Can You Manofy Your Girlfriend?

This is called a random walk. The given question is an example of a one dimensional random walk.

We are at position 3 on the $\mathbb{Z}$ axis.

Check this image for reference. I have used this image for explaining the answer.

Now, we want to get to $\mathbb{X}$ before $\mathbb{Y}$ . If we get to $\mathbb{Y}$ , we cannot play anymore because we have $0$ money to bet.

The probability to reach $\mathbb{X}$ before $\mathbb{Y}$ , that is to reach $9$ before $-3$ (refer to the image for this) is given as $\dfrac{dist(Y)}{dist(X) + dist(Y)}$ where $\text{dist(X)}$ is the distance from where we are to $\mathbb{X}$ and the same is true for $\text{dist(Y)}$ .

Check the wikipedia article on random walks for clarification on the above formula

Thus, our answer is $\dfrac{3}{ 9 + 3} = \dfrac{3}{12} = \dfrac{1}{4} = \boxed{0.250}$

You have equal chance of doubling your money or going bankrupt. So, you have a 0.5 probability of getting to $6. Once at $6, you again have a 0.5 probability of doubling your money to $12. The total probability that you can quadruple your money is then 0.5 * 0.5 == 0.25

For each $0 \le n \le 12$ , let $p(n)$ be the probability that you will get the money to buy a rose. Set $p(0) = 0$ and $p(12) = 1$ . Then we have, for $1 \le n \le 11$ , $p(n) = \frac12 p(n-1) + \frac12 p(n+1)$ which we rewrite as $p(n+1) - p(n) = p(n) - p(n-1)$ It follows that for all $1 \le n \le 12$ , $p(n) - p(n-1) = p(1) - p(0) = p(1)$ . Then in fact, $p(n) = \sum_{k=1}^n (p(k) - p(k-1)) = \sum_{k=1}^n p(1) = np(1)$ Now, $p(12) = 1$ implies $12 p(1) = 1$ , or $p(1) = 1/12$ . Thus, for each $k$ , $p(k) = k/12$ . In particular, $p(3) = 3/12 = \boxed{.25}.$

Once you start writing out the sample space for the probabilities to get from 3$ to 12$ you would notice infinite chains cropping up. There is always a chance of winning and losing alternatively an infinite number of times.

The simplest way to solve it is to break up the question into parts which have a finite sample space. It is given that there is an equal chance to win as well as loose. But the probability of getting 12$ and getting 0$ (loosing) isn't the same if you look at it analytically. Its more probable that you would loose 3$ than get 9$ more.

So we break it up equally. There's an equal chance of winning and loosing a dollar. Therefore there must be an equal chance of winning and loosing 2 dollars too and so on. Why? Provided that the player continues to gamble, all those infinite possibilities must end up to gaining a given number of dollars or loosing the same number of dollars.

To get our solution, we first find the probability of getting 6 dollars. It is 0.5. (Either loosing all 3$ or gaining 3$ more). Then we do it for twelve. It now becomes 0.25 (Either loosing all 6$ or gaining 6$ more)

Therefore the answer is 0.25

Let us consider a number line from 0 to 12. The game ends when we touch either 0 or 12.

Presently we are at 3. As the probability of winning and losing are same, its equally probable to reach either 0 or 6 from 3. That is, the probability of reaching 6 is 0.5

Now, with the same logic, its equally probable to reach 0 or 12 from 6. That is the probability of reaching 12 from 6 is 0.5

Hence, the net probability of reaching 12 from 3 is 0.5*0.5=0.25

Bhavin I told you dat its nothin special bout numbers 3 and 12 . You can see dat through other comments too. Try to search for gambler's ruin. It mit help.

Imagine a horizontal line marked with numbers 0 to 12. The problem above is equivalent to starting at 3, moving left or right at a step, with probability 0.5 each. The game ends when you reach 0 or 12.

The solution is a standard result of a symmetric one-dimensional random walk with end-points. You are 3 steps away from 0 and 9 steps away from 12. The probability of reaching 12 first would be 3/(3+9) = 0.25 while the probability of reaching 3 first would be 9/(3+9) = 0.75