Simple Probability Problems

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October 7th 2012, 07:04 AM
jegues

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?
October 7th 2012, 08:26 AM
Plato

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 $\left( {\frac{1}{6}} \right)$

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

So the answer is the sum of a geometric series.
October 7th 2012, 09:53 AM
jegues

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 $\left( {\frac{1}{6}} \right)$

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

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)?
October 7th 2012, 10:40 AM
Plato

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 $\sum\limits_{k = 1}^\infty {\left( {\frac{5}{6}} \right)^{2k - 2} \left( {\frac{1}{6}} \right)} = \frac{6}{{11}}$
October 9th 2012, 01:03 PM
jegues

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 $\sum\limits_{k = 1}^\infty {\left( {\frac{5}{6}} \right)^{2k - 2} \left( {\frac{1}{6}} \right)} = \frac{6}{{11}}$

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

It must simply down to $\frac{36}{11}$ , correct? How does it go from the summation to that?
October 9th 2012, 01:21 PM
Plato

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 $\frac{36}{11}$ , correct? How does it go from the summation to that?

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

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

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

Let $A=\frac{1}{6}~\&~r=\frac{5}{6}$
October 9th 2012, 02:11 PM
jegues

Re: Simple Probability Problems

Quote:

Originally Posted by Plato

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

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

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

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

The first one should be,

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

Correct?

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