Strategies for the Penny-Nickel-Dime-Quarter-Dollar Game

My second-grader brought home the following game:

Draw the following grid, one for each player.

Take turns rolling a single die. For each roll, a player must write the number rolled in one of the columns of the first row, and do the multiplication to determine his/her score for that roll.

Continue until each player has rolled five times.

Calculate the total value of the round, and write it in the first row of the "total" column.

Repeat for the remaining four rows. The player with the highest grand total, i.e. the sum of the row totals, is the winner.

The purpose of this game is obviously to help kids learn the basics of counting money. But I, naturally, got distracted by the strategy problem: what is the optimal response to rolling a "3" on the first roll (of a row)? A "5" on the second roll? And so on.

My solution is given below, with proof following. Note that the value of the game under a random strategy is simply 3.5 (the expected value of a die roll) times the sum of all the coins, or 3.5*$1.41 = $4.935. Playing my strategy has an expected value of $6.675, for an increase of about 35%. It enables the player to retrieve approximately half the difference between the naïve value, $4.935, and the maximum possible outcome (from rolling all 6's), $8.46.

The strategies are given below, with proof following:

Proof.

Proceeding from the end of the game back to the beginning:

FIFTH MOVE: You have no choice to make.

FOURTH MOVE: Let R be the current roll and A and B be the values of the uncovered coins, A < B. Clearly the value of applying R to A is RA + 3.5B, and R to B is 3.5A + RB, because 3.5 = (1 + 2 + 3 + 4 + 5 + 6)/6 is the expected value of the last die roll. For R given, consider the difference RA + 3.5B – (RB+3.5A) = R(A-B) – 3.5(A-B) = (R-3.5)(A-B). Because A < B, this expression is positive iff R < 3.5. QED. The value of the game at this juncture, before the die roll, is 0.5*(2A+3.5B) + 0.5*(5B+3.5A) = A + 1.75B + 2.5B + 1.75A = 2.75A + 4.25B (because E(R|R<3.5) = 2 and E(R|R>3.5) = 5).

THIRD MOVE: Now consider three uncovered coins, A < B < C. The value of the game when R is applied to A, B, or C respectively, is given by:

A: X = RA + 2.75B + 4.25C

B: Y = 2.75A + RB + 4.25C

C: Z = 2.75A + 4.25B + RC

A dominates when X>Y and X>Z, i.e. when (R-2.75)(A-B) > 0 and (R-2.75)A – 1.5B + (4.25-R)C = 4.25C – (2.75A + 1.5B) + R(A-C) > 0. The former condition requires R < 2.75, and the latter condition is decreasing in R (because A < C). It takes its lowest possible value meeting the requirement R < 2.75 when R = 2.75, or 1.5C – 1.5B. Clearly this is positive, because C > B. Therefore, apply R to A when R < 2.75. QED.

B dominates when Y>X and Y>Z. From above, Y>X when R > 2.75. For Y>Z, examine (R-4.25)(B-C), which is positive when R < 4.25 (because B < C). Therefore, apply R to B when 2.75<R<4.25. QED.

C dominates when Z>X and Z>Y. From above, Z>Y when R > 4.25. For Z>X, examine 2.75A + 4.25B + RC – RA – 2.75B – 4.25C = 2.75(A-B) + 4.25(B-C) + R(C-A). This expression is increasing in R, so it takes its smallest value consistent with R > 4.25 at R = 4.25, with the value 2.75(A-B) + 4.25(B-A) = 1.5(B-A) > 0. Therefore, apply R to C when R > 4.25. QED.

The value of the game at this point, before the die is rolled, is

[(1.5A + 2.75B + 4.25C) + (2.75A + 3.5B + 4.25C) + (2.75A + 4.25B + 5.5C)]/3 = (7A + 10.5B + 14C)/3.

SECOND MOVE: Now consider four uncovered coins, A < B < C < D. The value of the game when R is applied to A, B, C or D, respectively, is given by:

A: W = RA + (7B + 10.5C + 14D)/3

B: X = RB + (7A + 10.5C + 14D)/3

C: Y = RC + (7A + 10.5B + 14D)/3

D: Z = RD + (7A + 10.5B + 14C)/3

A dominates when W>X, W>Y, and W>Z, i.e., when

1) 3*RA + (7B + 10.5C + 14D) - [3*RB + (7A + 10.5C + 14D)] > 0, equivalently, when

3R(A-B) + 7(B-A) = (3R–7)(A-B) > 0, or R < 7/3.

2) 3*RA + (7B + 10.5C + 14D) - [3*RC + (7A + 10.5B + 14D)] > 0: If R = 7/3, then (2) reduces to

- 3.5B + 3.5C = 3.5(C-B) > 0, which is always true when B < C.

3) 3*RA + (7B + 10.5C + 14D) - [3*RD + (7A + 10.5B + 14C) > 0: If R = 7/3, then (3) reduces to

-3.5B -3.5C + 7D = 7(D – 0.5(B+C)) > 0. But 0.5(B+C) < D, because 0.5(B+C) is the midpoint or average of B and C, which are both less than D. QED.

[proceed similarly for the remaining cases]

The value of the game at this stage is

{A + (7B + 10.5C + 14D)/3 + 2A + (7B + 10.5C + 14D)/3 + 3B + (7A + 10.5C + 14D)/3 + 4C + (7A + 10.5B + 14D)/3 + 5D + (7A + 10.5B + 14C)/3 + 6D + (7A + 10.5B + 14C)/3}/6 = [37A + 54.5B + 71.5C + 89D]/18

FIRST MOVE: Finally, consider five (all) uncovered coins, A < B < C < D < E. The value of the game when R is applied to A, B, C, D or E, respectively, is given by:

A: V = RA + [37B + 54.5C + 71.5D + 89E]/18

B: W = RB + [37A + 54.5C + 71.5D + 89E]/18

C: X = RC + [37A + 54.5B + 71.5D + 89E]/18

D: Y = RD + [37A + 54.5B + 71.5C + 89E]/18

E: Z = RE + [37A + 54.5B + 71.5C + 89D]/18

Proceeding as above, the choice of coin A dominates when V>max(W,X,Y,Z); coin B dominates when W > max(V,X,Y,Z), and so forth. The algebra is tedious and repetitive, but because this is the first play, we know A = 1 cent, B = 5 cents, etc., so the conditions reduce to

A: V = $0.01R + $6.343

B: W = $0.05R + $6.261

C: X = $0.10R + $6.109

D: Y = $0.25R + $5.514

E: Z = $1.00R + $1.805

(Row) is greater than (column) when:

The optimal first moves follow directly from the table. Note that covering the dime first is optimal only when 3.03 < R < 3.97, i.e. never.

The overall value of the game is

{$0.01 + $6.343 + $0.01*2 + $6.343 + $0.05*3 + $6.261 + $0.25*4 + $5.514 + $1.00*5 + $1.805 + $1.00*6 + $1.805}/6

= $6.71

The minimum possible score, from rolling all ones, is $1.41. The maximum is $8.46. The value of a strategy of covering a coin at random, or equivalently of covering the coins in some predetermined sequence, is 3.5*1.41 = $4.93. Therefore, the optimal strategy yields about 36% more than the random or deterministic strategy. Using this strategy takes a player about halfway from the value of the random strategy to the value from rolling all 6's.