You have a basketball hoop and someone says that you can play one of two games.

Game1 : You get one shot to make the hoop.

Game2 : You get three shots and you have to make two of three shots.

If p is the probability of making a particular shot, for which values of p should you pick one game or the other?

**Answer:**

**Probability of winning the Game 1:**

The probability of winning game 1 is p, by definition.

**Probability of winning the Game 2:**

Let s(k, n) be the probability of making exactly k shots out of n.The probability of winning the Game 2 is the probability of making exactly two shots out of three OR making all three shots. In other words:

P(winning) = s(2, 3) + s(3, 3)

The probability of making all the three shots is:

s(3, 3) =

The probability of making exactly two shots is:

P(making 1 and 2, & missing 3) + P(making 1 and 3, & missing 2) + P(making 2 and 3, & missing 1)

= [p * p * (1-p)] + [p * (1-p) * p] + [(1-p) * p * p] = 3*(1-p)*

Adding these together, we get:

= + 3(1-p) = + 3 - 3 = 3 - 2

**Which game you should play?**

You should play Game 1 if P(Game 1) > P(Game 2):

p > 3 - 2 1 > 3p - 2 2 - 3p + 1 > 0 (2p - 1)(p - 1) > 0

Both terms must be positive, or both must be negative, But we know p < 1, so p – 1 < 0. This means both terms must be negative.

2p - 1 < 0 2p < 1 p < 0.5

So, we should play Game 1 if 0 < p < 0.5 and Game 2 if 0.5 < p < 1.

If p = 0, 0.5, or 1 then P(Game 1) = P(Game 2), so it doesn't matter which game we play.

This article is contributed by **Brahmani Sai**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.