Hardways
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?
Quote: FatGeezusI 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.
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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?
(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
tuttigym
Quote: tuttigymMission, 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?
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tuttigym
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.
Quote: Mission146(35/36)^212 = 0.00254856221
1/0.00254856221 = 1 in about 392.38
I agree.
I don’t 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.Quote: Mission146Quote: tuttigymMission, 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?
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tuttigym
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
Quote: Ace2I don’t 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 postQuote: Mission146Quote: tuttigymMission, 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?
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tuttigym
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
“Expected number,” 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.
You’d 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.
Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.Quote: tuttigymMission, 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?
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tuttigym
495 / 1,609 * 495 =~ 152.3 expected come out rolls in 495 rolls. As an estimate, 1 - (35/36)^152.3 = 98.6%
Quote: Ace2Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.Quote: tuttigymMission, 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?
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tuttigym
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.
Quote: tuttigymQuote: Ace2Using a Markov chain I get 98.21658% chance you will roll at least one 12 during comeout.Quote: tuttigymMission, 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?
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tuttigym
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?
Quote: tuttigymI 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.
Quote: ThatDonGuyIf you mean exactly once, it is 495 x (1/36) x (35/36)^494 = about 1 / 80,449.
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If you mean at most once, it is that plus (35/36)^495 = about 1 / 75,136.
I don’t think he means that. He means 12 on come outs in the next 495 rolls, starting roll 1 as a come out.
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.
Quote: Ace2Assuming the question is:
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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.
Yes I think that’s the question though great if tuttygim confirms. I think he picked 495 because that’s the reduced denominator in the chance of winning a pass line bet. So in some sense he thinks of that as a “craps cycle” maybe? But I could be guessing wrong as I don’t quite follow.
Quote: unJonQuote: Ace2Assuming the question is:
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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.
Yes I think that’s the question though great if tuttygim confirms. I think he picked 495 because that’s the reduced denominator in the chance of winning a pass line bet. So in some sense he thinks of that as a “craps cycle” maybe? But I could be guessing wrong as I don’t 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
Quote: MDawgSpeaking of dogs gambling, I have become aware of this cartoon. 2 Stupid Dogs.
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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.
In this cartoon I understand that the dogs call out for "Hard eight! Hard eight!" at the craps table.
