FatGeezus
FatGeezus
Joined: Jun 12, 2010
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September 23rd, 2021 at 9:21:27 AM permalink
I sat down to play the Bubble Craps game at the Tropicana AC casino. I noticed that the Hard 4 had not been thrown in quite a while. The Bubble Craps display board keeps track of the last time a Hardway 4, 6, 8, 10 was thrown. I kept watching to see how long the streak would go.

FINALLY. The last time a Hard 4 was thrown was 212 rolls ago. On the 213th roll a Hard 4 was finally thrown.

What are the odds of not throwing a Hard 4 - 212 times?
Mission146
Mission146
Joined: May 15, 2012
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September 23rd, 2021 at 9:28:54 AM permalink
Quote: FatGeezus

I sat down to play the Bubble Craps game at the Tropicana AC casino. I noticed that the Hard 4 had not been thrown in quite a while. The Bubble Craps display board keeps track of the last time a Hardway 4, 6, 8, 10 was thrown. I kept watching to see how long the streak would go.

FINALLY. The last time a Hard 4 was thrown was 212 rolls ago. On the 213th roll a Hard 4 was finally thrown.

What are the odds of not throwing a Hard 4 - 212 times?

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    (35/36)^212 = 0.00254856221

    1/0.00254856221 = 1 in about 392.38

    It's pretty unlikely, but unfortunately not quite as remarkable as one might think it should be. What would be remarkable is if the dice were to avoid every single hard way, Snake Eyes and Midnight for 212 consecutive rolls:

    (30/36)^212 = 1.6352187e-17 or 0.000000000000000016352187

    1/0.000000000000000016352187 = 1 in about 61,153,899,000,000,000
    https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
    tuttigym
    tuttigym
    Joined: Feb 12, 2010
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    September 23rd, 2021 at 10:41:22 AM permalink
    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym
    Mission146
    Mission146
    Joined: May 15, 2012
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    September 23rd, 2021 at 11:38:15 AM permalink
    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post



    Yes, you could calculate the expected number of Come Out rolls in that sample by just dividing it (it's on WoO somewhere) from the 495, then just determine what the probability is of having that many rolls without a 12.
    https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
    Wizard
    Administrator
    Wizard
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    September 23rd, 2021 at 1:38:32 PM permalink
    Quote: Mission146

    (35/36)^212 = 0.00254856221

    1/0.00254856221 = 1 in about 392.38



    I agree.
    It's not whether you win or lose; it's whether or not you had a good bet.
    Ace2
    Ace2
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    September 23rd, 2021 at 1:41:41 PM permalink
    Quote: Mission146

    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post



    Yes, you could calculate the expected number of Come Out rolls in that sample by just dividing it (it's on WoO somewhere) from the 495, then just determine what the probability is of having that many rolls without a 12.
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    I donít think that would work. The expected number of something is usually a totally different calculation than the probability of something. Also, I think he is asking the probability of exactly one, not the probability of more than zero.
    Itís all about making that GTA
    Mission146
    Mission146
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    September 23rd, 2021 at 1:56:18 PM permalink
    Quote: Ace2

    Quote: Mission146

    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post



    Yes, you could calculate the expected number of Come Out rolls in that sample by just dividing it (it's on WoO somewhere) from the 495, then just determine what the probability is of having that many rolls without a 12.
  • link to original post

    I donít think that would work. The expected number of something is usually a totally different calculation than the probability of something. Also, I think he is asking the probability of exactly one, not the probability of more than zero.
  • link to original post



    ďExpected number,Ē in this case just means how many Come Out rolls (relative to total) based on how many rolls (on average) an initial Come Out roll takes to resolve.

    Youíd have to use something, otherwise, too many PSO would often result in more CO than you expect to see on average and even one uncharacteristically long roll could result in less.
    https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
    Ace2
    Ace2
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    September 23rd, 2021 at 4:16:31 PM permalink
    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post

    Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.

    495 / 1,609 * 495 =~ 152.3 expected come out rolls in 495 rolls. As an estimate, 1 - (35/36)^152.3 = 98.6%
    Last edited by: Ace2 on Sep 23, 2021
    Itís all about making that GTA
    tuttigym
    tuttigym
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    September 27th, 2021 at 12:02:56 PM permalink
    Quote: Ace2

    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post

    Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.

    495 / 1,609 * 495 =~ 152.3 expected come out rolls in 495 rolls. As an estimate, 1 - (35/36)^152.3 = 98.6%
  • link to original post


    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. I apologize for any confusion. I can see where there would be a 98+% chance of throwing a 12 in 495 CO's, but that was not my intent.
    unJon
    unJon
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    September 27th, 2021 at 4:02:41 PM permalink
    Quote: tuttigym

    Quote: Ace2

    Quote: tuttigym

    Mission, Could one calculate the odds of throwing a 12 at come out only once in 495 tosses? Since come out tosses are not consecutive and the actual number of tosses in between can never be known, is such a calculation even possible?

    tuttigym

  • link to original post

    Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.

    495 / 1,609 * 495 =~ 152.3 expected come out rolls in 495 rolls. As an estimate, 1 - (35/36)^152.3 = 98.6%
  • link to original post


    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. I apologize for any confusion. I can see where there would be a 98+% chance of throwing a 12 in 495 CO's, but that was not my intent.
  • link to original post



    495 come out rolls? Or 495 sequential rolls, some of which are come outs?
    The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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