Well, I’ve come up with what I believe can be logically proven (SEE THE LINK TO MY PROOF ABOVE) as the theoretically shortest possible (in terms of total number of rolls) game of Monopoly (assuming no trading, quitting, or cheating of any kind). To be fair, it builds upon the work done by the folks over at http://scatter.wordpress.com, but it certainly is shorter and much more likely mathematically! That said, don’t hold your breath…it’s still pretty much impossible to ever occur in one of your home games.

Without further ado, here it is (note that there are several variations not covered by the example, but they are accounted for in my calculations). I leave it to the reader to figure out what they all are, or of course you can just message me:

Player 1, Turn 1:

Rolls 5,5 -> just visiting ($1500 liquid)

Rolls 6,6 -> Chance -> advances to nearest utility and purchases it ($1425 liquid)

Rolls 5,4 -> purchases park place ($1075 liquid**)

Player 2, turn 1:

Rolls 3,1 -> pays 10% income tax ($1350)

Player 1, Turn 2:

Rolls 1,1 -> purchases boardwalk ($675 liquid)

Rolls 2,1 -> com chest -> Bank error in your favor +200 ($1075 liquid)

Player 1 builds three houses on Boardwalk, two houses on Park Place ($75 liquid)

Player 2, Turn 2:

2,1 -> Advanced to Boardwalk (GAME OVER)

** Note that my definition of “liquid” includes mortgage value of Water Works as well as cash. It does NOT include mortgage value of park place and boardwalk, since we need to build on them.

__________________________________________________

According to my calculations, the probability of any seven-roll game of monopoly occuring (assuming that it can only be done with PP and BW) is to follow. BIG PROPS TO BLUE MONSTER for finding a way to make seven rolls work with the oranges!!! New post to follow:

(1/36)*(1/36)*(1/16)*(1/9)*(1/18)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/15) + (1/9)*(2/15)*(1/18)*(1/14)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/9)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/15) + (1/9)*(2/15)*(1/18)*(1/14)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/18)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/15) + (1/9)*(1/15)*(1/18)*(1/14)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/9)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/15) + (1/9)*(1/15)*(1/18)*(1/14)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(1/15)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/14) + (1/9)*(1/7)*(1/18)*(1/13)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(1/15)*(1/36)*[(1/18)*(1/8)*(1/18)*(1/14) + (1/9)*(1/14)*(1/18)*(1/13)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(2/15)*(1/36)*[(1/18)*(1/8)*(1/6)*(1/14) + (1/9)*(1/7)*(1/6)*(1/13)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(2/15)*(1/36)*[(1/18)*(1/8)*(1/6)*(1/14) + (1/9)*(1/14)*(1/6)*(1/13)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(1/15)*(1/36)*[(1/18)*(1/8)*(1/36)*(1/14) + (1/9)*(1/7)*(1/36)*(1/13)]

+

(1/36)*(1/36)*(1/16)*(1/9)*(1/6)*(1/15)*(1/36)*[(1/18)*(1/8)*(1/36)*(1/14) + (1/9)*(1/14)*(1/36)*(1/13)]

+

(1/6)*(1/16)*(1/36)*(1/18)*(1/15)*(1/36)*(1/14)*(1/16)*(1/36)*(1/18)*(1/13)*(1/18)

+

(1/6)*(1/16)*(1/36)*(1/18)*(1/15)*(1/36)*(1/14)*(1/36)*(1/13)*(1/16)*(1/18)*(1/18)

~=

184,804,377,911 to 1 |

So a bit better than 185 billion to 1. 🙂

PS – I would like to give a special thanks to my best friend and fellow Monopoly nerd, Stephen Priester, for helping me point out the “sneaky” first move possibility for player two of rolling a seven and going back three!! “The problem is dead, Jim.”

Nice work!

You also need to take into account the probability that the Chance cards come up in the order you describe.

Assuming there are 16 Chance cards, the probability the three cards you need are on top are:

1/(16 choose 3) = 1/560

And the probability those cards are in the right order are:

1/6

Combine this with your original calculation:

1 / (184,864,262,675 * 560 * 6)

or

1 in 621,143,922,588,000

See my reply below about how my calculations take the cards into effect and are certainly valid. I like how you approached permutations by backing out of combinations though…very cool.

1 / (16 Choose 3) = (3*2*1)/(16*15*14)

Dividing by 6 just gives 1/(16*15*14), which can be seen spread out throughout my formulas. I think I may do a new post that breaks down my numbers in detail.

Or, whoops, now that I look at it, your calculations already take the chance cards into account! Please feel free to remove my comment.

No problem! I know it’s a little messy, but everything has been accounted for in the calculations. I may do a follow-up post if there is demand for it proving that not only have we accounted for all possible 7-roll games, but also that it is impossible to do it shorter.

I’m sorry I haven’t been able to comment earlier–this is truly awesome! We used a couple of your approaches and came up with a couple of other 7-roll games.

No problem! I was starting to think you were ignoring us. 😉

Excellent job coming up with the seven-roll variation involving the oranges!! My previous post stands corrected: it is possible to do it with something other than BW and PP.

I’ve very interested in your logic that proves that 7 is the minimum. I’ve been drifting there myself, but I haven’t quite fully convinced myself yet (although by writing the below, I think I might have gotten there!).

I have convinced myself that the minimum number of rolls necessary to buy a color group is 4 (Orange or BW/PP can be purchased in 4 rolls–none others (I think). That being the case, the minimum number of turns is 2 per player. And to build, player 1 must get a color group, so player 1 has a minimum of 4 rolls, player 2 has a minimum of 2 rolls.

But, if Player 2 is to be bankrupted by orange, it will take landing on two properties since even with hotels, the Oranges can’t bankrupt you with one property (given that the most you can lose on one roll is $200), so if it’s Oranges, player two must make at least three rolls. Four Rolls for Player 1, three rolls for player 2 = 7.

If it’s BW/PP, Player one can buy them in 4 rolls, and Player 2 can get there in 2 rolls. So that’s six. But then it gets a little more complicated, because to buy BW/PP in 4 rolls, you either have to use up the Chance card to send you to Boardwalk (which will then cause Player 2 to use more than 2 rolls to get there) or you have be on Park Place and then roll a 1-1 to get Boardwalk. Since it is impossible to get to Park Place on less than three rolls (I think), that means the 1-1 is the fourth roll, and because it is doubles, you are forced roll once more. In either case, you’ve got to add at least one roll to the total, and therefore, 7 becomes the minimum.

Does that square with your logic?

Your Park Place and Boardwalk logic is spot-on with mine, but check out my newly-added proof page for the rest. The way I went about doing it may not be very pretty, but I believe the underlying logic to be relatively elegant.

Whoops, as in your second post, you can also buy the light purple (St. Charles) group in four rolls. But that produces BOTH the 1-1 doubles problem and the having to land on two big rents issue. So, that route keeps you at 8 rolls minimum, I think.

Yup! The parity of States works out just right to make it impossible to complete in seven.

This comment might go better under the 8-roll light purples post, but it is slightly relevant to the 4-roll purchase discussion above, so I hope you don’t mind that I put it here. You probably thought of this yourself already, so it may be kind of moot – if so, I’m sorry. But I’m kind of bored today, so I thought I’d post it anyway.

It is actually possible, using a current board that has only “pay $200” on the Income Tax square (not the 10% option as on the older boards) to get the light blue properties in 4 rolls for Player 1 and bankrupt Player 2 within 8 rolls, via the following sequence:

Player 1, Turn 1:

Rolls 3-3: Oriental Avenue, buy for $100 (Player 1 now has $1400)

Rolls 1-1: Vermont Avenue, buy for $100 (Player 1 now has $1300)

Rolls 6-3 (or any other combination to get 9): Community Chest, Advance to Go (Player 1 now has $1500)

Player 2, Turn 1:

Rolls 2-2: Income Tax, pay $200 (Player 2 now has $1300)

Rolls 2-1: Chance, Go Back 3 Spaces -> Income Tax, pay $200 (Player 2 now has $1100)

Player 1, Turn 2:

Rolls 6-3 (or any other combination to get 9): Connecticut Avenue, buy for $120; then put hotels on Oriental, Vermont, and Connecticut for $750 in building costs (Player 1 now has $630)

Player 2, Turn 2:

Rolls 1-1: Oriental Avenue, with hotel is $550 (Player 1 now has $1180, Player 2 now has $550)

Rolls 2-1: Connecticut Avenue, with hotel is $600 – GAME

As far as I can tell, the only variation on this rolling sequence is to have Player 1 buy Oriental and Connecticut on turn 1, then on turn 2 roll an 8 with doubles (to get to the “Advance to Go” Community Chest) followed by any 8 that’s not a double to complete the monopoly.

On the older boards, where paying 10% is allowed for income tax, the win can be achieved in 9 rolls. Assuming best play/smartest decisions, player 2 only pays a total of $285 on his/her two landings on income tax. Player 2’s second turn needs to be the following on an old gameboard for the quickest game possible involving the light blues:

Rolls 1-1: Oriental Avenue, $550 rent (Player 2’s cash reserves go from $1215 to $665)

Rolls 1-1: Vermont Avenue, $550 rent (Player 2 now has $115)

Rolls 6-3 (or 9 in any other combination): Community Chest, Pay School Tax of $150 – GAME

For me , it was interesting to note that the newer gameboards (without the pay 10% option on Income Tax) come with a set of Community Chest cards that do not include one that requires the player to pay $150 – the pay $150 has been replaced by one that only makes the player pay $50 instead (I read that on another blog). I guess the extra money lost by landing on Income Tax early in the game on a newer board was balanced slightly by the changed Community Chest card.

I think that of all the color-group monopolies to lose with quickly, this would be the most annoying/heartbreaking for me. I mean, I expect to lose quickly when my opponent gets 3 houses on Boardwalk before I get a monopoly of my own, and among the serious players I know, the oranges (and to a lesser extent the light purples) are respected as being powerful monopolies. But it’s so easy to overlook the light blues that it would be completely shocking (even humiliating) to lose in the method I described above, as far as I’m concerned.

On another note, I think it would be interesting to find out what the total odds would be of winning/losing the game (assuming best play) within 2 turns for each player (max 12 rolls if the doubles fall just right), or something like that. From your website and the other one, it’s clear that BW/PP, the oranges, the light purples, and now the light blues can cause a quick end. I’m pretty sure that I could come up with an 8-roll win with the reds, too, and maybe the yellows (but I’m not sure). However, I can’t quite come up with all of the combinations of rolls and Chance/Community Chest cards you used above when calculating all the ways to win with BW/PP in 7 rolls. Is there any way you could publish more information about the possible rolls and cards on your blog? I’d be interested to know what they are if you have time. 🙂

Thanks for your comment! I will be looking into it tonight if possible and will give a well-deserved response. 🙂