# Simple Stat Problems (Probability)

• Oct 6th 2012, 12:31 PM
jegues
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,

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

but that is incorrect. Any ideas?
• Oct 6th 2012, 12:47 PM
MaxJasper
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]
• Oct 6th 2012, 01:05 PM
jegues
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]

• Oct 8th 2012, 10:17 AM
harish21
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,

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

but that is incorrect. Any ideas?

for the second question
$p=\dfrac{\dbinom{4}{1}\dbinom{13}{5}}{\dbinom{52}{ 5}}$

where $\dbinom{4}{1}$ gives the suit for the flush
$\dbinom{13}{5}$ gives 5 cards in that suit
$\dbinom{52}{5}$gives combination of possibilities for 5 cards
• Oct 8th 2012, 10:33 AM
harish21
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,

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

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:

$\dfrac{1}{6}+\bigg(\dfrac{5}{6}\bigg)^2 \dfrac{1}{6}+\bigg(\dfrac{5}{6}\bigg)^4 \dfrac{1}{6}+\cdots = \dfrac{6}{11}$
• Oct 8th 2012, 11:29 AM
awkward
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

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

Solve for p.