for this reason I recognize that if you role a typical pair the dice, your opportunities of acquiring Snake eyes (double 1s) is $1$ in $36$. What I"m not sure of is just how to carry out the tennis2007.orgematics to number out your opportunities of rolling snake Eyes at least once throughout a collection of rolls. I know if I role the dice $36$ times it won"t lead to a $100\%$ chance of rolling snake Eyes, and while i imagine it"s in the upper nineties, I"d favor to figure out exactly how i can not qualify it is.

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



The probability that hitting it at least once is $1$ minus the probabilty the never hitting it.

Every time you roll the dice, you have a $35/36$ opportunity of not hitting it. If you role the dice $n$ times, then the only case where you have actually never hit it, is once you have actually not hit it every single time.

The probabilty of no hitting v $2$ rolls is for this reason $35/36 imes 35/36$, the probabilty of not hitting v $3$ roll is $35/36 imes 35/36 imes 35/36=(35/36)^3$ and so ~ above till $(35/36)^n$.

Thus the probability the hitting the at the very least once is $1-(35/36)^n$ where $n$ is the number of throws.

After $164$ throws, the probability of hitting the at least once is $99\%$

point out
edited Oct 4 "18 at 13:00

J. M. Ain't a tennis2007.orgematician
70.7k55 gold badges184184 silver badges333333 bronze badges
reply Oct 3 "18 at 13:19

b00n heTb00n heT
13.8k11 yellow badge3030 silver- badges4141 bronze badges
add a comment |
The various other answers describe the basic formula because that the probability of never ever rolling snake eyes in a collection of $n$ rolls.

However, you additionally ask specifically about the instance $n=36$, i.e. If you have a $1$ in $k$ chance of success, what is your chance of gaining at the very least one success in $k$ trials? It turns out the the answer come this inquiry is quite similar for any reasonably large value that $k$.

It is $1-ig(1-frac1kig)^k$, and also $ig(1-frac1kig)^k$ converges come $e^-1$. So the probability will certainly be about $1-e^-1approx 63.2\%$, and also this approximation will certainly get better the larger $k$ is. (For $k=36$ the genuine answer is $63.7\%$.)

answered Oct 3 "18 at 13:40

particularly LimeEspecially Lime
36.6k77 gold badges4343 silver- badges7777 bronze badges
include a comment |
If you role $n$ times, climate the probability the rolling snake eyes at least once is $1-left(frac3536 ight)^n$, together you either roll snake eyes at least once or not at every (so the probability of these two events should amount to $1$), and also the probability of never ever rolling line eyes is the same as requiring the you roll one of the other $35$ possible outcomes on every roll.

point out
answer Oct 3 "18 at 13:22
Sam StreeterSam Streeter
1,52477 silver badges1818 bronze badges
include a comment |

her Answer

Thanks for contributing an answer to Stack Exchange!

Please be certain to answer the question. Carry out details and also share her research!

But avoid

Asking because that help, clarification, or responding to other answers.Making statements based upon opinion; ago them increase with referrals or personal experience.

Use tennis2007.orgJax to style equations. tennis2007.orgJax reference.

To find out more, check out our advice on writing great answers.

See more: 10 Year Old Cat Was Doused With Liquid D Is Palmolive Dish Soap Safe For Cats ?

Draft saved
Draft discarded

Sign increase or log in in

authorize up using Google
sign up using Facebook
authorize up utilizing Email and Password

Post as a guest

email Required, but never shown

Post together a guest


Required, but never shown

article Your prize Discard

By clicking “Post your Answer”, girlfriend agree to our regards to service, privacy policy and also cookie policy

Not the prize you're spring for? Browse other questions tagged probability dice or ask your very own question.

Featured on Meta
What is the probability of rolling at the very least one $7$, $11$, or doubles in one experiment consist of of 2 rolls?
Yahtzee Bar game
What space the opportunities of drawing 2 shotguns in Zombie Dice in this situation?
how do you calculation the amount of combine of 1000 dice rolls?
Expected number of dice rolls for a succession of dice rolls ending at snake eyes
dice odds that a certain number roll at least once
chances of rojo doubles on 2d6 vs 1d4 and also 1d6.
Probability of rolling at the very least one line eyes (pair of 2 ones) with four dice, rolling 3 time
warm Network questions much more hot inquiries

question feed
subscribe to RSS
inquiry feed To i ordered it to this RSS feed, copy and also paste this URL right into your RSS reader.
stack Exchange Network
site style / logo © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.11.5.40661

tennis2007.orgematics stack Exchange works best with JavaScript enabled

her privacy

By click “Accept every cookies”, friend agree stack Exchange have the right to store cookies on your machine and disclose information in accordance v our Cookie Policy.