Question

How do you play Chicago? There are eleven rounds in the game, one for each combination that can be made by adding two dice, namely the numbers two through 12. Each round has a target combination starting with two and going up all the way to 12. Going clockwise, the players take turns to roll both dice one time. If players roll the target combination, then they score points equal to the target combination, otherwise they score zero. For example, if the round corresponds to target combination six, then the player scores six points if the two dice add up to six. Else, the player scores no points. The player with the highest score at the end of the eleventh round wins the game.

If the player wins a dollar for every point, he/she gets and losses three dollars for getting no points, what are the expected winnings or losses on each turn?

Answer 1

Total sample space size = 6×6 =12

Probability distribution table for getting a sum of Numbers on a pair of dice  is as follows.

Sum of dice	possible ways to get this sum	number of ways to get this sum	probability of getting this sum or winning	probability of losing or not getting this sum
2	(1,1)	1	1/36	35/36
3	(1,2),(2,1)	2	2/36	34/36
4	(1,3),(2,2)(3,1)	3	3/36	33/36
5	(1,4),(2,3),(3,2),(4,1)	4	4/36	32/36
6	(1,5)(2,4)(3,3)(4,2)(5,1)	5	5/36	31/36
7	(1,6),(2,5),(3,4),(4,3),(5,2)(6,1)	6	6/36	30/36
8	(2,6),(3,5),(4,4),(5,3),(6,2)	5	5/36	31/36
9	(6,3),(3,6),(4,5),(5,4)	4	4/36	32/36
10	(5,5)(6,4),(4,6)	3	3/36	33/36
11	(6,5),(5,6)	2	2/36	34/36
12	(6,6)	1	1/36	35/36
			Sum = 1

So,  If the player wins a dollar for every point, he/she gets and losses three dollars for getting no points

As we know that expected profit = probability × profit

So,  

The table for expected winnings and losses for each round is as follows..

Round number (sum to be obtained )	expected winnings (in dollars) = proabability of winning × winning amount	Expected losses = probability of losing × loss amount
2	1/36 × 2 = 2/36	35/36 × 3 = 105/36
3	2/36 × 3 = 6/36	34/36 × 3 = 102/36
4	3/36 × 4 = 12/36	33/36 × 3 = 99/36
5	4/36 ×5 = 20/36	32/36 × 3 = 96/36
6	5/36 ×6 = 30/36	31/36 ×3 = 93/96
7	6/36 ×7 = 42/36	30/36 × 3 = 90/36
8	5/36 × 8 = 40/36	31/36 × 3 = 93/36
9	4/36 ×9 = 1	32/36 ×3 = 96/36
10	3/36 × 10 =30/36	33/36×3 = 99/36
11	2/36 × 11=22/36	34/36 ×3 = 102/36
12	1/36 × 12= 12/36	35/36 ×3 = 105/36
	Sum of expected Winnings = 7 dollars	sum of expected losses = 30 dollars

So, final expected winnings /losses from this game of 11 rounds = expected winnings + expected losses

= 7 - 30 = -23 dollars

Negative sign shows losses

Hence, on average a player is expected to lose 23 dollars in this game (for 11 turns)

On each turn, you are expected to lose 23/11 = 2.09 dollars

(Note I have given the expected winning and losses for each turn separately in table 2 above so if you want individual expected winnings or losses you just have to subtract the two columns for each row separately . That will give you your individual expected value of each turn.  )