Simple Stat Problems (Probability)

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,

$\displaystyle (1)(\frac{11}{51})(\frac{10}{50})(\frac{9}{49}) ( \frac{8}{48} )$

but that is incorrect. Any ideas?

Re: Simple Stat Problems (Probability)

P(#1,1|6) = 1/6

P(#2,1|6)=(5/6)*(1/6)=5/36

P(#1,2|6)=(5/6)*(5/6)*(1/6)=25/216 [#1,2=1st person second rolling]

Re: Simple Stat Problems (Probability)

Quote:

Originally Posted by

**MaxJasper** P(#1,1|6) = 1/6

P(#2,1|6)=(5/6)*(1/6)=5/36

P(#1,2|6)=(5/6)*(5/6)*(1/6)=25/216 [#1,2=1st person second rolling]

The answer it lists is 0.55, I can't really follow your work though.

Re: Simple Stat Problems (Probability)

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?

**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,

$\displaystyle (1)(\frac{11}{51})(\frac{10}{50})(\frac{9}{49}) ( \frac{8}{48} )$

but that is incorrect. Any ideas?

for the second question

$\displaystyle p=\dfrac{\dbinom{4}{1}\dbinom{13}{5}}{\dbinom{52}{ 5}}$

where $\displaystyle \dbinom{4}{1}$ gives the suit for the flush

$\displaystyle \dbinom{13}{5}$ gives 5 cards in that suit

$\displaystyle \dbinom{52}{5}$gives combination of possibilities for 5 cards

Re: Simple Stat Problems (Probability)

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?

**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,

$\displaystyle (1)(\frac{11}{51})(\frac{10}{50})(\frac{9}{49}) ( \frac{8}{48} )$

but that is incorrect. Any ideas?

For the first problem, you have

P(Y)=probability of YOU winning by rolling a 6 = 1/6

P(F)=probability of your FRIEND winning by rolling a 6 = 1/6

you roll the dice as: Y, F, Y, F, Y, F,...until a 6 is rolled

You will win if you can roll a 6 on the first, third , fifth, seventh,... roll

You only get the third throw if both you and your friend dont throw a 6 on the first and second throws respectively.

so, P(you win on the first roll)=1/6

P(you win on the third roll) = 5/6*5/6*1/6=(5/6)^2*1/6

P(you win on the fifth roll) = 5/6*5/6*5/6*5/6*1/6=(5/6)^4*1/6

.

.

and so on

so the probaibility that you win is:

$\displaystyle \dfrac{1}{6}+\bigg(\dfrac{5}{6}\bigg)^2 \dfrac{1}{6}+\bigg(\dfrac{5}{6}\bigg)^4 \dfrac{1}{6}+\cdots = \dfrac{6}{11}$

Re: Simple Stat Problems (Probability)

Harish21 has already given an excellent solution of P1, but here is another approach which avoids summing infinite series.

Let's say the first player's chance of winning is p.

With probability 1/6, he rolls a 6 on his first roll and wins. With probability 5/6, he rolls something else. It's then the second player's turn, and his (the second player's) chance of winning is p, so the first player's chance of winning is the 1-p. So

$\displaystyle p = \frac{1}{6} + \frac{5}{6}(1-p)$

Solve for p.