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A person can guess the outcome of $6$ (cubical) dice rolled one by one, $1$ in $46656$ times. So, how much would it take for $46656$ people? Would it be every time, still $46656$, or something else?

ViHdzP
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Julian Costa
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    are you asking what the likelihood of 1 out of the 46656 people guessing it properly? – Sonny Da Silva-Peters Jun 14 '17 at 16:45
  • It should be clear that it's not every time. After all, there is a small but nonzero probability that every person guesses the same number on each roll. A single person has a $46655/46656$ chance of getting wrong, so assuming everybody chooses independently, what is the probability that everyone gets it wrong? – Jason Jun 14 '17 at 16:45
  • @SonnyDaSilva-Peters no,i mean if all of 46656 people start guessing at same time. – Julian Costa Jun 14 '17 at 16:46
  • Of course, this assumes everyone guesses independently. If everyone is allowed to collude, the optimal strategy will see exactly one person guessing all rolls correctly each time. – Jason Jun 14 '17 at 16:47
  • So statisticaly every time one of those 46656 should guess the outcome? @Jason – Julian Costa Jun 14 '17 at 16:49
  • @JulianCosta the expected number of correct guesses is one, however the probability that at least one person guesses correctly is $1-(\frac{46655}{46656})^{46656}$, which is not 100%. – JMoravitz Jun 14 '17 at 16:51
  • Statistically, no. I gave you a hint on how to work out the probability that at least one person guesses correctly, assuming everyone guesses independently. That would be what I call the "statistical" odds. If collusion is allowed, the number of people and total number of combinations are equal, so it's pretty obvious that with the right strategy someone will always "guess" correctly. – Jason Jun 14 '17 at 16:52
  • Thank you all for the answers! – Julian Costa Jun 14 '17 at 16:54
  • @JMoravitz So the sum of the probability of 1 person guessing correctly times one plus probability of 2 people guessing correctly times 2, etc. is 1 - which is the expected number of correct guesses? Whereas the probability of at least one person guessing correctly is 1 - probability of zero people guessing correctly? – Χpẘ Jun 14 '17 at 22:18
  • @Χpẘ that is correct... if you wish for more details, see the binomial theorem and the binomial distribution. It is well known that the expected value of a binomial distribution is $n\cdot p$. – JMoravitz Jun 14 '17 at 22:56
  • @JMoravitz that clears up something I've long wondered: why when the probability is 1 in $n$, performing the action $n$ times has less than a probability of 1. From experience I know flipping a coin twice or rolling a die 6 times doesn't guarantee a particular result. However it seemed counter intuitive to say you have a 1 in 36 chance of rolling snake eyes, but 36 rolls has a chance of yielding no snake eyes. But "36 rolls" has "hidden" within it the chance to roll two, three or more sets of snake eyes (in addition to none). – Χpẘ Jun 14 '17 at 23:20

2 Answers2

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The chance of $k$ successes is close to $1/(k!e)$. For example, the chance of no success, or exactly one success, are both $1$ in $e$, or about $37\%$

Empy2
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Probability that at least one person guesses right is one minus probability that everybody guesses wrong.

So:

  1. Probability that one person guesses wrong is $1-\frac{1}{46656}$
  2. Probability that everyone out of 46656 people guesses wrong is $\left(1-\frac{1}{46656}\right)^{46656}$
  3. Therefore probability that at least one of 46656 people guesses right is $1 - \left(1-\frac{1}{46656}\right)^{46656}$

Since it is not easy to compute this quantity exactly, we will use the following fact (proof): $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^{n} = \frac{1}{e}$$ Now we can approximate $$\left(1-\frac{1}{46656}\right)^{46656} \approx \frac{1}{e}$$ And hence probability that at least one of 46656 guesses right is $$1 - \left(1-\frac{1}{46656}\right)^{46656} \approx 1 - \frac{1}{e} \approx 63\%$$

emptysamurai
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