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Probabilistic problems in scams
There are four or five booths around the monument of Qingcheng Park in Hohhot. The rules of this prize-winning activity are: 1 * * 24 glass beads, 1 * * 3 colors, 8 in each color, and the winner should grab 12 glass beads from these 24 glass beads. If there are eight in one color, four in another color and none in one color, and you catch 543, that is, five in one color, four in one color and three in another color, you have to give the vendor 30. One yuan catch 15 times. The reporter stepped forward and caught four of the 12 glass balls that the reporter caught for the first time, that is, 444, and also won the 5 yuan, and caught 543 for the second time. In just two minutes, the reporter lost 60 yuan.

What's going on here? The reporter interviewed a mathematics major in Inner Mongolia University by telephone, and he explained the reason for doing so.

It seems that these winners skillfully used mathematical probability. No matter how hard you try, you always lose. We still advise you not to take part in such prize-winning activities.

First, calculate the probability of catching 543. What is the probability of catching 543 with eight balls of three colors and putting it in the bag?

* * * There are three kinds of balls. The final result does not need the order of colors, so the first grab can be arbitrary, and the probability is 1. When you grab the ball for the second time, there are 23 balls in the bag. The ball you want to grab is the same color as the ball you grabbed for the first time. There are only 7 such balls left in the bag, so the probability of grabbing the ball for the second time is 7/23. Similarly, the probability of catching the ball for the third time to the fifth time is 6/22, 5/2 1 and 4/20 respectively.

By the time you catch the ball for the sixth time, there are 19 balls in the bag, so the denominator is 19. The balls you want to catch must be different from the previous five times. There are 16 such balls in the bag, so the probability is 16/ 19. The probability of catching the ball for the seventh time to the ninth time is 7/ 18, 6/ 17, 5/ 16 respectively. By the tenth catch, there were 15 balls in the bag, and the denominator was 15. The ball you want to catch is not the same color as the one you caught nine times before. There are 8 such balls in the bag, so the probability is 8/ 15. The capture probabilities of 1 1 and 12 are 7/ 17 and 6/ 16 respectively. Finally, multiply the catch probability by 12 times, and the final data is the final probability.

The above is online for your reference. As for the probability algorithm, it can be calculated as follows: 543, such as red, white and black,

Specify who is 5, who is 4 and who is 3, and then take A33.

For example, red 5, white 4 and black 3. Then, the probability is: (c85× c84 * c83 * a33)/(c24/12) =1317120/2704156 = 48.

That is to say, if you catch 2, there is an average of 1, which is 543. Naturally, I lost miserably. Hmm. How interesting