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Two players, Adam and Eve, are throwing a die, and the first one to get a 6, will win.

Eve is throwing first, what is the probability that Adam can win?

$P(A)$ = probability Adam to win

$P(E)$ = probability Eve to win

I know the formula for conditional probability $ P(A|B) = \frac{P(A\cap B)}{P(B)}$

Also, I know that the conditional probability I need to calculate is $ P(A|\overline{E}) $

What I don't know is, what the intersection between these to events is.

$P(A\cap\overline{E}) = ?$

I know that the result should be $0,4545$

Also, what how do I calculate the probability for Adam to win, if the 6 was in the first 3 throws?

Thank you

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    $\begingroup$ You should not calculate $P(A|E^c)$. That would be the probability that Adam wins given the fact that Eve didn't win and it is essentially equal to 1 because almost surely either one of them wins. Try expressing the event that Adam wins as an infinite union of disjoint events whose probabilities will be easy to calculate. $\endgroup$ – Nocturne Nov 2 '14 at 2:44
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Let $e$ be the probability that Eve wins, and let $a$ be the probability that Adam wins. It is reasonably clear that $e+a=1$.

We condition on the result of the first throw. If Eve throws a $6$, then she wins. This has probability $\frac{1}{6}$.

Suppose that Eve throws a non-$6$ on her first throw. This has probability $\frac{5}{6}$. If that happens, the roles of Adam and Eve are reversed, and the probability Eve wins is $a$. It follows that $$e=\frac{1}{6}+\frac{5}{6}a=\frac{1}{6}+\frac{5}{6}(1-e).$$ We have obtained the linear equation $$e=\frac{1}{6}+\frac{5}{6}(1-e).$$ Solve. We get $e=\frac{6}{11}$ and therefore $a=\frac{5}{11}$.

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  • $\begingroup$ Much cleaner method than mine. I'll have to remember that one. Just goes to show that there can be several contrasting methods to solving a single question. $\endgroup$ – JMoravitz Nov 2 '14 at 17:47
  • $\begingroup$ It is really your Markov chain argument without the terminology and stripped of matrices. That part of your answer is very valuable, for it points OP, and other readers, forward. $\endgroup$ – André Nicolas Nov 2 '14 at 18:02
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If Eve wins, it happens on either the first throw, the third throw, the fifth throw, etc...

If Eve wins on the first throw: $\frac{1}{6}$

If Eve wins on the third throw (his second throw): She missed, he missed, she won: $\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{1}{6}$

If Eve wins on the fifth throw (her personal third throw): She miss, he miss, she miss, he miss, she hit: $(\frac{5}{6})^4\cdot\frac{1}{6}$

If Eve wins on the $(2n+1)$'th throw: $(\frac{5}{6})^{2n}\cdot\frac{1}{6}$

So, calculate $\sum\limits_{n=0}^{\infty}(\frac{5}{6})^{2n}\frac{1}{6}$ to add all of the possible outcomes where she wins together.

Alternatively, for Adam you have the same situation but offset a bit. You can either take 1 and subtract the previous result, or do it again using: $\sum\limits_{n=0}^{\infty}(\frac{5}{6})^{2n+1}\frac{1}{6}$

As for winning within the first three throws, it is probably easiest to draw a tree diagram.


For a completely different approach, consider the absorbing markov chain with four states: He won, His turn (hasn't won yet), Her turn (hasn't won yet), She won. There is probability of moving from His turn to He won with probability $\frac{1}{6}$, else probability $\frac{5}{6}$ to become her turn. If he won, it remains at he won with probability $1$. Similarly so for her probabilities.

We get then an absorbing matrix in standard form with order of entries as He won, She won, His turn, Her turn as the following:

\begin{bmatrix} 1 && 0 && \frac{1}{6} && 0\\ 0 && 1 && 0 && \frac{1}{6}\\ 0 && 0 && 0 && \frac{5}{6}\\ 0 && 0 && \frac{5}{6} && 0\end{bmatrix}

This is in the form: \begin{bmatrix} I & S\\0 & R\end{bmatrix} and the limiting matrix becomes \begin{bmatrix} I & S(I-R)^{-1}\\0 & 0\end{bmatrix}

Solving for $S(I-R)^{-1}$: $I-R = \begin{bmatrix}1&-\frac{5}{6}\\-\frac{5}{6}&1\end{bmatrix}$

$(I-R)^{-1} = \frac{36}{11}\cdot\begin{bmatrix}1&\frac{5}{6}\\ \frac{5}{6}&1\end{bmatrix}$

$S(I-R)^{-1} = \begin{bmatrix}\frac{6}{11} & \frac{5}{11} \\ \frac{5}{11} & \frac{6}{11}\end{bmatrix}$

So, the limiting matrix is then:

\begin{bmatrix} 1& 0 & \frac{6}{11} & \frac{5}{11} \\ 0 & 1 & \frac{5}{11} & \frac{6}{11} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}

To see if it is currently Eve's turn, if Adam wins, take the element corresponding to Adam win's row, and Eve's turn column, which was how I labeled it as the first row and fourth column. Thus, he has a $\frac{5}{11}$ chance to win if it is currently her turn.

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  • $\begingroup$ Maybe he will figure out using your well put answer, but according to the problem Eve is throwing first, so he will have to adjust for the $2n$ and not the $2n+1$ throwing. $\endgroup$ – Décio Soares Nov 2 '14 at 2:56
  • $\begingroup$ My mistake. I'll fix that now as I add a second solution as well. $\endgroup$ – JMoravitz Nov 2 '14 at 2:59

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