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A begins by throwing a dice until he gets $6$, then B does the same thing.
What is the probability that A throws more times than B?

I try to solve it, but I got 2 different answers:
1. We mark: $X=$How many times A throws, $Y$=How many times B throws. $$P(X>n|Y=n)=\frac{P(X>n)\cdot P(Y=n)}{P(Y=n)}=P(X>n)\\ \sum_{k=n}^\infty\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=\left(\frac65\right)^{1-n}=\left(\frac56\right)^{n-1}$$ 2. The second way is: We can say that "B throw less times than A", so let's assume that A throws the dice $n$ times. So: $$P(Y<n)=\sum_{k=1}^{n-1}\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=1-\left(\frac56\right)^{n}$$

They are opposite, and I don't know which one of them is correct or both of them are incorrect..

Please help with this.
Thank you!

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  • $\begingroup$ A more challenging question would be to compute the probability of B getting a 6 6(two consecutive sixes) sooner than A getting a 5 6. $\endgroup$
    – gar
    Mar 30, 2014 at 13:55

7 Answers 7

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A and B throwing dice are two independent events, so you need to multiply them and evaluate an infinite sum of geometric series.

$\displaystyle P(X>n)=\left(\dfrac{5}{6}\right)^n\\\\ P(Y=n)=\left(\dfrac{5}{6}\right)^{n-1}\cdot \left(\dfrac{1}{6}\right) $

and the required probability is:

\begin{align*} P(X>Y) &=\sum_{n\ge 1} P(X>n)\cdot P(Y=n)\\ &=\sum_{n= 1}^{\infty} \left(\dfrac{5}{6}\right)^{\!n}\cdot \left(\dfrac{5}{6}\right)^{\!n-1}\cdot\left(\dfrac{1}{6}\right)\\ &=\frac{6}{5}\cdot\sum_{n= 1}^{\infty} \left(\dfrac{5}{6}\right)^{\!2n}\cdot\left(\dfrac{1}{6}\right)\\ &=\frac{1}{5}\cdot\sum_{n= 1}^{\infty} \left(\dfrac{25}{36}\right)^{\!n}\\ &=\frac{1}{5}\cdot\frac{25/36}{11/36}\\ &=\dfrac{5}{11}\\ &= 0.45\overline{45}. \end{align*}

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In order to avoid the double sum, note that the problem is symmetric. The probability that A throws more times than B is equal to the probability that B throws more times than A; call this probability $p$.

Now let's calculate the probability that both A and B throw the same number of times; this will be $1-2p$. This is

$P(A=B) = \sum_k P(A=k, B=k) = \sum_k P(A=k) P(B=k) = \sum_k P(A=k)^2$

and we can write out the sum explicilty as

$\sum_{k=1}^\infty ((5/6)^{k-1} (1/6))^2$

or, after some simplification,

$(1/36) \sum_{j=0}^\infty (25/36)^j$.

Summing the series, this is $1/11$. So $1-2p = 1/11$ and therefore $p = 5/11$.

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enter image description here I am a new user and I really need 5 points. Hope this helps.

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As has been pointed out, if we find the probability $p$ of a tie, we will know the answer. The probability of a tie is readily found by summing an infinite series. We calculate it another way.

A tie can happen in two ways: (i) both A and B get a $6$ on their first throw or (ii) neither does, but they ultimately tie. The probability of (1) is $\frac{1}{36}$. The probability of (ii) is $\frac{25}{36}p$. Thus $$p=\frac{1}{36}+\frac{25}{36}p.$$ Solve this linear equation for $p$. We get $p=\frac{1}{11}$.

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Note that if A doesn't get a 6 first roll (probability 5/6), then the probability of B getting a 6 first is the same as the probability of A getting a 6 first was before the first roll.

If $p_A $ is the probability of A getting a 6 first, $p_B$, of B first. Then we have

$$p_A + p_B = 1\text{, and }\;p_B = \dfrac{5}{6} p_A$$

So

$$\dfrac{6}{5} p_B + p_B = 1$$

hence

$$p_B = \dfrac{5}{11}$$

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You need to sum over $n$ and multiply by the probability that $Y$ throws $n$.

Let $X=$number of throws until A gets a 6, $Y$=number of throws until B gets a 6 Then $A$ wins if \begin{align*} &\sum_{n=1}^{\infty} P(X>n|Y=n)\cdot P(Y=n) \\ &=\sum_{n=1}^{\infty}\frac{P(X>n)\cdot P(Y=n)}{P(Y=n)} \cdot P(Y=n) =\sum_{n=1}^{\infty} P(X>n) \cdot P(Y=n)\\ &= \sum_{n=1}^{\infty} \left(\left(\frac{5}{6}\right)^{n}\left( \frac{1}{6}\right) + \left(\frac{5}{6}\right)^{n+1}\left( \frac{1}{6}\right) +\ldots\right) \cdot \left(\frac{5}{6}\right)^{n-1}\left( \frac{1}{6}\right)\\ &=\frac{5}{11} \end{align*}

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To find the probability that A throws more times than B, we need to sum the probabilities that A throws two or more times given that B throws once, A throws three or more times given that B throws twice and so on.

So we need to evaluate

$\sum_{k=1}^{\infty}P(A_{throws} > k,B_{throws}=k)$

Now for B throwing once, and A throwing two or more times, we have as probability (using the sum to infinity of a geometric progression)

$\large \frac{1}{6}\times\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2\left(\frac{\frac{5}{6}}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)$

For B throwing twice, and A throwing three or more times, we have as probability (using the sum to infinity of a geometric progression as before)

$\large\left(\frac{5}{6}\right) \frac{1}{6}\times\sum_{k=2}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^2}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^3$

For B throwing three times, and A four or more times we have

$\large\left(\frac{5}{6}\right)^2 \frac{1}{6}\times\sum_{k=3}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^3}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^5$

Generalising for B throwing $m$ times and A throwing $m+1$ or more times, the probability is given by

$\large \frac{1}{6}\left(\frac{5}{6}\right)^{2m-1}$

Thus, the probability that A throws more times than B is given by the following scaled sum to infinity of a geometric progression with ratio $\left(\frac{5}{6}\right)^2$

$\large \sum_{k=1}^{\infty}P(A_{throws}>k,B_{throws}=k)=\frac{1}{6}\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^{2k-1}=\frac{1}{6}\frac{\frac{5}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{5}{11}$

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