Probability Of Winning In Noughts And Crosses

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Ever since I was a child I would have this playground near the swimming pool within my neighborhood, I and a few other children about my age would play different games every week. One day, a friend of mine introduced me to Noughts and Crosses, although where I come from, we call them ‘TIC TAC TOE’. When my friend started drawing 3×3 boxes, I was confused to what he was doing but when I observed further, he drew a circle on one of the boxes. I was confused so I asked him what he was doing, he then told me how to play. It was a simple yet fun and competitive game. I then began playing with friends and my brother more frequently as a child.

The game works with the first player placing a circle or a cross on any of the 3×3 boxes and need to make a line of 3 circles or crosses the opposite team also needs to do the same thing, whoever gets a line of 3 first is the winner. Players may also block the opposing player’s line by placing their own piece to try to stop the player. This deceptively simple game is an ancient game played in Britain for centuries (also called Tick Tack Toe / Tic Tac Toe in America and some countries). However, there is evidence that Noughts and Crosses go way back, back to ancient Rome, they were called Terni Lapilli. There is also evidence that the game goes way back to ancient Egypt although it is not as the Tic Tac Toe we know of today, the evidence supports the claim that Tic Tac Toe originates from Egypt.

Back to topic, since I am going to use Noughts and Crosses as my basis research, there are a few math topics we can cover, but I choose probability, and my Topic is the probability of winning in Noughts and Crosses. However, since finding the chance of winning tic tac toe is quite simplistic, we are therefore going to try on other ways to make the calculations more interesting, such as the use of 4×4 or 5×5 boxes. We’ll also look at the chances of the first player winning and the second player winning. Formulas Used Probability : P(A)=n(A)/n(U) Mean : x̄ = (Σ -xi) / n Calculations Before we start with our calculations, we are going take a look at the probability of winning as a whole. Since there are only 2 players involved, therefore the chance of winning or losing is ½ or a 50:50. However, this is just the probability of either winning or losing, what we’re going to discuss is the chances of winning under different conditions. 1 2 3 4 5 6 7 8 9 Since there are 9 different possible positions we can start with, therefore to begin the calculations we need to start with the 9 possible places to pick.

Therefore the number of possible game positions is represented by 3^9 (19,683). This means the possibility of any 3×3 square being filled with an X, O, or empty/blank. There are so many different ways that the squares can be filled, and the term goes as 9! (Nine factorial) which is equal to 362 880. This is the number of different positions that you can fill on the grid. How did we get 9! Or where did it come from? Since firstly you have 9 choices of positions, then once a player places a cross or a circle you’ll be left with 8 box left, then 7, the 6, etc. however when we take a look, all the initial points are just reflections or rotations of different positions, therefore we can simplify it as 3 different initial points. In this essay, we’re also going to be looking at the Combinatorial Game Theory or CGT for short.

A Combinatorial Game Theory is a mixture of mathematics and Theoretical Computer Science (TCS) that studies Sequential games; games like chess, checkers, backgammon, or Noughts and Crosses. Tic Tac Toe gives each player with a number of alternative moves. While it is often clear that a move is a win, lose, or a draw near the end of a game, it is usually much harder to assess where a particular move will lead in the beginning. Tree Diagram (Incomplete tree diagram) Source: The tree diagram was incomplete because it goes on under even longer and takes up too much space. But from the diagram above, we have a better idea of completing the chart. Keep in mind that positions that can be achieved from rotating or reflecting the position are merged narrowing down the numbers. The tree diagram above is not of all possibilities since that would take up way too much space and time, so instead of using all possible positions, anything that could be mirrored or rotated is put as one so the possibilities are minimized.

There are 25 ways to fill up the tic tac toe board with O and X but since there are only 2 probable outcomes, win or lose, it makes the chances of winning to 50:50 or a 50% chance. But that’s too simple. So here are some of the chances of other different outcomes to the game, like perhaps the probability of winning a game with just 5 to 6 moves, or the probability of getting a draw, or the probability of the first player winning the game. (Here are also documentation necessary for the IA) Start off with the probability of winning with just 5 to 6 moves. For the calculation of probability, we’ll be using: P(A)=(n(A))/(n(U)) Probability=(number of thepossible outcome of a certain event)/(Total number of all possible outcome) Total Number of all possible outcome in Tic Tac Toe: 9! (From the 9 different places to pick) Here is the calculations of probability of winning with 5 moves x=(8×3!×6×5)/9! 8 comes from the 8 possible lines that is available to create a 3-in-a-row (3 vertical, 3 horizontal and 2 diagonals).

Meaning there are only 8 possible lies for a row to be present at, thus leaves us with 8 to start with. 3! is achieved from the 3 moves for the first player has to win the game in just 5 moves total, hence it is 3×2×1 (3!). Assuming X goes first and O second, there would be 3 Xs and 2 Os involved. And with 3 boxes filled with X that leaves us with 6 other boxes to fill the 2 Os, therefore leaving 6 in the equation. 5 was achieved by the total number of movements used. The whole equation is then divided by the total number of possible moves which is 9! And when we work on the equation, we are left with 1440 possible games ending in a win on the fifth move This leads to the probability value of: x=(8×(3×2×1)×6×5)/((9×8×7×6×5×4×3×2×1)) x=(8×6×6×5)/3628800 x=(8×36×5)/3628800 x=1440/3628800 =0.00396825396 When we turn this into percentage form, we are left with only about 0.0039…  100= 0.39% chance of winning in the 5th move.

Now with the 6 movements, the equation gets altered a bit. For the equation of probability of winning on the 6th move being the first player is: x=(8×3!×6×5×4)/9! x=(8×6×6×5×4)/3628800 =(8×36×20)/3628800 =5760/3628800 The answer to the equation is that there are 5760 possible outcomes. However, though, this is still crude, meaning it has not removed the possibilities that the X and O may simultaneously have a 3-in-a-row. We need to exclude all cases where there are 3 Os and 3 Xs in a row (both can’t be diagonal), so the formula to exclude them is 6×3!×2×3! = 432 possibilities. 8×3!×6×54 = 5760, now we need to subtract the 432 from the 5760 which leaves us with 5328 possible outcomes of the first player winning with a total of 6 moves. Decimal wise the probability is: x=5328/3628800 x=0.00146825396 Which in percentage is translated as around 0.00146…100= 0.15% chance of winning in the 6th move.

Probability of a draw: For the number of different possible outcomes resulting in a draw, there are 3 basic draw patterns, if we include rotations and reflections that would result in 8+4+4= 16 different patterns, and the formula that is used is 16×5!×4! = 46080 possible outcomes of a draw In result, what we get is: x=46080/3628800 =0.01269841269 Which is about 1.27% chance of a game ending in a draw, which is bigger as compared to the other percentages before it. When testing out the probability’s validity by going up against a bot for 20 times, here is the result I’ve got: Column trials 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 results D D D W D D L W D D L L L W W W D D W W From the table above here are some Analysis I made: 9/20 Draws 7/20 Wins 4/20 Loses With a Mode of Draw and most of the wins end in the 6th move. Evaluation: Although we only calculated the probability of winning on the 5th and 6th move, we were given a very small number in percentage which would probably mean that there are more room for a probability on other outcomes and other conditions.

I also used a bot or an Artificial Intelligence instead of someone else to play tic ta toe with, which might affect the end result of the findings. However, since I used reference of different sources, there should be a better validity to the findings. According to the Calculations, Draw had a bigger percentage of outcome than the winning at the 5th move or 6th move, and despite the minimum support, The Percentage chance of ‘Draw’ is higher than the others, and the results of the table show that there are higher draw amounts compared to the wins and losses. When we look at it however, any plaer can have different results in their trials than I do since the game involves subjectivity, meaning a table can be all lose, all win, or all draw rather than the distribution I have. With that, I think that the probability in Tic Tac Toe is one of the most inaccurate one, meaning the probability is not that reliable when playing tic tac toe, although for sure if we play accordingly, the number of draws may be fitted to the probability.

Conclusion To conclude all the findings above, we can be sure that the outcome of the game is not strongly based upon the probability or chances since it is not randomized, but rather the opposite. Tic Tac Toe is strongly based upon the strategy and turn based game aspect to achieve a win, lose, or draw. Other Combinatorial games like chess, checkers and such are very heavily based upon the turn based and strategic aspect of the game, which would make the player think harder to win, Therefore Probability doesn’t really help much in Tic Tac Toe, although there are still uses to the probability, it helps predict, just how probability is commonly used. Although it does not predict very well but it still helps.

To me, I think that there is not that much use in counting probability in Tic Tac Toe, however it does help me to think on the move I Should do and another couple moves ahead. That concludes all the findings I have gathered from the different sources, here are some of the pictures of the process for this IA. (History of noughts and crosses, n.d.) (Sternal-Johnson, 2013) (Bottomley, 2001) Bibliography Bottomley, H. (2001). How many Tic-Tac-Toe (noughts and crosses games are possible? .

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