Odds of winning

Question

Hello

how to callculate odds of winning one of two events of 40% chances to win?

Answer 1

As I understand your question, there are two events, which I'll call First Event and Second Event. The probability of winning First Event, by itself, is 40%, or 4 in 10, or 4/10. The probability of winning Second Event is the same, and it's unrelated to having won First Event. We want to know what the probability is that a player may win either First Event or Second Event or both.

Think of this as the player rolling a die with ten sides, numbered 1-10. The player wins the event if she rolls a 1, 2, 3, or 4, but not if she rolls any other number. She has a 40% possibility of a winning roll in First Event. When that event is done, she now has a 40% probability of a winning roll in Second Event.

This isn't always feasible, but let's make a table to show the possible rolls. The First Event roll will go in rows, from 1-10. The Second Event Roll will go in columns, from 1-10. We'll put a 'W' in a space if either the First Event roll or the Second Event roll would win, i.e., if the roll for either event is ≤ 4

⇾ First Event Roll / 1 2 3 4 5 6 7 8 9 10

↓ Second Event Roll

1 W W W W W W W W W W

2 W W W W W W W W W W

3 W W W W W W W W W W

4 W W W W W W W W W W

5 W W W W L L L L L L

6 W W W W L L L L L L

7 W W W W L L L L L L

8 W W W W L L L L L L

9 W W W W L L L L L L

10 W W W W L L L L L L

If you count those (10 × 10 = ) 100 cells, you'll see a win in 64, and a loss in 36. What's happening is that the losses are dependent. Our player can only lose if she loses First Event and if she loses second event. So if we look at that dependency, we can flip the question and say "What is the chance she loses both events?"

When we ask the question that way, we can bring out the interdependency of the variables and ask, what's the probability of losing both events? That becomes easier. The chance to lose First Event is 6/10, and so for Second Event. Since they're related, since she has to lose both, we can use the multiplicative property of probabilities, and 6/10 * 6/10 gives us the 36/100 we saw in the table. Therefore, the result has to be 36/100.

Alternately, we can see that there are 24 spots where she wins First Event but not Second. 24 spots where she wins Second but not First. And 16 where she wins both. So, since winning First does not affect winning Second, since winning one does not affect winning both, we use the additive property again, and we find a 100 - (24 + 24 + 16) == 100 - (loss probability = 36) = 64% chance.

Statistics is often less about the math than defining the dependencies.