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

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
$\noindent \textbf{Hint.} If the probability that the sum of the results of two die is 6 (respectively 8) is a (respectively b), the probability 6 occurs first is \frac{a}{a+b} (make sure you can show this!).$

Firstly Integrand....... 4 AM mafffsss

pls ignore thr higlights.

Anyway.

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

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

Originally Posted by HeroWise
Firstly Integrand....... 4 AM mafffsss

pls ignore thr higlights.

Anyway.

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

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

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

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

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.

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

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•