split - Probability - six sided die

[NeedMathHalp]

I really need help with this....

Ken has a six sided die. He rolls the die, and if the result is not even, he rolls the die one more time. Find the probability that he ends up with an even number.

I thought at first it would be 2/3.

 

[Dr.Peterson]

Please post this as a new thread, rather than hijacking another.

And please show how you got your result, so we have something to discuss, to correct your thinking if necessary.

 

[Deleted member 4993]

I really need help with this....

Ken has a six sided die. He rolls the die, and if the result is not even, he rolls the die one more time. Find the probability that he ends up with an even number.

I thought at first it would be 2/3.

What is the probability that Ken will get an even number in first roll?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

 

[pka]

I really need help with this....Ken has a six sided die. He rolls the die, and if the result is not even, he rolls the die one more time. Find the probability that he ends up with an even number.

Here is a table of outcomes in terms or odd or even tossing a die twice:
\(\begin{array}{*{20}{c}} E&E \\ E&O \\ O&E \\ O&O \end{array}\)
There are only four. How many result in an even under the rules?

 

[Dr.Peterson]

Here is a table of outcomes in terms or odd or even tossing a die twice:
\(\begin{array}{*{20}{c}} E&E \\ E&O \\ O&E \\ O&O \end{array}\)
There are only four. How many result in an even under the rules?

Technically, in this scenario, the outcomes are

E​

O E​

O O​

since the second die isn't rolled when the first is even. I'm sure that's why the OP thought it was 2/3. The problem is that these are not equally likely outcomes. So you need either to make a tree (or equivalent) and keep track of the probability of each outcome, or just think of the second roll as if it were done but not looked at, like this:

E (E)​

E (O)​

O E​

O O​

These are equally likely.

 

[pka]

Technically, in this scenario, the outcomes are

E​

O E​

O O​

​

Have you ever studied the problem of points? Pascal and Fermat both found that is solution is in the games not yet played.