Question 3: For the unfinished game, let us consider a four—person game based on "two—up". More specifically, two coins are placed on a wooden stick, which is used to toss the two fair coins in the air. Therefore, we can get four sequences of tosses: HT, TH, HH and TI [H for Head and T for Tail). There are four players, Alice, Bob, Candy and David, to play the game. a The pot is 100 dollars. 0 In each round, they play the two— up game and the Wilmer gets one point. HT: Alice wins one point; TH: Bob wins one point; HH: Candy wins one point; 'IT: David wins one point. I The player who gets 5 points is the winner and gets the whole pot of $100. Suppose they decide to end the game early when Alice has 3 points, Bob has 2 points, Candy has 4 points and David has 4 point. How should they divide the pot?

The probability of each outcome here is computed as: 0.5*0.5 = 0.25 as both the coins here are fair. The number of rounds which have already been done here is computed as:

Step-by-step explanation

The probability of each outcome here is computed as: 0.5*0.5 = 0.25 as both the coins here are fair. The number of rounds which have already been done here is computed as:

= 3 + 2 + 4 + 4 = 13
The probability for Candy to win here is computed as:
= P(C) + P(BC) + P(BBC) + P(AC) + P(BAC) + P(ABC) + P(BBAC) + P(ABBC) + P(BABC)
= 0.25 + 2*0.252 + 3*0.253 + 3*0.254
= 0.4336
Now as David and Candy are at the same situation here, therefore, the probability for David to win would also be given as: 0.4336
The probability for Alice to win here is computed as:
= P(AA) + P(BAA) + P(ABA) + P(BBAA) + P(BABA) + P(ABBA)
= 0.252 + 2*0.253 + 3*0.254 = 0.10546875
Therefore, the probability for Bob now is computed here as:
= 1 - 2*0.4336 - 0.10546875
= 0.0273
Therefore, the pot amount distribution here is given as:
For Alice, the amount given here is: 0.10546875*100 = 10.55 dollars
For Bob: 100*0.0273 = 2.73 dollars
For Candy and David: 43.36 dollars each.