How to calculate the probability?

Question

Mathematics

Hello!
There are 5 cards in front of me [6, 7, 8, 9, 10]. I choose one of them. Then my opponent chooses any other card (all cards are face down). Then we show the cards. Whoever has the higher number wins.
Are there any formulas for calculating the probability of winning? Scenarios are also of interest (i.e. several games in a row [for example, after how many games in a row the probability of winning will approach 100%]).
Thank you :) I think it's interesting.
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Since my understanding of this topic is at the level of a dog, I can assume that:
only 5 cards. I take 1. => the chance that I need this card is 0.2
but with the connection of the opponent there are problems. I don't know in advance which card he will take. If he draws a card with a face value of 9, then to win I need to draw 10 (and the probability of this card is 0.2).
However, if he draws a 6, then the probability of winning is 100%.

Answer 1

1/2. Several games in a row (provided that the initial state is the same, i.e. all cards are returned to the place, which card is where - unknown) - 1/2 in each game. The probability of winning at least once in n attempts is 1 - (1/2)^n.
Elementary logic dictates that neither player has an advantage over the other.
With t.z. theorvera probability of winning from the first one = 1/5 (took the highest card) + 1/5 * 3/4 ​​(took the second highest card, and the opponent took any of the 3 lowest cards) + 1/5 * 2/4 + 1/5 * 1/4 = (4 + 3 + 2 + 1)/20 = 1/2.

Answer 2

You can go through all the options. That is, you drew 6 opponent 6, you drew 6 opponent 7, etc. Then count the number of options favorable to you and the number of possible options. Then we find the probability, the number of favorable events is divided by the number of possible ones. That's all. Well, the probability of 100% winning is easy to find. Divide 1 by the resulting probability. Round up to integers. This is the maximum number of attempts, 1 of them will be winning for you.
I think I explained clearly.

Answer 3

The probability of winning in this scenario is always 1/2.
Since the outcome of the next game does not depend on the outcome of the previous one, no matter how many games you play, the probability is ALWAYS 1/2.
This is what casinos use because it seems that, for example, waiting in roulette for several black ones in a row and betting on red, you will increase the probability of winning - nothing like that :-)

Answer 4

The idea of ​​counting favorable and unfavorable options, which Misha7 proposes, is very fruitful - the probability will turn out by definition.
But you don't have to count, because for every outcome you win, there will be exactly one outcome where you lose. It is clear which one: the cards are the same, but the hands fell differently. At the same time, different outcomes correspond to different ones.
Therefore, without any calculations, we can immediately say that favorable outcomes make up exactly half of all . Regardless of the number of cards, by the way - the main thing is that there are no repetitions. It would be possible to pull 1000 cards with numbers from 1 to 1000, the answer would still be the same.
jcmvbkbc's solution is also completely correct, although the symmetry considerations (none of the players have an advantage) seem a bit non-obvious to me.
But for icelaba, the correct answer is crowned with a completely wrong reasoning. So he and the probability of red or black in roulette will be equal to 1/2, because "the outcome of the next game does not depend on the outcome of the previous one." In fact, these probabilities are less than 1/2, because there is a "zero" sector (or even two sectors - "zero" and "double zero"), which is neither red nor black. The fact that the game has no memory has nothing to do with the probability of winning. It just doesn't change from game to game.