Simple coin probability problem, detailed in it.

I have thought about the result of the first round of coin toss. Let's extrapolate from your results and see if the reference answer is correct. My conclusion is that the answer is wrong. Two thirds are correct. Two ways to prove my conclusion.

Let's assume that the probability of C playing in the first round is P, and then continue our analysis.

First of all, in the first round of eight cases, the probability of 1/4 continues, and the probability of 3/4 is coin toss, of which 2/3 is playing. Suppose that in the case of 1/4, the probability of throwing a coin c behind is p'.

Ok, assuming p' is known, how to calculate p? If the landlord hasn't studied conditional probability, let me briefly explain that, like the tree path diagram, it is possible to continue to vote with 1/4 at first, and then continue to vote with probability p', so this road 1/4 provides a probability of 0.25 P '；; Then the opponent's 3/4 directly gives the incident probability of 2/3, which may provide the incident probability of 3/4 * 2/3 =1/2. These two paths are parallel choices, so the final sum of these two values is the total probability of the event, namely:

p=0.25p'+0.5

Now consider the value of p'. When we finish the last round and find that we will continue to throw, we will find that all the situations we are facing are exactly the same as the last round, and there is a probability of 1/4 to continue, 3/4 to produce results, and the probability of producing results is 2/3. In other words, P' is actually the same as the P we just calculated.

So the question is simple: p = 0.25 p+0.5 =>；; p=0.5/0.75=2/3

This idea needs a little calculation. In fact, intuition has told 1/2 that this answer is wrong. Why? Because the three teams A, B and C only have different names, other conditions are the same, that is to say, the probability of participating in the first round is the same, if they are all 1/2. The probability of not participating in the first round is also 1/2. The probability of C not participating in the first round is 0.5, which means that the probability of AB playing together in the first round is 0.5. Then, if C participates in the first round, he can't play alone, and A and B must also play, then the probability of at least one team of A or B playing is greater than 0.5, which obviously contradicts ABC's basic understanding of mutual symmetry.

Now let's briefly talk about the method that basically does not need calculation, or consider it along the route of three teams symmetry. No matter what the whole thing is, in fact, it is necessary to choose one team from the three teams not to participate in the competition, and the remaining two will participate. Then three teams are selected with equal probability. In fact, if you regard the events that determine the order of appearance as a whole and ignore the details of coin toss, this question is no different from drawing lots from three paper balls with ABC written on them. The probability of winning, that is, the probability of not participating, is 1/3, so the probability of participating is of course 2/3.

In fact, the mistakes in reference answers are also easy to understand. Dividing 4 directly by 8, the mistake behind it is to include the situation that C can't play in the first round and needs to be re-cast.

In fact, the answer may be right and the question is wrong. There is always a problem.