You may also like

problem icon

Gambling at Monte Carlo

A man went to Monte Carlo to try and make his fortune. Whilst he was there he had an opportunity to bet on the outcome of rolling dice. He was offered the same odds for each of the following outcomes: At least 1 six with 6 dice. At least 2 sixes with 12 dice. At least 3 sixes with 18 dice.

problem icon

Balls and Bags

Two bags contain different numbers of red and blue balls. A ball is removed from one of the bags. The ball is blue. What is the probability that it was removed from bag A?

problem icon

Coin Tossing Games

You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?

Thank Your Lucky Stars

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Thank you for this solution Andrei (Andrei Lazanu, School 205 Bucharest) and for the link to the useful site:

For a 2x2 grid I need to make 2 moves.

There are 4 possible paths:

left, left
left, up
up, left
up, up


Only 2 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$ {{1 \over{2^2}}\times2} = {2 \over4} = {1 \over2} $$

For a 3x3 grid I need to make 4 moves.

There are 16 possible paths.

Only 6 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$ {{1 \over{2^4}}\times6} = {6 \over16} = {3 \over8} $$

For a 4x4 grid I need to make 6 moves.

There are 64 possible paths.

Only 20 of these take me to the top left-hand corner of the grid, so the probability of getting to the opposite corner is: $$ {{1 \over{2^6}}\times20} = {20 \over64} = {5 \over16} $$

I found on the Internet, at the Math Forum, the formula together with the explanation.
The address is: http://mathforum.org/library/drmath/view/54218.html

The formula generating the number of ways to go from one corner to another is: $$ {[2(n-1)]!} \over{[(n-1)!]^2} $$

The formula generating the probability of landing in the opposite corner in a n x n grid is: $$ {{1 \over{2^{2(n-1)}}}} \times{{[2(n-1)]!} \over{[(n-1)!]^2}}. $$

I verified my results and they worked for n = 2, 3 and 4.