Drawing from EC's jar

The maximum number of marbles we could have without getting a total of $8$ marbles from any pile is $7$ from each, and then we must add $1$ to get the answer of $7\times4+1=29$ . Generalization : With $n$ piles of $k$ items, then the number of ways to be sure to pick $i$ items, with $i$ less than $k$ , from the $n$ piles is $n(i-1)+1$ .

This is the perfect problem for the Pigeon Hole Principle. The Pigeon Hole Principle states that if there are A balls and B boxes, where A is greater than B then at least one box will have more than one ball. Here, we have our balls, and our colors as our boxes. If we wanted to have at least two balls of the same color, we would pull out 5 socks, the first four being white, red, blue, and green. The 5th would have to be one of the colors, thus making two balls the same color. Similarly, we pick out seven of each color, knowing that the next ( 29th ) ball picked out would have to be one of the four colors, making the eighth ball in the color group.

Since we need to draw 8 balls of same colour, the maximum number of balls which we can draw and fail to do that, is to draw 7 balls of each colour, which amounts to 28 balls. So, the minimum number of balls for success should be 29.

Consider the worst possible case, where we draw 7 balls of each of the 4 colours - 28 balls. This means that no matter what ball you draw next, this 29th ball will be the eighth ball of whatever colour it happens to be. So you need to draw at minimum 29 balls to guarantee you have taken 8 of the same colour.

In the worst case, assume that we chose 1 ball from different color i.e., 1 from white , 1 from red , 1 from blue , 1 from green ......Now this can take place for 7 times i.e., 7 x 4 balls = 28 balls ....Now, in the 8th case we can choose any ball as addition of any ball of any color will lead us to at least 8 balls of same color ..... So , in the case where our luck is dead for a while , we will have to choose 28 + 1 = 29 balls to ensure that we have at least 8 balls of same color....

By the generalized pigeonhole principle, we have that we must choose $4\cdot(8-1) + 1 = 29$ .

The worst case is that we keep taking balls, but each consecutive ball is a different colour. That is you take white , then * red , then * blue , then green , then white again...

This repeats until one takes 7 of each colour, or 4 $\times7$ = 28 balls.

Now, no matter what the colour of the next ball is, we will have 8 balls with the same colour.

Thus we take 28+1 = *29 balls * in total.

by PHP, (7+7+7+7+1)=29

para resolver esse tipo de problema devemos pensar na pior das hipóteses que nesse caso seria pegar 28 pedras sendo 7 de cada cor assim a próxima pedra so pode ser a oitava de mesma cor

Na pior da hipóteses, você vai tirar 7 pedras brancas, 7 vermelhas, 7 azuis e 7 verdes. Isso contabiliza 28 pedras. Na próxima jogada tu vai ter que tirar a oitava pedra de qualquer maneira, contabilizando 29 pedras.

If there must be at least 8 of the same color, the worst case would be 7 of any color, 7 of anohter, 7 of another and 8 of another, and that makes $7 + 7 + 7 + 8 = 29$

We can choose 7balls of each colour then we have to choose 1 ball of any colour. Then we are sure that we will get 8 balls of 1 type. Hence we can say that this the minimum (7*4 +1)=29 balls has to be taken out in order to be sure of getting 8 balls of any colour

Consider the worst case scenario when we have 7 balls of the same color . So we would have taken 28 balls. Taking another one would guarantee that we have 8 balls of the same color. So the minimum number of balls we have to take is $28+1=29$ .

we took 7 balls of each colour then take any ball. it will be 8 balls of same colour..