# box counter problem

• May 10th 2010, 07:16 PM
kingman
box counter problem
Dear Sir,
I need help some help with the below problem and would appreciate very much if can offer some help.
thanks
Kingman

A box contains 15 counters numbered 1 to 15. Counters are drawn successively from the box, one at a time with replacement. A "Win" is obtained when the first number is drawn again.
Given that the first "win' occurs on the 3rd draw, find the probability that the second "win" is obtained on or before the 7th draw.

• May 10th 2010, 08:51 PM
undefined
Quote:

Originally Posted by kingman
Dear Sir,
I need help some help with the below problem and would appreciate very much if can offer some help.
thanks
Kingman

A box contains 15 counters numbered 1 to 15. Counters are drawn successively from the box, one at a time with replacement. A "Win" is obtained when the first number is drawn again.
Given that the first "win' occurs on the 3rd draw, find the probability that the second "win" is obtained on or before the 7th draw.

I don't know what a counter is, so I'll just use object.

The question is the same as asking "Five objects are drawn. What is the probability that there is a win among the 2nd through 5th draws?" This is because a given draw is independent from any preceding draws.

So using my equivalent problem, it's the probability that a win is obtained on the 2nd draw, plus the probability of a win on the 3rd draw given that the 2nd draw was a loss, plus the probability of a win on the 4th draw given that the previous draws were losses, plus the probability that the 5th draw is a win given that the previous draws were losses.

$\displaystyle \frac{1}{15}+\left(\frac{14}{15}\right)\left(\frac {1}{15}\right)+\left(\frac{14}{15}\right)^2\left(\ frac{1}{15}\right)+\left(\frac{14}{15}\right)^3\le ft(\frac{1}{15}\right)=\frac{12209}{50625}$
• May 10th 2010, 11:07 PM
Soroban
Hello, Kingman!

I solved it the long way . . . very long!

This is conditional probability.
We need Bayes' Theorem: .$\displaystyle P(A\,|\,B) \;=\;\frac{P(A \wedge B)}{P(B)}$

Quote:

A box contains 15 counters numbered 1 to 15.
Counters are drawn successively from the box, one at a time with replacement.
A "win" is obtained when the first number is drawn again.

Given that the first "win" occurs on the 3rd draw, find the probability
that the second "win" is obtained on or before the 7th draw.

Answer: .$\displaystyle \frac{12,\!209}{50,\!625}$

We want: .$\displaystyle P(\text{2nd win on or before 7th draw}\,|\,\text{1st win on 3rd draw})$

. . . . . . . $\displaystyle =\;\frac{P(\text{2nd win on or before 7th draw} \wedge \text{1st win on 3rd draw})} {P(\text{1st win on 3rd draw})}$ .[1]

The denominator is: .$\displaystyle P(\text{win on 3rd draw})$
. . The first draw can be any of the 15 numbers: .$\displaystyle \frac{15}{15} \:=\:1$
. . The second must some Other number: .$\displaystyle \frac{14}{15}$
. . The third must be a win: .$\displaystyle \frac{1}{15}$
Hence: .$\displaystyle P(\text{win on 3rd draw}) \:=\:1\cdot\frac{14}{15}\cdot\frac{1}{15} \:=\:\frac{14}{15^2}$ .[2]

The numerator is: .$\displaystyle P(\text{1st on 3rd draw} \wedge \text{2nd win on 4th, 5th, 6th or 7th draw})$

. . 2nd win on 4th draw: .$\displaystyle \begin{array}{ccccccc} \text{any} & \text{other} & \text{win} & \text{win} \\ 1 & \frac{14}{15} & \frac{1}{15} & \frac{1}{15} \end{array} \quad=\quad\frac{14}{15^3}$

. . 2nd win on 5th draw: .$\displaystyle \begin{array}{ccccccc} \text{any} & \text{other} & \text{win} & \text{other} & \text{win} \\ 1 & \frac{14}{15} & \frac{1}{15} & \frac{14}{15} & \frac{1}{15} \end{array} \quad=\quad \frac{14^2}{15^4}$

. . 2nd win on 6th draw: .$\displaystyle \begin{array}{ccccccc} \text{any} & \text{other} & \text{win} & \text{other} & \text{other} & \text{win} \\ 1 & \frac{14}{15} & \frac{1}{15} & \frac{14}{15} & \frac{14}{15} & \frac{1}{15} \end{array} \quad = \quad \frac{14^3}{15^5}$

. . 2nd win on 7th draw: .$\displaystyle \begin{array}{ccccccc} \text{any} & \text{other} & \text{win} & \text{other} & \text{other} & \text{other} & \text{win} \\ 1 & \frac{14}{15} & \frac{1}{15} & \frac{14}{15} & \frac{14}{15} & \frac{14}{15} & \frac{1}{15} \end{array} \quad=\quad \frac{14^4}{15^6}$

Hence, the numerator is: .$\displaystyle \frac{14}{15^3} + \frac{14^2}{15^4} + \frac{14^3}{15^3} + \frac{14^4}{15^6} \;=\;\frac{170,926}{15^6}$ .[3]

Substitute [2] and [3] into [1]:
. . $\displaystyle P(\text{2nd win on or before 7th draw}\,|\,\text{1st win on 3rd draw}) \;=\;\frac{\dfrac{170,926}{15^6}}{\dfrac{14}{15^2} } \;=\;\frac{12,209}{50,625}$

• May 11th 2010, 02:00 AM
kingman
Big Thank you
Dear Soroban,
Thanks very much for your detailed solution but I have
some hiccup which I hope you can shed some light to me.
Thanks very much
Kingman

The first draw can be any of the 15 numbers: .15/15=1

I believe you are referring to the sum of the 15 outcomes

eg.(1,other,1), (2,other,2),(3,other,3),( 4,other,4)
.............(15,other,15) from which the sum of the probability equals to 15*(1/15*14/15*1/15) Is this what you mean?

Also I cannot see how the outcome of the numerator of
you conditional probability equation (1)

[IMG]file:///C:/DOCUME%7E1/User/LOCALS%7E1/Temp/msohtml1/01/clip_image002.gif[/IMG].
can be a subset of the denominator when the outcome of the denominator has 3 element ,whereas the outcome of the numerator has 4,5,6 and 7 elements
Eg. The outcomes of the denominator looks like
.(1,other,1), (2,other,2),(3,other,3),( 4,other,4)
.............(15,other,15)
Whereas the numerator looks like (1,other,1,1)…. (15, others,15,15)
And (1, others,1, other , 1), ……..(15, others,15, other , 15)
And (1, others,1, other ,other, 1), ……..(15, others,15, other , other,15)
And (1, others,1, other ,other, other 1), ……..(15, others,15, other , other, other15)
• May 11th 2010, 02:17 AM
kingman
Thank very much but..
Dear undefined,
Thanks very much for your kind reply but there is something which I don't quite get it when you say:

"The question is the same as asking "Five objects are drawn. What is the probability that there is a win among the 2nd through 5th draws?" This is because a given draw is independent from any preceding draws."

Can you please elaborate why we can use Five objects instead of 15 objects and change the question to " find the probability that there is a win among the 2nd through 5th draws?"
Also I don't quite understand when you explain that "This is because a given draw is independent from any preceding draws'.
Sorry that I must have missed some very important subtle point in this question.

Also what if I don't change the 15 objects to 5 object: howthe solution should look like?

thanks
Kingman

• May 15th 2010, 06:47 AM
kingman
Thanks Undefined
Dear undefined,
Thanks very much for your kind reply but there is something which I don't quite get it when you say:

"The question is the same as asking "Five objects are drawn. What is the probability that there is a win among the 2nd through 5th draws?" This is because a given draw is independent from any preceding draws."

Can you please elaborate why we can use Five objects instead of 15 objects and change the question to " find the probability that there is a win among the 2nd through 5th draws?"
Also I don't quite understand when you explain that "This is because a given draw is independent from any preceding draws'.
Sorry that I must have missed some very important subtle point in this question.

Also what if I don't change the 15 objects to 5 object: howthe solution should look like?

thanks
Kingman
• May 15th 2010, 07:54 AM
undefined
Quote:

Originally Posted by kingman
Dear undefined,
Thanks very much for your kind reply but there is something which I don't quite get it when you say:

"The question is the same as asking "Five objects are drawn. What is the probability that there is a win among the 2nd through 5th draws?" This is because a given draw is independent from any preceding draws."

Can you please elaborate why we can use Five objects instead of 15 objects and change the question to " find the probability that there is a win among the 2nd through 5th draws?"
Also I don't quite understand when you explain that "This is because a given draw is independent from any preceding draws'.
Sorry that I must have missed some very important subtle point in this question.

Also what if I don't change the 15 objects to 5 object: howthe solution should look like?

thanks
Kingman

Okay, so I didn't change 15 to 5. Notice that there are a bunch of 15s in my answer.

Consider that after the first object is drawn, then each draw afterwards either is that object (with probability 1/15) or isn't that object (with probability 14/15). So the probability of a win for any particular draw after the first draw is always 1/15. It is impossible to win with only one draw, because you have to draw the first object "again." We clear so far?

The meaning of "This is because a given draw is independent from any preceding draws" is exactly what the words mean.

Consider another example, rolling a 6-sided die. Suppose we say, "We are successively rolling a 6-sided die with the goal of rolling a 4. Rolling a 4 is considered a 'win.' Given that the first 'win' occurs on the 21st roll, what is the probability that the second win is obtained on the 22nd roll?"

Now, two dice roll events are independent from each other, so we don't have to consider the first 21 rolls, and the answer is simply 1/6.

Understand?