You are playing a game with three cards. At the start, all three cards are face down. You know that a different real number is written on every card, but you do not know what the numbers are. The rules of the game are as follows: You can choose any card and turn it over (assuming that you have not already turned it over). You can either keep the card, or discard it. If you discard a card, then you cannot go back to it; you must choose a new card. If the card you have kept has the largest real number on it, then you win the game. If you use the optimal strategy, then what is the probability that you win the game

concluding, the idea is discarding cards until we get,at least, a 6, in that case, the smallest probability of winning the game is 0.59, which is more than in half of the cases.

How to find the probability?

Now, we know that there are 52 cards in a normal deck, if we only keep the ones with real numbers, there are 40.

These numbers go from 1 to 10.

5.5 is the average real number in the cards, so, having a 6 or more in a card is what we expect.

This means that if the card that we turn up is smaller than 6, we discard it.

If the card has the value 6 or larger, we keep it.

In this case, the probability of losing is if one of the other cards has a value larger than 6.

The probability that a random card has a value larger than 6 is:

p = (number of cards with a value larger than 6)/(total number of cards)

p = (16/39)

(The quotient is 39 instead of 40, because we already know one of the cards)

This means that if keep the 6, the probability of winning is:

1 - 16/39 = 0.59

If instead, we got a 7, obviously we keep it, in that case the probability of winning is:

1 - 12/39 = 0.69

And so on, concluding, the idea is discarding cards until we get,at least, a 6, in that case, the smallest probability of winning the game is 0.59, which is more than in half of the cases.

if you want to learn more about probability:

https://brainly.com/question/25870256

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