I understand somewhere on among the Wizard"s sites I experienced the answer come this question, however after looking for 2 hrs I cannot discover the answer.

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So... I think I read that the odds(probability) of rolling the exact same number twice in a row(any combination) was 10.67%... Any type of two sixes in a row.... Any type of two nines in a row... Etc...

The odds of roll a pair room one in six.The odds of rolling the exact same pair double in a heat is (1/6) * (1/36) = 1/(36*6) = 1/216The odds of roll a no pair is 5/6. The odds of rojo the exact same non-pair twice is (5/6) * (1/18) = 5/(6*18) = 5/108 = 1/21.6
times in a rowPair (1/x) no pair (1/x)
2216 21.6
31,296 388.8
446,656 6,998.4
51,679,616 125,971.2



I think by, "Same number," that meant, "Same outcome," come the extent that a six can be 5-1 or 3-3, doesn"t matter. I can be wrong, though, however he go mention, "Two nines," so that"s why I construed it that way.

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You need the linked probability of roll 2 adhered to by 2, 3 complied with by 3, ..., ..., ..., 12 adhered to by 12.1 means to role a 2 for this reason P(2 followed by 2) = 1/36 X 1/362 methods to role a 3 so P(3 followed by 3) = 2/36 X 2/363 means to roll a 4 ... = 3/36 X 3/364 methods to roll a 5 ... = 4/36 X 4/365 " " 66 " " 7 and so on etc and so on So you finish up with:-(1x1 + 2x2 + 3x3 + ... ... ... + 6x6 + 5x5 + ... ... 1x1) / (36 x 36)= (1 + 4 + 9 + 16 + 25 + 36 + 25 + 16 + 9 + 4 + 1) / 1296= 146/1296 = 11.265432%edit: Mission win me come it. At least we obtained the exact same answer :-)

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Thank you for the fast replies !!When ns said any type of number, I supposed the exact same number in any combination. Any mix of nines, back to back.For some reason, I had 10.67% grounding in my mind cells. I use this knowledge while betting the dark side and also laying odds top top the first occurrence the a number to stoke the lay-odds. I then regress the odds come a "comfortable-for-me" amount ~ the second roll. For example(I to be a cheap bettor, playing a $5 table), when the 9 comes up ~ above the an initial roll, i lay $15. When the following number appears and also it is not a 9 or 7, ns regress to $9 and also leave it for the remainder of the shooter"s turn. I hope that renders sense... If not, you re welcome feel complimentary to lamb-baste me and also show me the errors in mine ways. Thanks in advance... JeffreyEdit to include : provided the math above by the well reply-ers, can I deduce the following odds for each in-play number for the Don"t side? : 4 = 16 / 1296 = 0.0123456795 = 25 / 1296 = 0.01929012356 = 36 / 1296 = 0.02777777788 = 36 / 1296 = 0.02777777789 = 25 / 1296 = 0.019290123510 = 16 / 1296 = 0.012345679TOT = 154 154/1296 = 11.88 %