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.

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