# Waldegrave Problemi: Minmax Oyununda Karma Strateji

## Waldegrave Problem

İçindekiler

In the fifth chapter of Pierre R’emond de Montmort’s book, published in 1713, “Essay d’analyse sur les jeux de hazard”, there is a section with correspondence between Montmort and Nicolaus Bernoulli about the calculation of probability problems. These correspondences started in 1710 and continued until 1713. Among these correspondences, Montmort tells Nicolas Bernoulli about his friend Francis Waldegrave’s analysis of the French card game “Le Her”. In the letter dated November 13, 1713, Montmort provides a mixed strategy solution for Waldegrave’s “Le Her”, a two-player minmax game.

The question that Waldegrave is looking for is the following: Is it possible that the dealer in the “Le Her” game with two players will win the game? Is the person who is the buyer, more likely to win?

## Le Her Game

Let’s talk about the game ” **Le Her** “. Le Her is played with a deck of 52 playing cards. Papers take different values from ace to king. As, one of the other cards, respectively, 2,3,4 ……., Male “Jack” takes the value of 10, Girl “Quenn” takes the value of 12 and Pastor “King” takes the value of 13.

There are two players in the game. Each player may be the card issuer (the player who set up the game), the **Dealer** (such as the dealer) or the **Buyer** who accepts the game . The rules are simple:

**1.**Each player receives one card face down. Players can only see their own paper. Other players cannot see this.**2 a.**The buyer can replace the paper he has received with the distributor if he wishes.**2b**. If the dealer’s hand in the card**chaplain**, the exchange becomes invalid. Both players continue to hold their own cards.**3 a.**Then, if the dealer wishes, he can leave the paper in his own hands for a third card (the first paper or the paper he replaced with the recipient).**3b**. If the new card is a**Pastor**, the card the dealer has received will be void and must keep the previous card.

After the cards have been dealt out, the card with **the highest** value **wins the game** . If both people have the same card, the **dealer is** considered the winner.

How do you play this game? Is it a distributor? Is it more reasonable to be a buyer?

## Solution of the Waldegrave Problem

The solution of the game requires a great probability calculation. Since each player has to make a “continue” or “OK” decision for 13 cards that can be obtained and there are two players in the game, a solution and return matrix of 13 over 2 X 13 over 2 is formed. Many studies refer to Melvin Dresher’s book ” **The Mathematics of Games of Strategy*** ” for the* long solution of the game *. **(Those who are curious about the long solution can investigate)*

The steps of the solution after some of the probability calculations are made are as follows:

The return matrix is narrowed using dominant strategies.

“When a

rational person gets three in his hand, he does not want to change paper, but in another hand he does not want to change the paper when he gets 4. In the case of a rational one’s choice, 4 will always be preferred to 3.“

### Rational Decision of the Buyer

After eliminating the precisely dominated strategies and some mathematical operations , it is concluded that the most rational decision for the ” **Buyer** ” is to request a new card in case of 8 or more cards. In case of a lower or lower card, the buyer should request a new card.

### Rational Decision of the Distributor

**In the case of a “dealer”** , if a card of nine or more is received (if the buyer wants or does not want to change the card), he / she should not request a new card; If a card is seven or under, it must request a new card.

So far, the question of what strategies the players will do when the cards are answered has been answered. However, it is a mystery to see what to do when seven comes for the buyer and eight for the distributor. In short, when these cards come, there is no pure strategy that can take the solution of the game to the Nash equilibrium. Both players have to apply a mixed strategy when the card comes in at these values (it involves probability calculations)

On a MinMax game **of LIC** ‘s payoff matrix for the game becomes a 2 x 2 matrix.

Since it is a zero-sum game, the matrix of the game is the same for the **Distributor** , but with the **Buyer it** is inverted. That is, if the buyer wins 16, the distributor receives -16 returns

**The** right strategy for the **Buyer and Distributor** is achieved with a mixed strategy. **When the dispenser** receives **eight,** he will remain on the paper with a probability of 3/8, and will change it with a probability of 5/8.

**The recipient** holds seven at a probability of 5/8 when he receives seven; Changes the paper with a probability of 3/8. By adding these possibilities, the game’s return matrix is recalculated. When probability calculations are made about who is the superior party in the game, it is concluded that the **Distributor** ‘s probability of winning the game is **48.7 percent and the ** **Buyer** ‘s probability of winning the game is 51.3 percent. **Why is the Buyer likely to win? **Here is why the **Buyer** is more likely to win; Due to the nature of the game, the **first move ** stems from the **Buyer’s choice** . Therefore, it partially gains superiority in the game.

If you want to play this game with your friends, keep in mind that the **Receiver** is superior in the game, and this superiority is due to his first move 🙂

Bellhouse, DR, & Fillion, N. (2015). Le Her and other problems in probability discussed by Bernoulli, Montmort and Waldegrave. Statistical Science, 30 (1), 26-39. Also Visit: Zoom Property