What Is the Probability of Winning the Game of Three with a Shuffled Deck?

[countzander]

Consider the game of Three."

Homework Statement

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?

Homework Equations

n
Permutation?

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!.

 

[HallsofIvy]

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

 

[Ibix]

I don't think your result is correct. Perhaps you could explain a little about your reasoning for your figures?

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.

 

[Dick]

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

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

 

[HallsofIvy]

you draw cards at random without replacement until your total points are 3 or more.

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

 

[rude man]

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

Do we have a winner?

 

[countzander]

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

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

 

[LCKurtz]

Homework Statement

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?

Homework Equations

n
Permutation?

The Attempt at a Solution

I think the desirable outcomes are given by ₃P₁.

No, the number of points in the sample space (outcomes) is 6

The sample space is 3!.

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.

So the chance of winning would be (₃P₁)/3!.

Why?

 

[rude man]

Homework Statement

I think the desirable outcomes are given by ₃P₁. The sample space is 3!. So the chance of winning would be (₃P₁)/3!.

Homework Statement

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

 

[Dick]

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

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.

 

[rude man]

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.

We're not all probabilists.

I agree, that answer is wrong.

 

[HallsofIvy]

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?

 

[rude man]

Ah, I smell a winner from H of I!

 

[LCKurtz]

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.

 

[rude man]

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.

I agree.