# Homework Help: Consider the game of Three.

1. Feb 12, 2013

### countzander

Consider the game of Three."

1. The problem statement, all variables and given/known data
You shuffle a deck of three cards: ace, 2, 3. With the ace worth 1 point, you draw cards at random without replacement until your total points are 3 or more. You win if your total points are exactly 3. What is the probability that you win?

2. Relevant equationsn
Permutation?

3. The attempt at a solution
I think the desirable outcomes are given by 3P1. The sample space is 3!. So the chance of winning would be (3P1)/3!.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 12, 2013

### HallsofIvy

Re: Consider the game of Three."

When you say "With the ace worth 1 point", do you mean the other cards are worth 0?

3. Feb 12, 2013

### Ibix

Re: Consider the game of Three."

It's worth noting that this is a very small sample space. There are only three cards and you are drawing without replacement so enumerating every possible sequence ad simply counting the outcomes is a perfectly reasonable approach.

4. Feb 12, 2013

### Dick

Re: Consider the game of Three."

If so, how would you get to more than 3?

5. Feb 12, 2013

### HallsofIvy

Re: Consider the game of Three."

Ah- I missed the "without replacement". Thanks.

6. Feb 12, 2013

### rude man

Re: Consider the game of Three."

Do we have a winner?

7. Feb 12, 2013

### countzander

Re: Consider the game of Three."

The cards are worth 1, 2, and 3 respectively.

8. Feb 12, 2013

### LCKurtz

Re: Consider the game of Three."

No, the number of points in the sample space (outcomes) is 6
No. The sample space could be viewed as {12, 13, 21, 23, 3, 3} or {123, 132, 213, 231, 312, 321}, depending on how you want to view it.

Why?

9. Feb 12, 2013

### rude man

Re: Consider the game of Three."

So what is that number? So we can vote if it's right or not ...

10. Feb 12, 2013

### Dick

Re: Consider the game of Three."

3P1 is 3!/(3-1)!=3 is fairly standard notation. 3/3!=1/2. You can vote if you want. But it's still wrong.

11. Feb 12, 2013

### rude man

Re: Consider the game of Three."

We're not all probabilists.

I agree, that answer is wrong.

12. Feb 13, 2013

### HallsofIvy

Re: Consider the game of Three."

The probability of winning on the first card, that is, drawing a 3, is 1/3.

If the first card is not a 3 it might be an ace, with probability 1/3, in which case, to win you must next draw the 2, not the three. What is the probability of that?

If the first card is not a 3 it might be a 2, with probability 1/3, in which case, to win you must next draw the ace, not the three. What is the probability of that?

13. Feb 13, 2013

### rude man

Re: Consider the game of Three."

Ah, I smell a winner from H of I!

14. Feb 13, 2013

### LCKurtz

Re: Consider the game of Three."

I expect everyone here except countzander knows the correct answer. We're waiting for an argument from him...something more than just a wrong answer.

15. Feb 13, 2013

### rude man

Re: Consider the game of Three."

I agree.