1. Probability and statistics

A person plays an infinite sequence of games. He wins the th game with probability {\displaystyle 1/{\sqrt {n}}}, independently of the other games.
(i) Prove that for any , the probability is one that the player will accumulate [FONT=inherit dollars if he gets a dollar each time he wins two games in a row.[/FONT]
(ii) Does the claim in part (i) hold true if the player gets a dollar only if he wins three games in a row? Prove or disprove it.
I am posting again this problem.

2. Re: Probability and statistics

A person plays an infinite sequence of games. He wins the ${\displaystyle n}$th game with probability ${\displaystyle 1/{\sqrt {n}}}$ independently of other games.
(1) Prove that for any ${\displaystyle A}$, the probability is one that the player will accumulate ${\displaystyle A}$ dollars if he gets a dollar each time he wins two games in a row.

2) Does the claim in part (i) hold true if the player gets a dollar only if he wins three games in a row? Prove or disprove it.

3. Re: Probability and statistics

Ok, so for part (1), calculate the probability that after the $n$th game, the player has accumulated exactly $A$ dollars. At $A+1$ games, this is the first time the probability is non-zero.

4. Re: Probability and statistics

Hello, If the player is playing 49th game, his probability of winning that game is$\frac{1}{7}$.The probability of winning two games in a row is $\frac{1}{\sqrt{n}*\sqrt{n+1}}$.Now how to compute next steps?

5. Re: Probability and statistics

The probability that the player will accumulate $A$ dollars after $A+1$ games is the probability that the player has won all $A+1$ games. That has probability:
$$\prod_{n=1}^{A+1}\dfrac{1}{\sqrt{n}}$$

The probability that the player will accumulate $A$ dollars after $A+2$ games is the probability that the player has won the first $A+1$ games plus the probability that the player loses the first game, but wins the last $A+1$ games. If the player loses a game in the middle, there is no way to get to $A$ dollars by the end of the $A+2$nd game.

The probability that the player will accumulate $A$ dollars after $A+3$ games is the probability that the player wins the first $A+1$ games, then loses the next two; or loses the first two, then wins the last $A+1$; or wins the first $k$ games (for $k-1$ dollars), loses one, then wins the last $A-k+2$ games (for $A-k+1$ dollars), giving a total of $k-1+A-k+1 = A$ dollars at the end of the $A+3$rd game.

Note: The player can never lose the first game, since he wins with probability $\dfrac{1}{\sqrt{1}} = 1$.

So, the third one would look like this:
$$P(\text{wins }A\text{ dollars after }A+3\text{rd game}) = \left( 1-\dfrac{1}{\sqrt{A+2}} \right) \left( 1-\dfrac{1}{\sqrt{A+3}} \right) \prod_{n=1}^{A+1} \dfrac{1}{\sqrt{n}} + \sum_{k=2}^{A+2} (\sqrt{k}-1)\prod_{n=1}^{A+3} \dfrac{1}{\sqrt{n}}$$

etc.

Now, change the probabilities to $$P(\text{wins at least }A\text{ dollars after }n\text{th game})$$

This means figure out the probability that the player loses key games that will prevent the player from winning at least $A$ dollars.

6. Re: Probability and statistics

Hello, I didn't understand this term $\displaystyle\sum_{k=2}^{A+2}(\sqrt{k}-1)$. Would you explain it?

7. Re: Probability and statistics

Ooh, this is probably a good application of the pigeonhole principle! For $A+1$ games, the player must win all games to win at least $A$ dollars.
For $A+2$ games, the player must win all games to win at least $A$ dollars.
For $A+3$ games, the player may lose up to any 1 game and still win at least $A$ dollars.
For $A+4$ games, the player may lose up to any 1 game and still win at least $A$ dollars.
For $A+5$ games, the player may lose up to any 2 games and still win at least $A$ dollars.
In general, after $A+1+n$ games, the player may lose any $\left\lfloor \dfrac{n}{2} \right\rfloor$ games and still win at least $A$ dollars. Prove this with the pigeonhole principle. Then show that the probability that the player will lose at least $\left\lfloor \dfrac{n}{2} \right\rfloor +1$ games approaches zero as $n \to \infty$.

8. Re: Probability and statistics

Here's another possibility. Maybe try a recursive relation:

\begin{align*}P(\text{win at least }A\text{ dollars after }n\text{ games}) & = P(\text{win at least }A\text{ dollars after }n-1\text{ games}) + P(\text{win exactly }A-1\text{ dollars after }n-1\text{ games AND win }n-1\text{st game})P(\text{win }n\text{th game}) \\ & = P(\text{win at least }A\text{ dollars after }n-1\text{ games}) + P(\text{win exactly }A-1\text{ dollars after }n-1\text{ games AND win }n-1\text{st game})\dfrac{1}{\sqrt{n}}\end{align*}

9. Re: Probability and statistics

Hello, I got the answer on internet as follows:-
Define the person's game as the infinite sequence ${\displaystyle \omega =\{\omega _{1},\omega _{2},...\}}$ where each ${\displaystyle \omega _{k}}$ equals either 1 (corresponding to a win) or 0 (corresponding to a loss).

Define the random variable ${\displaystyle \tau _{k}:\Omega \to \mathbb {N} }$ by

${\displaystyle \tau _{k}(\omega )=\{\#{\text{ times }}\omega _{j}=\omega _{j-1}=1,\,\forall 2\leq j\leq k\}}$ that is,${\displaystyle \tau _{k}}$counts how many times the player received two consecutive wins in his first ${\displaystyle k}$ games. Thus, the player will win ${\displaystyle \tau _{k}}$ dollars in the first${\displaystyle k}$ games. Clearly,${\displaystyle \tau _{k}}$is measurable. Moreover, we can compute the expectation:

${\displaystyle E[\tau _{k}(\omega )]=\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}.}$

Now observe what happens as we send${\displaystyle k\to \infty }$:

${\displaystyle E[\lim _{k\to \infty }\tau _{k}(\omega )]=\lim _{k\to \infty }\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}=\infty }$

Hence the expected winnings of the infinite game is also infinite. This implies that the player will surpass dollar ${\displaystyle A}$ in winnings almost surely.

(ii): Define everything as before except this time ${\displaystyle \tau _{k}(\omega )=\{\#{\text{ times }}\omega _{j}=\omega _{j-1}=\omega _{j-2}=1,\,\forall 3\leq j\leq k\}.}$

Then${\displaystyle E[\tau _{k}(\omega )]=\sum _{j=2}^{k}{\frac {1}{\sqrt {j}}}{\frac {1}{\sqrt {j-1}}}{\frac {1}{\sqrt {j-2}}}.}$ which gives ${\displaystyle E[\lim _{k\to \infty }\tau _{k}(\omega )]<\infty .}$ Thus we cannot assert that the probability of surpassing any given winnings will equal 1.