Question

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn - one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins

Answer 1

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn - one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins

Answer 2

The probability that, at the tip of the fourth round, each of the players has four coins is 5/192.

Given that game consists of 4 rounds and every round, four balls are placed in an urn one green, one red, and two white.

It amounts to filling in an exceedingly 4×4 matrix. Columns C₁-C₄ are random draws each round; row of every player.

Also, let $\%R_{A}$ be the quantity of nonzero elements in $R_{A}.$

Let $C_{1}=\left(\begin{array}{l}1\\ -1\\ 0\\ 0\end{array}\right)$.

Parity demands that $\%R_{A}$ and$\%R_{B}$ must equal 2 or 4.

Case 1: $\%R_{A}$=4 and $\%R_B$=4. There are $\left(\begin{array}{l}3\\ 2\end{array}\right)$=3 ways to put 2-1's in $R_A$, so there are 3 ways.

Case 2: $\%R_{A}$=2 and $\%R_B$=4. There are 3 ways to position the -1 in $R_A$, 2 ways to put the remaining -1 in $R_B$ (just don't put it under the -1 on top of it!), and a pair of ways for one among the opposite two players to draw the green ball. (We know it's green because Bernardo drew the red one.) we are able to just double to hide the case of $\%R_{A}=4,\%R_{B}=2$ for a complete of 24 ways.

Case 3: $\%R_A=\%R_B=2$. There are 3 ways to put the -1 in $R_{A}$. Now, there are two cases on what happens next.

• The 1 in $R_B$ goes directly under the -1 in$R_A$. There's obviously 1 way for that to happen. Then, there are 2 ways to permute the 2 pairs of 1,-1 in $R_C$ and$R_D$. (Either the 1 comes first in$R_C$ or the 1 comes first in $R_D$.)
• The 1 in $R_B$ doesn't go directly under the -1 in $R_A$. There are 2 ways to put the 1, and a couple of ways to try and do the identical permutation as within the above case.

Hence, there are 3(2+2×2)=18 ways for this case. There's a grand total of 45 ways for this to happen, together with 12³ total cases. The probability we're soliciting for is thus 45/(12³)=5/192

Hence, at the top of the fourth round, each of the players has four coins probability is 5/192.

Learn more about probability and combination is brainly.com/question/3435109

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