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.

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?link to original post

(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?

tuttigymlink 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.

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?

tuttigymlink 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

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.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?

tuttigymlink 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 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:tuttigym

tuttigymlink to original post

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:tuttigym

tuttigymlink to original post

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:tuttigym

tuttigymlink to original post

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?