Extra-events cannot be avoided in learning probability.That's how easy it is to use and how great it is.
Where is the difference from ordinary probability?What is the best way to demonstrate power?This time, I would like to introduce the scene of calculating the normal probability and the scene of using the extra event, and I would like to explain how to use the extra event and how to distinguish it.
Example of not using co-events
First, let's look at an example that uses ordinary probabilities for better understanding.This time, we will use a lottery like the one in the image as an example.
example
``Here is a gacha-gacha with 100 capsules in it. 99 are lost and 1 is a hit. ”
Somehow, if you draw first, there are many losers, and I feel that it is better to draw later.However, if you draw later, if the person in front of you wins, when you draw, there may be a situation where you only lose.Well, what is the highest probability to draw a hit?Let us guide you through the math.
To conclude, the answer isThe probability of drawing a hit no matter how many times you draw is \(\frac{1}{100}\).
CommentaryI'm going to do it!
The probability that the first player wins \(P_1\) is \(\frac{1}{100}\) because it is "the probability that the first player wins".This is easy to understand. \(\frac{100}{1}\) is the probability of drawing 1 win out of 100 lotteries.
The probability that the second person wins \(P_2\) is "the probability that the first person wins and the second person wins", so
$$P_2=\frac{99}{100}\times \frac{1}{99}=\frac{1}{100}$$
It becomes.
The probability that the third person wins \(P_3\) is ``the probability that the first and second people draw a loss and the third person wins'', so
$$P_3=\frac{99}{100}\times \frac{98}{99}\times \frac{1}{98}=\frac{1}{100}$$
It becomes.
I will omit the fourth and subsequent players, but the probability of winning is constant at \(\frac{1}{100}\) no matter how many people draw.
If you don't use extra events |
If there are 100 gacha-gacha contents and 1 win, what is the probability of winning 100 times?
If you don't return the draw lottery, the answer is 100%.In other words, if you draw XNUMX times, you can definitely draw a hit.If you plot the number of draws on the horizontal axis and the probability of winning on the vertical axis, it will look like the graph below.
this isSince the drawn capsules are not returned to the vending machine, you can definitely draw a win by drawing 100 times.On the other hand, what about Soshage Gacha?Most of the systems return the lottery, but to find the probability of winning in that caseneed to use extra eventsThere is.
Example of using extra events (Gacha)
Earlier, we saw an example of not returning the drawn gacha.Here we will consider the case of returning the drawn gacha.When you need to find probabilities using co-eventsIt will be.
Earlier, when I pulled the gacha 100 times, I got a 100% hit.So, what is the probability that you will win if you pull back the gacha you pulled 100 times?Let's do the math.
The desired probability isProbability of winning at least once when drawing 1 gachas with a 100% chance of winningis.In other words, if you get a hit even once, it's OK.
The reason why this probability is calculated is that the drawn gacha is returned even if there is a win or a loss.
So how do you ask?
As a way of thinking,Subtract from 100% to find the probability that you won't get a hit even if you draw 1 times.For example, if there is a 100% chance that you will not win even if you draw 1 times, the chance that you will win once is 30%.
Now let's do the actual calculation.
The probability of not winning even if you draw 100 times \(P_n\) is the probability of drawing \(\frac{100}{99}\) 100 times in a row.
$$P_n=(\frac{99}{100})^{100}\simeq0.366$$
In other words, if you draw a gacha with a 1% chance of winning 100 times, there is a 36.6% chance that you will not win.
Then,Probability of winning at least onceLet's ask forThis is easy.
$$1-0.366=0.634$$
becomes.Isn't the result lower than expected?
By the way, the graph is as follows, and the probability of winning is not 500% even if you draw 100 times. (Actually, no matter how many times you pull it, it will never be 100%.)
For reference, the probability \(P_{500}\) of drawing at least once when drawing 1 times is
$$P_{500}=1-(\frac{99}{100})^{500}=0.9934\ \rightarrow\ 99.34\%$$
It becomes.
why use extra event
Why do you do such troublesome calculations?You may have wondered.I will explain it.
To find the probability of winning at least once,
1 time chance + 2 chances + 3 chances + ... + 100 chances
You don't have to do the calculation.Rather than doing this calculation, it's easier to use the residual event.
By the way, if you calculate the probability \(P_1\) of winning only once, it will be as follows.
$$P_1=(\frac{99}{100})^{99}+\frac{1}{100}\simeq 0.38$$
So it will be about 38%.
Summary of events
- The other event is a method of finding the probability of the opposite event and subtracting it from 100% when the probability you want to find is the sum of several probabilities.
- Used because it is more efficient than finding the sum of all and adding
- When returning the drawn gacha, the probability of drawing a win at least once when drawing a gacha with a win of 1% 100 times is about 1%, which is lower than expected
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