ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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September 27th, 2021 at 4:14:45 PM permalink
Quote: tuttigym

I reviewed my question and noted that I worded it poorly. The 12 is to appear or rolled only once in 495 CO. So the question revolves around NOT tossing more than one 12 in 495 CO's.


If you mean exactly once, it is 495 x (1/36) x (35/36)^494 = about 1 / 80,449.
If you mean at most once, it is that plus (35/36)^495 = about 1 / 75,136.
unJon
unJon
Joined: Jul 1, 2018
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September 27th, 2021 at 5:05:35 PM permalink
Quote: ThatDonGuy

If you mean exactly once, it is 495 x (1/36) x (35/36)^494 = about 1 / 80,449.
If you mean at most once, it is that plus (35/36)^495 = about 1 / 75,136.

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    I donít think he means that. He means 12 on come outs in the next 495 rolls, starting roll 1 as a come out.
    The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
    Ace2
    Ace2
    Joined: Oct 2, 2017
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    September 28th, 2021 at 1:24:39 PM permalink
    Assuming the question is:

    Starting with a comeout roll, what is the probability of rolling exactly one 12 during comeout in 495 total rolls (the 495 total is irrespective of comeout state).?

    The probability of >0 is easy to calculate with a Markov chain but I could not calculate the probability of >1 using Markov since it would require two absorbing states, which I think is impossible. The probability of exactly 1 would be the probability of >0 minus the probability of >1.
    Itís all about making that GTA
    unJon
    unJon
    Joined: Jul 1, 2018
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    September 28th, 2021 at 3:51:43 PM permalink
    Quote: Ace2

    Assuming the question is:

    Starting with a comeout roll, what is the probability of rolling exactly one 12 during comeout in 495 total rolls (the 495 total is irrespective of comeout state).?

    The probability of >0 is easy to calculate with a Markov chain but I could not calculate the probability of >1 using Markov since it would require two absorbing states, which I think is impossible. The probability of exactly 1 would be the probability of >0 minus the probability of >1.

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    Yes I think thatís the question though great if tuttygim confirms. I think he picked 495 because thatís the reduced denominator in the chance of winning a pass line bet. So in some sense he thinks of that as a ďcraps cycleĒ maybe? But I could be guessing wrong as I donít quite follow.
    The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
    tuttigym
    tuttigym
    Joined: Feb 12, 2010
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    October 3rd, 2021 at 8:59:13 AM permalink
    Quote: unJon

    Quote: Ace2

    Assuming the question is:

    Starting with a comeout roll, what is the probability of rolling exactly one 12 during comeout in 495 total rolls (the 495 total is irrespective of comeout state).?

    The probability of >0 is easy to calculate with a Markov chain but I could not calculate the probability of >1 using Markov since it would require two absorbing states, which I think is impossible. The probability of exactly 1 would be the probability of >0 minus the probability of >1.

  • link to original post



    Yes I think thatís the question though great if tuttygim confirms. I think he picked 495 because thatís the reduced denominator in the chance of winning a pass line bet. So in some sense he thinks of that as a ďcraps cycleĒ maybe? But I could be guessing wrong as I donít quite follow.
  • link to original post


    I am sorry if I have created some confusion, and I am grateful for the responsive posts.

    I will try to clarify more specifically the question although I believe TDG gave me the answer.

    If one were noting 495 consecutive hands and tracking only the come out rolls, what are the odds of throwing only one 12?
    tuttigym
    MDawg
    MDawg
    Joined: Sep 27, 2018
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    October 3rd, 2021 at 9:08:44 AM permalink
    Quote: MDawg

    Speaking of dogs gambling, I have become aware of this cartoon. 2 Stupid Dogs.

    However I cannot find it, or more especially, this episode, anywhere for viewing.
    https://fmovie.watch/tv-show/2-stupid-dogs/season/1/episode/5
    [Clicking the Play button in the above link doesn't seem to lead to anything productive.]

    VEGAS BUFFET
    The dogs are headed to the Lucky Nugget to partake of the 'Super Cheap Economy Style One Pound Hot Dog Buffet'. But they're early, so they wait. The Big Dog finds a quarter and uses it in a slot machine. By stroke of luck, they win the jackpot, attracting the attention of casino owner Hollywood. Wanting to win his money back, Hollywood takes the dogs around to the other games. But the dogs just want the buffet, so they keep betting everything, only to win more.

    The genre being dawgs, and gambling, and WINNING, naturally I am curious.

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    In this cartoon I understand that the dogs call out for "Hard eight! Hard eight!" at the craps table.
    I tell you itís wonderful to be here, man. I donít give a damn who wins or loses. Itís just wonderful to be here with you people. https://wizardofvegas.com/forum/gambling/betting-systems/33908-the-adventures-of-mdawg/

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