##### Member
Shadia has invented a game for one person. she throws two dice repeatedly until the sum of the two numbers shown is either six or eight. if the sum is six, she wins. if the sum is eight, she loses. if the sum is any other number, she continues to throw until the sum is six or eight. calculate the probability that Shadia wins the gameHey

#### InteGrand

##### Well-Known Member
Shadia has invented a game for one person. she throws two dice repeatedly until the sum of the two numbers shown is either six or eight. if the sum is six, she wins. if the sum is eight, she loses. if the sum is any other number, she continues to throw until the sum is six or eight. calculate the probability that Shadia wins the gameHey
$\bg_white \noindent \textbf{Hint.} If the probability that the sum of the results of two die is 6 (respectively 8) is a (respectively b), the probability 6 occurs first is \frac{a}{a+b} (make sure you can show this!).$

#### HeroWise

##### Active Member
Firstly Integrand....... 4 AM mafffsss

pls ignore thr higlights.

Anyway.

$\bg_white \text{P(Shadia wins)} = \frac{5}{36}$
$\bg_white \text{P(Shadia continues)} = \frac{25}{36}$
$\bg_white \text{P(Shadia wins first game)} = \frac{5}{36}$
$\bg_white \text{P(Shadia wins second game)} = \bigg(\frac{25}{36} \bigg) \bigg(\frac{5}{36} \bigg) = \bigg(\frac{125}{1296} \bigg)$
$\bg_white \text{P(Shadia wins second game)} = \bigg(\frac{25}{36} \bigg)\bigg(\frac{25}{36} \bigg) \bigg(\frac{5}{36} \bigg) = \bigg(\frac{3125}{56656} \bigg)$

$\bg_white \text{P(shadia wins)} = \bigg(\frac{5}{36} \bigg) + \bigg(\frac{125}{1296} \bigg) + \bigg(\frac{3125}{56656} \bigg) + ...$
$\bg_white a=\bigg(\frac{5}{36}\bigg) , r = \frac{25}{36} \qquad S_\infty=\frac{a}{1-r}$
$\bg_white S_\infty=\frac{\bigg(\frac{5}{36}\bigg)}{1-\bigg(\frac{25}{36}\bigg)} = \frac{5}{11}$

##### Member
Firstly Integrand....... 4 AM mafffsss

pls ignore thr higlights.

Anyway.

$\bg_white \text{P(Shadia wins)} = \frac{5}{36}$
$\bg_white \text{P(Shadia continues)} = \frac{25}{36}$
$\bg_white \text{P(Shadia wins first game)} = \frac{5}{36}$
$\bg_white \text{P(Shadia wins second game)} = \bigg(\frac{25}{36} \bigg) \bigg(\frac{5}{36} \bigg) = \bigg(\frac{125}{1296} \bigg)$
$\bg_white \text{P(Shadia wins second game)} = \bigg(\frac{25}{36} \bigg)\bigg(\frac{25}{36} \bigg) \bigg(\frac{5}{36} \bigg) = \bigg(\frac{3125}{56656} \bigg)$

$\bg_white \text{P(shadia wins)} = \bigg(\frac{5}{36} \bigg) + \bigg(\frac{125}{1296} \bigg) + \bigg(\frac{3125}{56656} \bigg) + ...$
$\bg_white a=\bigg(\frac{5}{36}\bigg) , r = \frac{25}{36} \qquad S_\infty=\frac{a}{1-r}$
$\bg_white S_\infty=\frac{\bigg(\frac{5}{36}\bigg)}{1-\bigg(\frac{25}{36}\bigg)} = \frac{5}{11}$
The solution in the book is 1/2??

##### Member
The solution in the book is 1/2??
Okay I figured it out. Shadia continues for 26/36

#### HeroWise

##### Active Member
$\bg_white You mean \, \frac{25}{36} and how is it half? Like it means she eventually wins so we dont include her losses but continues and wins in different situations$

#### kawaiipotato

##### Well-Known Member
The solution in the book is 1/2??
Just a note that you should expect 1/2 because the number of desired outcomes are equal in both cases.

#### KAIO7

##### Member
$\bg_white You mean \, \frac{25}{36} and how is it half? Like it means she eventually wins so we dont include her losses but continues and wins in different situations$
You've made a silly mistake in evaluating the probability of continuing the game; it should be 26/36 instead of 25/36. When using this, the limiting sum will be equal to 1/2, hence the probability of winning the game is 1/2

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