What is the probability that the game stops on the:

a. First draw.

b. Second draw.

c. Third draw.

d. Fourth draw.

e. Can the game go on "forever?" If not, what is the maximum number of draws before the game will end?

Tutors, sign in to answer this question.

Because of replacement, each time the probability of getting an Ace is 1/13, and the probability of not getting an Ace is 12/13.

a) 1/13

b) (12/13)(1/13), since the first try failed to get an Ace.

c) (12/13)^2(1/13), since the first two tries failed.

d) (12/13)^3(1/13), since the first three tries failed.

e) Theoretically, the draw can go forever, but the probability becomes smaller and smaller. lim{n->oo}(12/13)^n (1/13) = 0.

Hi Chris! There are three items to consider as you answer these questions.

First, what's the probability that the first card you draw will be an ace? To solve, we need to know the number of aces in the deck, and how many total cards are in the deck. Let's look at an equivalent problem. Say you have four billiard balls, two solids and two that are striped The chance of picking a solid is 50/50 (i.e., 50%), because two (solids) divided by four (total balls) is 1/2. In your case, perhaps you can presume you're working with a standard deck of 52 cards with 4 aces, so you can do the math to determine the probability of selecting an ace on the first draw.

Second, assuming you didn't draw an ace and the game proceeds to a second draw: The card you drew must go back into the deck before you draw a card again. Is the deck of cards you're working with now (before draw #2) any different than the deck of cards you originally started with (before draw #1)? Are there more or fewer cards in the deck? Are there more or fewer aces? If the conditions have changed, then the probability has changed. If the conditions remain the same, then the probability remains the same.

Lastly, if you flip a coin 100 times, you could conceivably get heads every time. Not very likely, but it's possible, because the odds of getting heads is never guaranteed at 100% (unless both sides of the coin are heads). This should help you respond to question e.

Hope this helps!

Bill

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.