You are watching: Odds of rolling snake eyes twice in a row

**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

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.

See more: Driving Distance From Wilmington Nc To Savannah Ga 303 Miles

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 %

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 row | Pair (1/x) | no pair (1/x) |

2 | 216 | 21.6 |

3 | 1,296 | 388.8 |

4 | 46,656 | 6,998.4 |

5 | 1,679,616 | 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.

See more: Driving Distance From Wilmington Nc To Savannah Ga 303 Miles

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 :-)

It might be an amazing side-bet. The crapless come gambling on the hop! If you don"t hit the come bet ideal away, friend lose! If you execute hit it, you victory 11 because that 1.

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 %