Is Monopoly Fair?

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Ever since we have purchased the Monopoly board game, it has become a weekend ritual for us. Almost every Friday/Saturday night Jo would pull out the board, currency, wooden dice, small houses and deed cards and spread them.

We are in for a surprise after playing the game for few weeks. As kids we thought the game is a GAME, ie random to high extent but fun. But as we develop an eye for the detail, the game does come out to be rather unfair. How else can you explain that one of us is paying rents through their noses to the other one owning reds and oranges and how else can you explain that the one who purchased properties next to GO seldom gets a visitor, including themselves. We started questioning, is monopoly fair? And thats where I set out to find it myself.

For starters, The monopoly board has 40 cells, and every player starts at GO. The moves are decided by sum of the faces on a 2 dice throw. If you get a double you get to throw again.

I have tried to simulate the game using excel. I have observed all the rules like community chest / chance cards, Jail and go to jail. But I have not observed doubles rule for I thought it had little impact on the outcome. I wanted to see if the expected probabilities of each cell (which is 1/40 or 2.5%) are close to the actual probabilities.

When I ran the simulation for 10000 dice throws for 4 players, the absolute difference between expected probabilities of each cell, color group are more or less near to the actual probabilities as you can see in the below charts. (click on them for bigger versions)

The maximum deviation is to the tune of 6% for individual cells and 1.93% for a color group (for there will be cancellations in color groups, as one cell gets more visitors the other would get less)

But that is not we experience when we play the game. We end up landing in Jail or a chance an awful lot of times more than we expect to land there. Thats because, no one ever plays a 10000 throw per person game, not in day to day versions. Its more like 200-400 throws. So when I ran the same experiment for 200 turns for 4 players, the results were more interesting as you can see them below.

As you can see the deviation in the actual probability and expected probability is huge. Some times 60% more for individual cells and 20% for color groups. This I guess explains the reason behind the Monopoly game strategies like buy anything in Orange and Red groups etc.

Obviously one fault of this experiment is if you run it again the actual probabilities are going to change in favor of something else. But almost always far from the expected results.

Write to me just in case you want to play around with my monopoly board game simulation excel sheet, I can mail it. The file is rather huge for upload.

Related links: Monopoly Wiki, Monopoly Fun Facts & Strategies, Similar Monopoly Simulations [1, 2]

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26 Responses to “Get busy this weekend, with OR XOR AND [Excel Homework]”

  1. Stef@n says:

    first solution for AND
    The two numbers are in A1 and B1
    = SUBSTITUTE (SUBSTITUTE (A1+B1*9*9, 9, 1), 8, 0)
    regards
    Stef@n

  2. Stef@n says:

    next solution for OR
    =1*SUBSTITUTE (A1+A2;2;1)

    regards
    Stef@n

  3. Stef@n says:

    last solution for XOR
    =1*SUBSTITUTE (A1+A2;2;0)

    regards
    Stef@n

  4. Irvine says:

    Or you could make use of the VBA logical operators!

    Define the following as custom functions
    Public Function BITXOR(x As Long, y As Long)
    BITXOR = x Xor y
    End Function

    Public Function BITAND(x As Long, y As Long)
    BITAND = x And y
    End Function

    Public Function BITOR(x As Long, y As Long)
    BITOR = x Or y
    End Function

    and then use them such:
    A B =BITOR(A,B) =BITAND(A,B) =BITXOR(A,B)
    0101 0100 0101 0100 0001

  5. Stef@n says:

    an another solution for AND
    =1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)

    note:
    the binary numbers are in A1 and A2 !
    regards
    Stef@n

  6. Shair says:

    I was obviously playing hooky at the beach during the bit-wise math lesson – you lost me at “Understanding bit-wise operations” 🙂

  7. Anup Agarwal says:

    After looking at the above solutions, I find my solution silly, but still:

    For the following formulae,
    Row 1: headers,
    Row 2: OR
    Row 3: AND
    Row 4: XOR

    Column 1: Input 1
    Column 2: Input 2
    Column 3: Result

    OR
    {=SUM(IF(MID(A2,ROW(OFFSET($A$1,0,0,LEN(A2),1)),1)+MID(B2,ROW(OFFSET($A$1,0,0,LEN(B2),1)),1)>0,1,0)*10^(LEN(A2)-ROW(OFFSET($A$1,0,0,LEN(B2),1))))}

    AND
    {=SUM(IF(MID(A3,ROW(OFFSET($A$1,0,0,LEN(A3),1)),1)+MID(B3,ROW(OFFSET($A$1,0,0,LEN(B3),1)),1)=2,1,0)*10^(LEN(A3)-ROW(OFFSET($A$1,0,0,LEN(B3),1))))}

    XOR
    {=SUM(IF(MID(A4,ROW(OFFSET($A$1,0,0,LEN(A4),1)),1)+MID(B4,ROW(OFFSET($A$1,0,0,LEN(B4),1)),1)=1,1,0)*10^(LEN(A4)-ROW(OFFSET($A$1,0,0,LEN(B4),1))))}

  8. Hui... says:

    @Anup
    Please don't consider your solution silly
    Firstly, You are the 3rd person to submit an answer
    Secondly, The best formula/function is the one that you know and understand.

  9. Michael Azer says:

    I think I have a very tedious solution, which people won't have the patience to do except in small numbers.

    I used the same problem setup as "Anup Agarwal"

    AND =IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)

    OR =IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)

    =IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)

  10. Michael Azer says:

    Sorry my last post was totally messed up

    AND
    =IF(AND(MID(B2,1,1)="1",MID(C2,1,1)="1"),1,0)&IF(AND(MID(B2,2,1)="1",MID(C2,2,1)="1"),1,0)&IF(AND(MID(B2,3,1)="1",MID(C2,3,1)="1"),1,0)&IF(AND(MID(B2,4,1)="1",MID(C2,4,1)="1"),1,0)
    OR
    =IF(OR(MID(B3,1,1)="1",MID(C3,1,1)="1"),1,0)&IF(OR(MID(B3,2,1)="1",MID(C3,2,1)="1"),1,0)&IF(OR(MID(B3,3,1)="1",MID(C3,3,1)="1"),1,0)&IF(OR(MID(B3,4,1)="1",MID(C3,4,1)="1"),1,0)
    XOR
    =IF(OR(AND(MID(B4,1,1)="1",MID(C4,1,1)="0"),AND(MID(B4,1,1)="0",MID(C4,1,1)="1")),1,0)&IF(OR(AND(MID(B4,2,1)="1",MID(C4,2,1)="0"),AND(MID(B4,2,1)="0",MID(C4,2,1)="1")),1,0)&IF(OR(AND(MID(B4,3,1)="1",MID(C4,3,1)="0"),AND(MID(B4,3,1)="0",MID(C4,3,1)="1")),1,0)&IF(OR(AND(MID(B4,4,1)="1",MID(C4,4,1)="0"),AND(MID(B4,4,1)="0",MID(C4,4,1)="1")),1,0)

  11. Stephen says:

    @stefan,
    I just couldn't get your solutions to work.
    01010101010 + 01010101110 = 02020210120
    what am i doing wrong?

  12. Stephen says:

    @anup
    ...I got yours to work!

  13. Rich says:

    @Stephen - I get the same, but Stef@an's second solution for AND does work (at least for the test cases I used)

  14. Stef@n says:

    @ Stephen / Rich
    yes , you are right ! - only this works:

    OR
    =1*SUBSTITUTE (A1+A2;2;1)
    XOR
    =1*SUBSTITUTE (A1+A2;2;0)
    AND
    =1*SUBSTITUTE (SUBSTITUTE (A1+A2;1;0);2;1)

  15. Michael Azer says:

    @Stef@n - You're answer is really smart, I never knew about the substitute function before. Great Work!

  16. Stef@n says:

    Thx Michael 🙂
    yes - it is simply easy 😉
    if you add 1 and 1 - excel calculate 2
    and then you have to substitute the 2 - new = 0 respectively 1

  17. Rob says:

    Here is a good resource for people wanting to learn binary and hexadecimal.
    http://justwebware.com/bitwise/bitwise.html

  18. Kyle McGhee says:

    Three that weren't asked for:
    NOT
    =SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),1,0),3,1)

    EQV
    =SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1+A2,0,3),2,3),1,0),3,1)

    IMP
    =SUBSTITUTE(SUBSTITUTE(A1+SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A2,0,3),1,0),3,1),0,1),2,0)

    (was using Daniel Ferry's bitwise file to verify against)

  19. Q says:

    @ Kyle

    Not only takes one parameter and inverts 0 -1 and 1-0

    Took out the +A2
    =SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,0,3),1,0),3,1)

  20. Great solutions!

    I'll add two:

    NAND =1*SUBSTITUTE (A1+A2,2,0)

    NOR=1*SUBSTITUTE(SUBSTITUTE (SUBSTITUTE(A1+A2,0,2),1,0),2,1)

  21. John Fuller says:

    This will work for binary numbers of any size (although the text format mask will have to have as many zeroes as there are digits in the longest addend)

    Assume binary #s are in C35 & C36, then add and format as text in C37:
    =TEXT(C36+C35,"000000000000")
    -sum- = 101112211112

    AND - SUBSTITUTE 0s for 1s in -sum-, then sub 1s for 2s
    =SUBSTITUTE(SUBSTITUTE(C37,"1","0"),"2","1")

    OR - sub 1s for 2s in -sum-
    =SUBSTITUTE(C37,"2","1")

    XOR - sub 0s for 2s in -sum-
    =SUBSTITUTE(C37,"2","0")

  22. Shane says:

    Just wandered by:

    AND:

    =SUBSTITUTE(A1+A2,1,0)/2

  23. Ronald Vonk says:

    Hi Chandoo,

    I am not (yet) really into bitwise calculation, but I am looking for a way to speed up my vba calculation with very big numbers. Would is ben convenient to use bitwise notation for this?

    Best regards,

    Ronald (the Netherlands)
    p.s. love your country!

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