Thread: Probability question please help

1. Probability question please help

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

2. Re: Probability question please help

Originally Posted by kpad5991
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
$\noindent \textbf{Hint.} If the probability that the sum of the results of two die is 6 (respectively 8) is a (respectively b), the probability 6 occurs first is \frac{a}{a+b} (make sure you can show this!).$

3. Re: Probability question please help

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

pls ignore thr higlights.

Anyway.

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

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

4. Re: Probability question please help

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??

5. Re: Probability question please help

Originally Posted by kpad5991
The solution in the book is 1/2??
Okay I figured it out. Shadia continues for 26/36

6. Re: Probability question please help

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

7. Re: Probability question please help

Originally Posted by kpad5991
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.

8. Re: Probability question please help

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

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