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Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $ m$ and $ n$ be relatively prime positive integers such that $ \frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $ m+n$.

If you roll a die $r$ times, the probability of getting a six on the $r$th try is:

$\frac{1}{6} \cdot \frac{5^{r-1}}{6^{r-1}}$

Now I believe casework for Dave so:

1 roll: $ \frac{1}{6} $, 2 rolls: $\frac{1}{6} \frac{5}{6}$, 3 rolls: $\frac{1}{6} \frac{5^2}{6^2}$, .... , $r$ rolls, $\frac{1}{6} \frac{5^{r-1}}{6^{r-1}}$

I would add this up, but still, we never found $r$?

HINTS ONLY

EDIT

As pointed by 5xum, $L=1$ isn't possible with $D = L - 1$. So I got:

$P = 1 \cdot (1/6)\sum_{r=2}^{\infty} (5/6)^{r-2} + 1 + ...$

But $P > 1$ already, which is impossible?

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  • $\begingroup$ I can't figure out how you constructed your series sum. You might consider looking at my response and writing out the first few terms...? $\endgroup$ – Brian Tung Aug 26 '15 at 17:37
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Let $D$ denote the random variable telling how many rolls Dave took to roll a $6$, and let $L$ denote the variable telling how many rolls Linda took.

Then what you calculated were $P(D=k)$ for certain values of $k$. For example, the probability that Dave took $5$ rolls to roll a $6$ is

$$\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac16$$

What you need to calculate is the probability that $D=L\pm1$ or $D=L$, which is the probability

$$P(D=1\wedge (L=1\vee L=2)) + P(D=2\wedge (L=1\vee L=2\vee L=3)) + \dots$$

and the sum is actually infinite. I think writing it down may simplify it a bit, and you will be summing power series, which is not that hard...

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  • $\begingroup$ Shouldn't be $D= L $ or $D = L \pm 1$? $\endgroup$ – Hetebrij Aug 25 '15 at 13:20
  • $\begingroup$ What is: $/vee$ and $/wedge$? $\endgroup$ – Amad27 Aug 25 '15 at 13:23
  • $\begingroup$ @Amad27 Symbols for or and and. $\endgroup$ – 5xum Aug 25 '15 at 13:31
  • $\begingroup$ $P_1 = \frac{1}{6}( \frac{1}{6} (1 + \frac{5}{6})$. and then, $P_2 = \frac{5}{6}\frac{1}{6}...$. But then $ L = 1$ isnt possible in that case! $\endgroup$ – Amad27 Aug 25 '15 at 14:53
  • $\begingroup$ @Amad27 What is $P_1$? What is $P_2$? In what case is $L=1$ not possible? $\endgroup$ – 5xum Aug 25 '15 at 16:57
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Let $D, L$ be the obvious random variables. Then the desired probability is

\begin{align} P & = P(D = 1, L = 1) + P(D = 1, L = 2) + P(D = 2, L = 1) \\ & + P(D = 2, L = 2) + P(D = 2, L = 3) + P(D = 3, L = 2) \\ & + P(D = 3, L = 3) + P(D = 3, L = 4) + P(D = 4, L = 3) \\ & + \cdots \end{align}

Each row constitutes an element in a geometric series. Find the first row sum and the ratio, and the rest is trivial.

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If $D$ and $L$ denote the number of rolls needed by Dave and Linda and $E$ denotes the event then:$$P(E)=P(D=L)+P(D=L-1)+P(D=L+1)$$

This with $P(D=L-1)=P(D=L+1)$ on base of symmetry, and with:

$$P(D=L)=\sum_{k=1}^{\infty}P(D=k\wedge L=k)=\sum_{k=1}^{\infty}P(D=k)P(L=k)$$For $P(D=L-1)$ you can also find such an expression.


edit:

$$P\left(D=L\right)=\sum_{k=1}^{\infty}P\left(D=k\right)P\left(L=k\right)=\sum_{k=1}^{\infty}\left[\left(\frac{5}{6}\right)^{k-1}\frac{1}{6}\right]^{2}=$$$$\frac{1}{36}\sum_{k=1}^{\infty}\left(\frac{25}{36}\right)^{k-1}=\frac{1}{36}\frac{1}{1-\frac{25}{36}}=\frac{1}{11}$$

As said: likewise you can find $P(D=L-1)$.

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  • $\begingroup$ $L + 1$ isnt allowed? Also, what about $L -2, L - 3, etc...$? +1 $\endgroup$ – Amad27 Aug 25 '15 at 15:35
  • $\begingroup$ If you have found $P(D=L-1)$ then you have also found $P(D=L+1)$ because (as said) the probabilities are equal by symmetry. So you are ready then. I don't really understand your second question. Expressions like $L-2$ etc. are irrelevant in this context. $\endgroup$ – drhab Aug 25 '15 at 15:39
  • $\begingroup$ The problem says it can be any number less than Linda's. so $L - 2$ can be a case as well $\endgroup$ – Amad27 Aug 25 '15 at 16:01
  • $\begingroup$ "equal to or within one of the number of times Linda rolls" sounds very much like $D=L$ or $[D$ is one more than, or one less than $L]$. Do you think that $P(D\leq L)$ must be calculated? $\endgroup$ – drhab Aug 25 '15 at 17:19
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    $\begingroup$ @Amad27 it's an infinite series because $D$ and $L$ are bother Geometric Random Variables. $D$ is the count of tries until a six, which can realise values of $\{1, 2, 3, 4, ..., \to \infty\}$. They asked to find the probability that $L$ is within $1$ step of $D$, so they're after $\mathsf P(L-1\leq D\leq L+1)$ $\endgroup$ – Graham Kemp Aug 25 '15 at 23:31

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