Simple Probability Problems

P1) You and your friend will play the following game. You will take turns rolling a fair six-sided die. You will roll first. As soon as one player rolls a 6, the game is over and that player wins. What is the probability that you win?

**Attempt:**

If we both rolled at the same time it would be a 50% chance that I would win, but since I roll first I can win before the other player even has a chance at rolling so I have an advantage. I just can't figure out how to quantify that advantage. Anyone know?

P2)In a game of poker, you are dealt five cards from a deck of 52 cards. What is probability that you get a flush (five cards of the same suit)?

**Attempt:**

I thought this would be me the answer,

http://latex.codecogs.com/png.latex?...%7B48%7D%20%29

but that is incorrect. Any ideas?

Re: Simple Probability Problems

Quote:

Originally Posted by

**jegues** P1) You and your friend will play the following game. You will take turns rolling a fair six-sided die. You will roll first. As soon as one player rolls a 6, the game is over and that player wins. What is the probability that you win?

The problem clearly states *You will roll first*. So you both do **not roll at the same time**.

The probability you win on the first roll is $\displaystyle \left( {\frac{1}{6}} \right)$

The probability you win on the ninth roll is $\displaystyle \left( {\frac{5}{6}} \right)^8\left( {\frac{1}{6}} \right)$.

So the answer is the sum of a geometric series.

Re: Simple Probability Problems

Quote:

Originally Posted by

**Plato** The problem clearly states *You will roll first*. So you both do **not roll at the same time**.

The probability you win on the first roll is $\displaystyle \left( {\frac{1}{6}} \right)$

The probability you win on the tenth roll is $\displaystyle \left( {\frac{5}{6}} \right)^9\left( {\frac{1}{6}} \right)$.

So the answer is the sum of a geometric series.

I agree with everything you wrote, but how does my problem set obtain an answer of 0.55 then? Can you explain?

Also, could you give me a hint for P2)?

Re: Simple Probability Problems

Quote:

Originally Posted by

**jegues** I agree with everything you wrote, but how does my problem set obtain an answer of 0.55 then? Can you explain?

Also, could you give me a hint for P2)?

Look at my edit. If you roll first then you win on an odd numbered roll.

The probability of your winning is $\displaystyle \sum\limits_{k = 1}^\infty {\left( {\frac{5}{6}} \right)^{2k - 2} \left( {\frac{1}{6}} \right)} = \frac{6}{{11}}$

Re: Simple Probability Problems

Quote:

Originally Posted by

**Plato** Look at my edit. If you roll first then you win on an odd numbered roll.

The probability of your winning is $\displaystyle \sum\limits_{k = 1}^\infty {\left( {\frac{5}{6}} \right)^{2k - 2} \left( {\frac{1}{6}} \right)} = \frac{6}{{11}}$

Can you explain how to simply that summation to one term?

It must simply down to $\displaystyle \frac{36}{11}$, correct? How does it go from the summation to that?

Re: Simple Probability Problems

Quote:

Originally Posted by

**jegues** Can you explain how to simply that summation to one term?

It must simply down to $\displaystyle \frac{36}{11}$, correct? How does it go from the summation to that?

The answer to your question is $\displaystyle \frac{6}{11}$, **period**.

$\displaystyle \left| r \right| < 1 \Rightarrow \sum\limits_{n = 1}^\infty {Ar^n } = \frac{{Ar}}{{1 - r}}$.

$\displaystyle \left| r \right| < 1 \Rightarrow \sum\limits_{n = 1}^\infty {Ar^{2n - 2} } = \frac{A}{{1 - r^2 }}$

Let $\displaystyle A=\frac{1}{6}~\&~r=\frac{5}{6}$

Re: Simple Probability Problems

Quote:

Originally Posted by

**Plato** The answer to your question is $\displaystyle \frac{6}{11}$, **period**.

$\displaystyle \left| r \right| < 1 \Rightarrow \sum\limits_{n = 1}^\infty {Ar^n } = \frac{{Ar}}{{1 - r}}$.

$\displaystyle \left| r \right| < 1 \Rightarrow \sum\limits_{n = 1}^\infty {Ar^{2n - 2} } = \frac{A}{{1 - r^2 }}$

Let $\displaystyle A=\frac{1}{6}~\&~r=\frac{5}{6}$

The first one should be,

$\displaystyle \left| r \right| < 1 \Rightarrow \sum\limits_{n = 1}^\infty {Ar^n } = \frac{{A}}{{1 - r}}$.

Correct?

Also, where do these equations come from? I have never seen them before...