I am so confused; wouldn't it just be 1/5(5)? The probability of rolling a six-sided dice five times and getting a 1 at least one time is 5/6, but how would this work for a five-sided dice?
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1$\begingroup$ Compute the probability of not getting a $1$ in the $5$ rolls, and then compute $1-$ that probability.\ $\endgroup$– Rushabh MehtaCommented Feb 27, 2022 at 17:49
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$\begingroup$ What does 1/5(5) mean? Actually hard to think of a more confusing way to write something so short... $\endgroup$– luluCommented Feb 27, 2022 at 17:50
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$\begingroup$ It appears the probability that you are computing is the probability of rolling any one five different outcomes. i.e. $(1/5)*5$.Of course this should come out to $1$ as this is a guarantee on a five-sided die with five different outcomes $\endgroup$– WaveXCommented Feb 27, 2022 at 17:52
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$\begingroup$ Since no one else has pointed it out, the probability of rolling a six-sided die five times and getting a 1 at least one time is not 5/6. You're starting with a false premise, so that's why you're finding it difficult to generalize to a different case. $\endgroup$– Alex JonesCommented Feb 28, 2022 at 2:06
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3 Answers
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Suppose the die has $S$ sides and is thrown $n$ times. At any given throw, the chance of getting a 1 is $\frac{1}{S}$ and the chance of getting something else than 1 is $\frac{S-1}{S} = 1 - 1/S$. Let $E$ denote the event where at least a 1 is thrown after $n$ throws. The complement of $E$ is the event where every single throw is not a 1. Hence
$$P(E) = 1- P(E^c) = 1 - (1-1/S)^n$$
Setting $S = 5$ and $n=5$, we get $P(E) = 1 - (1-1/5)^5 = 1 - (4/5)^5$
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Your dice has $5$ possible outcomes, each with a probability of $\frac{1}{5}$. So on each throw $P(1) = \frac{1}{5}$. The probability of rolling the dice $5$ times and getting $1$ at least one time can be easily calculated if you notice that the probabilities fit a binomial distribution.
$P(x = 0, n=5, p=0.2) = \binom{5}{0}0.2^0(1-0.2)^5 = 0.8^5 = 0.32768$
This means the probability of not getting any $1$ is $0.327$. Therefore the probability of getting $1$ at least one time is $1-0.327 = 0.67232 \approx 67.2\%$
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1$\begingroup$ A very approachable explanation of the binomial distribution and its uses in probability can be found here: istat924158936.wordpress.com/2022/02/16/… If you are looking for something less beginner friendly and more technical, just check the Wikipedia page. $\endgroup$– lafinurCommented Feb 27, 2022 at 18:12
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The probability of rolling a six-sided dice five times and getting a 1 at least one time is 5/6
Okay, so the probability of rolling a six-sided die seven times and getting a 1 at least one time is $7/6$?
But probabilities can't be larger than $1$.
You cannot add probabilities unless events are disjoint (i.e. can't both happen at once). Here, it can happen that you get a 1 on more than one roll. Your calculation is double counting sequences with more than one $1$ in them.
In the simple case where we roll the six-sided die twice, write out each possible pair of rolls explicitly. There are $36$ options. $10$ include exactly one 1. There is $1$ option where both rolls land on 1. So the probability of getting a $1$ at least once is $11/36\neq 2/6$.
This generalises to the inclusion-exclusion principle.
