If a die is thrown $5$ times, what is the probability of throwing at least $3$ ones continuously?

Question

The Question is:

Throw a die $5$ times, what is the probability of throwing at least $3$ ones continuously.

I wanted to verify my solution in which I let X denote the positions which the ones could be to count the number of total cases in which there was at least $3$ consecutive items.

Three in a row:

XXX_ _

_ _ XXX

_ XXX _

Four in a row:

_ XXXX

XXXX _

Five in a row:

XXXXX

With this, I assumed that the total number of possible outcomes would be $6$ and the total sample space would be $6^5$ giving us a probability of $1/6^4$.

Additionally, if the question just asked for at least three of any number would that be $36/6^5$ as there are $6$ numbers and $6$ ways each can satisfy the criteria?

Thank you!

Answer 1

Let $Y$ denote not getting a 1. Let $X$ denote getting a 1. Being careful to avoid double counting.

$N(XXXY\_ )=N(\_YXXX)=5*6$

$N(YXXXY)=5*5$

$N(YXXXX)=N(XXXXY)=5$

$N(XXXXX)=1$

Which gives a total of 96 agreeing with drhab's answer. And so the probability is $\frac{96}{6^5}$. If you want the probability of getting at least three consecutive numbers. Then it would be this probability multiplied by 6. Because getting a number say x, at least 3 times, precludes getting any other number y at least 3 times. And the events are disjoint.

Answer 2

If I understand well then the dice are thrown one by one.

Give the dice chronological numbers $1,2,3,4,5$.

• Let $E$ denote the event that the dice with numbers $1,2,3$ show face 1.
• Let $F$ denote the event that the dice with numbers $2,3,4$ show face 1.
• Let $G$ denote the event that the dice with numbers $3,4,5$ show face 1.

Then to be found is $P(E\cup F\cup G)$ and applying the principle of inclusion/exclusion we find:$$P\left(E\cup F\cup G\right)=P\left(E\right)+P\left(F\right)+P\left(G\right)-P\left(E\cap F\right)-P\left(F\cap G\right)-P\left(E\cap G\right)+P\left(E\cap F\cap G\right)=$$$$3\times6^{-3}-2\times6^{-4}-6^{-5}+6^{-5}=16\times6^{-4}$$

This tells us also that there are $16\times6=96$ possibilities.

Answer 3

Let $\sf X$ be a result of 1, $\sf Y$ be a result of something else, and $\sf Z$ be a result of anything. The disjoint outcome patterns for "at least three consecutive 1 among five rolls" are: $$\sf XXXZZ~, YXXXZ~,ZYXXX$$ So the probability is $$\dfrac{1^3\cdotp6^2+5\cdotp1^3\cdotp6+6\cdotp5\cdotp1^3}{6^5}=\dfrac{1}{81}$$