Question

Tic-tac-toe: In the game of tic-tac-toe, if all moves are performed randomly the probability that the game will end in a draw is 0.127. Supposed seven random games of tic-tac-toe are played. What is the probability that at least one of them will end in a draw

Answer 1

Answer:

$P(X \geq 1) = 1-P(X$

And if we find the individual probability we got:

$P(X=0)=(7C0)(0.127)^0 (1-0.127)^{7-0}=0.3865$

And replacing we got:

$P(X \geq 1) = 1-P(X$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=0.127)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

$P(X \geq 1) = 1-P(X$

And if we find the individual probability we got:

$P(X=0)=(7C0)(0.127)^0 (1-0.127)^{7-0}=0.3865$

And replacing we got:

$P(X \geq 1) = 1-P(X$

Answer 2

Answer

Draw game: 16/126 = 0.126875

Explanation:

Assuming X as first player, O as second player, there are C(9,5) = 126  

different ways to arrange five X's (or four O's) within nine positions.

Among these, 16 are draw games: (XXO,OXX,XOO), (XOX,XXO,OXO), (XXO,OOX,XOX),  (XOX,XOX,OXO) together with their horizontal and vertical reflections.

There are 12 configurations with a sure win for O: three O's on one diagonal,

for a total of 2 diagonals times 6 choices for the fourth O.

There are 36 "undecided" configurations, i.e. configurations with both three

X's and three O's on a line, according to the following count:

three O's on one side line, that is 4 sides times 6 choices for the  

fourth O = 24.  three O's on one middle line, that is 2 middle lines times 6 choices   for the fourth O  = 12.

This leaves 126 − 16 − 12 − 36 = 62 configurations   with a sure X win.

The win in an undecided configuration will depend on which side  

completes first a triad on a line. Assume a player completes a triad at  

his 3rd draw: then the two remaining X's or O's must have appeared in  

the first two draws, for a total of C(2,2) =1 possibility each. If the  

triad is completed at the 4th draw, then the two remaining X's or O's  

had to be distributed within the first three draws, giving C(3,2) = 3  

possibilities each. Finally, if the first players completes his triad  

at the 5th draw, the two remaining X?s must have appeared within four  

draws, giving C(4,2) = 6 possibilities. In accordance, there are four  

different possibilities:

X completes at the 3rd draw, O at the 3rd or 4th: X wins, with 1x(1+3) =

4 configurations

O completes at the 3rd draw, X at the 4th or 5th: O wins, with 1x(3+6) =

9 configurations

X completes at the 4th draw, O at the 4th: X wins, with 3x3 = 9 configurations

O completes at the 4th draw, X at the 5th: O wins, with 3x6 = 18 configurations

Thus, the probability that an undecided configuration results in an X win

is (4+9)/40 = 13/40, whilst the probability of an O win is (9+18)/40 = 27/40.

The final probabilities can now be computed:

First player: (62 + 36*13/40)/126 = 0.584850

Second player: (12 + 36*27/40)/126 = 0.288275

Draw game: 16/126 = 0.126875