight: 480

    ..which is not right as ...[text shortened]... but doesn't. Probably a permutation where there should be combination somewhere, or vice versa.
    Full house: C(5,3)*6*5*1 = 300
    pair: C(5,2)*6*5*4*3 = 3600
    Straight: 6*5*4*3*2 = 720

    which when removing that "gap straight" adds to 7776
  6. Standard member talzamir
    Art, not a Toil
    22 Feb '13 15:42 / 1 edit
    So.. fixed. Seems we posted at the same time. =)

    five-of-a-kind: 6
    four-of-a-kind: 150
    full house: 300
    three-of-a-kind: 1200
    two pairs: 1800
    one pair: 3600
    straight: 240
    no combination, gap straight: 480
  7. Subscriber joe shmo On Vacation
    Strange Egg
    22 Feb '13 16:43 / 1 edit
    Originally posted by talzamir
    So.. fixed. Seems we posted at the same time. =)

    five-of-a-kind: 6
    four-of-a-kind: 150
    full house: 300
    three-of-a-kind: 1200
    two pairs: 1800
    one pair: 3600
    straight: 240
    no combination, gap straight: 480
    Can you walk me through the two pair calculation? I'm struggling to come up with it.

    I can't get away from 3600, the number for 1 pair which is obviously wrong, so what am I double counting?

    C(5,2)*6*C(3,2)*5*4=3600
  8. 23 Feb '13 01:09
    are aces wild?
  9. Standard member talzamir
    Art, not a Toil
    23 Feb '13 01:27 / 1 edit
    I went by yazzy rules. So a straigth is 1-2-3-4-5 or 2-3-4-5-6, but not 4-5-6-1-2. And for aces or deuces wild.. *grins* we'd get a whole different set of probabilities.
  10. Standard member talzamir
    Art, not a Toil
    23 Feb '13 01:39 / 1 edit
    Joe wrote: Can you walk me through the two pair calculation? I'm struggling to come up with it.


    Actually.. I got annoyed at the mistakes I did in the first calculation, so I did it the easy way, and wrote a bit of code that went over the 7,776 combinations and checked what they are.

    One pair is indeed in 3,600 combinations. In the first go I did it like this.

    Pick the number for the pair; six possibilities. They can be in (5,2) = 10 different places in the sequence.

    The other numbers that go into the gaps are one of five, one of four, and one of three possibilities.

    6 x 10 x 5 x 4 x 3 = 3,600.
  11. Subscriber joe shmo On Vacation
    Strange Egg
    23 Feb '13 17:29
    Originally posted by talzamir
    Joe wrote: Can you walk me through the two pair calculation? I'm struggling to come up with it.


    Actually.. I got annoyed at the mistakes I did in the first calculation, so I did it the easy way, and wrote a bit of code that went over the 7,776 combinations and checked what they are.

    One pair is indeed in 3,600 combinations. In the first go I did it ...[text shortened]... ps are one of five, one of four, and one of three possibilities.

    6 x 10 x 5 x 4 x 3 = 3,600.
    Ok, I was asking how you calculated "two pair's" number of calculations, but don't worry about it. I did some research and found why I was twice as high, because in each one of the combinations the order of the pairs can be switched (e.g. 1,1,2,2,5 has a compliment combination where the order of the pairs are switched i.e. 2,2,1,1,5), I missed that entirely.
  12. 23 Mar '13 21:25
    Its more complicate with aces wild, as in perudo.

    Here is a puzzle - in perudo you throw your dice and score for the largest number that are all the same, except with aces wild, eg, 1,1,3,5,5,6 would be four fives.

    I threw N dice which, with aces wild were just enough dice to make it more likely I would throw three of a kind than a pair. How many dice did I throw?

    Note three of the kind has to be the largest set to count, the above throw is four fives, and is NOT a pair OR three of a kind.