for this reason I recognize that if you role a traditional pair that dice, your possibilities of acquiring Snake eye (double 1s) is \$1\$ in \$36\$. What I"m not certain of is how to carry out the londonchinatown.orgematics to figure out your possibilities of rolling snake Eyes at least once throughout a collection of rolls. I know if I role the dice \$36\$ time it won"t result in a \$100\%\$ possibility of rolling line Eyes, and also while i imagine it"s in the upper nineties, I"d choose to figure out exactly how i can not qualify it is.

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

Every time you role the dice, you have actually a \$35/36\$ chance of not hitting it. If you role the dice \$n\$ times, climate the only situation where you have never struggle it, is as soon as you have not hit it every solitary time.

The probabilty of not hitting through \$2\$ rolls is hence \$35/36 imes 35/36\$, the probabilty of no hitting through \$3\$ rolls is \$35/36 imes 35/36 imes 35/36=(35/36)^3\$ and also so top top till \$(35/36)^n\$.

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

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

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edited Oct 4 "18 at 13:00

J. M. Ain't a londonchinatown.orgematician
answered Oct 3 "18 at 13:19

b00n heTb00n heT
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The various other answers explain the basic formula for the probability of never ever rolling snake eyes in a collection of \$n\$ rolls.

However, you additionally ask specifically around the situation \$n=36\$, i.e. If you have actually a \$1\$ in \$k\$ possibility of success, what is your opportunity of acquiring at the very least one success in \$k\$ trials? It transforms out that the answer to this concern is quite similar for any kind of reasonably big value of \$k\$.

It is \$1-ig(1-frac1kig)^k\$, and also \$ig(1-frac1kig)^k\$ converges come \$e^-1\$. So the probability will certainly be about \$1-e^-1approx 63.2\%\$, and this approximation will get much better the bigger \$k\$ is. (For \$k=36\$ the genuine answer is \$63.7\%\$.)

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answer Oct 3 "18 at 13:40

specifically LimeEspecially Lime
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If you roll \$n\$ times, then the probability of rolling line eyes at the very least once is \$1-left(frac3536 ight)^n\$, together you either role snake eyes at the very least once or no at every (so the probability of these two events should sum to \$1\$), and also the probability of never ever rolling snake eyes is the very same as requiring the you roll among the various other \$35\$ feasible outcomes on every roll.

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reply Oct 3 "18 in ~ 13:22
Sam StreeterSam Streeter
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