# Coin Toss Project

### By : Jonathan Laks

This project is going to tell if a specific game is fair or not. Then, I will also invent two more games. One of them is fair, while the other is not. In addition, I am going to give the probability that player 1 or playing 2 will win. Finally, I will summarize my experience.

This is the first game : Two players each toss 3 coins. If the result involves 2 heads, Player One wins. If the result is anything else, Player Two wins.

Is this game unfair or fair? to show it, I made a word document showing the odds of which could be seen above. This will also be explained in the tag below.

As you could see in the diagram, there are 8 outcomes, but only outcomes numbers 4, 6, and 7 had the outcome of only two heads. 3/8 is 38%, and therefore, this game is not fair. Also, player 2 has a 62% chance of winning.

My unfair game involved a randomizer (on the computer) and a 3*3 square. On the 3*3 square, write the numbers 1-9. An example of a randomizer would be https://www.random.org/lists/. An example of how the square looks like is below.

If the randomizer takes the same number twice in a row, player 1 wins and crosses out the repeated number. If the randomizer shows a different number twice, player two wins and can cross out the second number. There are examples later. Below there is a video of how to create the squares.

If you get 4 once again, then player one gets the point. However if you get a different number in the next randomization (for example, after this 4 you get a 7), player two gets the point.

Above is a box diagram. The blue sets are the numbers player one can win with while the black circled sets are everything player two wins. As you can see, if the first random picks out 1, there is a 1/9 chance that it will pick 1 again. This happens to all the other numbers, so 1 over 9 multiplied by 9 (both the denominator and numerator) is 9/81, or simply back to 1/9. This means that player 1 has a 11.1% chance of winning while player 2 has a 88.9% chance of winning, therefore, proving this game is

My fair game involves two dice. Here are the instructions : First, player one throws two dice simultaneously. Then, he adds the difference between the numbers on the dice (the bigger number minus the smaller number). Player 2 does the same thing, and whoever has at the end a higher score, wins. Also, if someone gets a double, for example 3 and 3, he subtracts 3 from 3.

Above is a diagram of the results. (NA stands for not applicable because it will always be the bigger minus the smaller number). The results show what each player can get with each roll. This game is fair because both of them have the same dice, and the same dice means that there isn't a higher or lower chance player 1 will get 2 numbers, for example a 6-1, than player 2 will. The same example percentagewise means that there is a 1/21, or a 4.8% chance, that player 1 OR player 2 will get 6-1. This, again, proves this game is

When I created the math game, I felt it was challenging and was kind of desperate so I looked up ideas. Then after making two simple ones, I was told they were too simple and had to re-do them. It took me some time to think about and finish the two more complicated games, but I think I did a good job with them.