My Math Forum Prob. of Rolling a Dice 5 Times and it Landing on 1 at least 3 times

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 May 22nd, 2019, 01:09 PM #1 Newbie   Joined: Oct 2018 From: UK Posts: 8 Thanks: 0 Prob. of Rolling a Dice 5 Times and it Landing on 1 at least 3 times I posted a question a few months ago regarding the probability of a Coin landing on Heads 6 times or more out of a total of 10 Throws (assuming the probability is exactly 50/50). Someone was helpful enough to break down a formula which I had a hard time understanding into an easy to understand example (see below): ------------------------------------------------------------ The probability of a Coin landing on Heads 6 times or more is: $\dfrac{1}{1024} * \left ( \dfrac{10!}{6! * 4!} + \dfrac{10!}{7! * 3!} + \dfrac{10!}{8! * 2!} + \dfrac{10!}{9! * 1!} + \dfrac{10!}{10! * 0!} \right ) = \\ \dfrac{1}{1024} * (210 + 120 + 45 + 10 + 1) \approx 37.70\%.$ ------------------------------------------------------------ I created a computer program that could work out the probability for other scenarios based on the formula above using factorials. However it did not work for anything other than coin throws. For example the formula did not work for dice throws where the probability is 1/6. Could someone please provide a breakdown of a formula (like the one above) to work out the probability of rolling a dice Five times and it landing on the number 1 Three times or more (assuming the probability is exactly 1/6. I can then extrapolate the example to work for very different scenarios. Thanks
 May 22nd, 2019, 01:40 PM #2 Global Moderator   Joined: May 2007 Posts: 6,770 Thanks: 700 The formula is based on the binomial distribution. Let $n$ be the number of trials and let $p$ be the probability of success and $q=1-p$ the probability of failure. Then the probability of exactly $k$ successes in $n$ trials is $\binom{n}{k}p^kq^{n-k}$ In your case $n=5$ and $p=1/6$ so what you want is$\sum_{k=3}^5\binom{5}{k}(\frac{1}{6})^k(\frac{5 }{6})^{5-k}$. Note $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is standard notation for binomial term.
May 22nd, 2019, 02:42 PM   #3
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 Originally Posted by mathman The formula is based on the binomial distribution. Let $n$ be the number of trials and let $p$ be the probability of success and $q=1-p$ the probability of failure. Then the probability of exactly $k$ successes in $n$ trials is $\binom{n}{k}p^kq^{n-k}$ In your case $n=5$ and $p=1/6$ so what you want is$\sum_{k=3}^5\binom{5}{k}(\frac{1}{6})^k(\frac{5 }{6})^{5-k}$. Note $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ is standard notation for binomial term.
Thanks for your response but I was hoping for an answer broken down to its most simple form - like the example I provided. Preferably one that includes factorials (if applicable)

My maths relating to probability is pretty basic so i'm not even sure what the symbol means. I can't google it as I don't know it's name.

May 22nd, 2019, 02:50 PM   #4
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 Originally Posted by enigmatic Thanks for your response but I was hoping for an answer broken down to its most simple form - like the example I provided. Preferably one that includes factorials (if applicable) My maths relating to probability is pretty basic so i'm not even sure what the symbol means. I can't google it as I don't know it's name.
$\dfrac{5!}{3!2!}\left(\dfrac 1 6\right)^3\left(\dfrac 5 6\right)^2 + \dfrac{5!}{4!}\left(\dfrac 1 6\right)^4\left(\dfrac 5 6\right) + \left(\dfrac 1 6\right)^5$

May 22nd, 2019, 03:49 PM   #5
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 Originally Posted by romsek $\dfrac{5!}{3!2!}\left(\dfrac 1 6\right)^3\left(\dfrac 5 6\right)^2 + \dfrac{5!}{4!}\left(\dfrac 1 6\right)^4\left(\dfrac 5 6\right) + \left(\dfrac 1 6\right)^5$
Thank you. I almost understand how to extrapolate that formula to more complex scenarios. However I don't understand where the 2! in the first part of the equation (5!/3!2!) and the 4! in 5!/4! come from since they weren't numbers I mentioned in the problem. (I suspect 2! represents the failures).

Also do the powers of 3, 2, 4 and 5 remain the same regardless of the number of successes, trials and probability?

Is it possible to provide a very similar formula that includes say 7 roles of the dice where 4 or more rolls successfully land on 1. I could then extrapolate from your previous example how the formula would work for more complex scenarios?

Last edited by enigmatic; May 22nd, 2019 at 03:52 PM. Reason: clarification regarding the question

May 22nd, 2019, 06:24 PM   #6
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 Originally Posted by enigmatic Thank you. I almost understand how to extrapolate that formula to more complex scenarios. However I don't understand where the 2! in the first part of the equation (5!/3!2!) and the 4! in 5!/4! come from since they weren't numbers I mentioned in the problem. (I suspect 2! represents the failures). Also do the powers of 3, 2, 4 and 5 remain the same regardless of the number of successes, trials and probability? Is it possible to provide a very similar formula that includes say 7 roles of the dice where 4 or more rolls successfully land on 1. I could then extrapolate from your previous example how the formula would work for more complex scenarios?
Spend a few minutes reading the wiki on the binomial distribution and see if you still have any questions.

Also review the definition of $\dbinom{n}{k} = \dfrac{n!}{k!(n-k)!}$

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