Problem 084

[JamieCamardelle]

I am trying to do problem 84, and my first step is seeing how to obtain probability 6.24% of landing in jail with 6-sided dice. I thought my reasoning was very thorough yet I did not get the right answer (I got 6.34273%). Can anyone give me a hint on where I went wrong here? I put in comments so you can follow my reasoning. Is my assumption that they could have been in any square previously with the same likelihood an error? That assumption doesn't sit well with me, but I can't figure any other starting point.

(* Chance 1 and Community Chest 1 each have probability 1/16 of pulling Go To \
Jail.
In addition, the probability of rolling a third double (1/216) and the \
probability of not
rolling a third double but just visiting because of the dice outcome \
are accounted for here
From CH1, the probability of rolling 3 is 2/36.
From CC1, the probability of rolling 8 is 5/36 *)

CH1 = 1/16 + (15/16)*(1/216 + (215/216)*(2/36));
CC1 = 1/16 + (15/16)*(1/216 + (215/216)*(5/36));

(* strikerange are the squares within landing distance by dice. It
includes squares B2, B1, R1, T1, A2, A1, GO, H2, and T2. It does not \
include CH1 or CC1 because those probabilities are already listed above *)

strikerange = (215/216)*(1 + 3 + 4 + 5 + 6 + 4 + 3 + 2 + 1)/36 + 9/216;

(* You are certain to go to jail if you came from G2J *)

G2J = 1;

(* There are six CC and CH
squares combined, and two
of them have been accounted for,
leaving the four that are
not within striking distance. Their probabilities are all \
the same *)
CHCC = 4*(1/16 + (15/16)*(1/216));

(* Now 16 squares are accounted for, leaving 24 where the only chance of \
landing in jail is rolling a third double. *)
DUBS3 = 24/216;

(* Now the sum of the probabilities is divided by 40 (assuming each square \
had the same probability of being occupied before the roll *)
N[(
DUBS3 + CHCC + G2J + CH1 + CC1 + strikerange)/40]

[daniel.is.fischer]

Well, not all squares are equally probable to be the starting point on that turn, e.g. G2J is never a starting point.
The probability to end your turn in jail is the sum of the probability of ending your turn in jail given that you start on square s times the probability of having ended your previous turn on square s. The different probabilities for having ended your previous turn on different squares account for the difference of 6.24% and 6.34%.

[JamieCamardelle]

Thanks for your reply. I am still trying to wrap my mind around this one, but I don't give up easily. I guess the initial starting point has to be GO, and there is some kind of recursion involved. I am not familiar with recursion techniques. Perhaps this problem will make me learn.

[daniel.is.fischer]

Although the initial starting point is indeed GO, that is immaterial. After a few dozen turns, the probabilities are distributed over the whole board and we consider the limit distribution after an infinite number of turns. If you do it via recursion, you can take any starting point and will always approach the same limit.

[JamieCamardelle]

Your reply is very helpful - I had begun to suspect so about recursion being unneccessary (maybe by the method-of-wishful-thinking).
I made some changes based on what you said, and got probability 06.2386% now, but as my thinking is fuzzy I do not trust this figure even though it rounds to 6.4%
The changes I made were:
- the probabilities, of landing on CC or CH and not pulling GoToJail are 14/16 and 5/16 respectively
- DUBS3 changed from 24/216 to 23/216
- the resulting sum is divided by 39 instead of 40

I need to go away from this problem for awhile and come back when my thinking is more fresh. Thank you very much for your help.

[btilly]

First a note. I don't think I've played monopoly since the 80s, so I don't remember the rules at all. While most can rely on their knowing the game, I can't, and the word description offered was not entirely clear to me.

What are the rules with doubles? I remember something about going again if you have a double. Do we count only the last square you land on after a sequence of doubles (with that possibly being jail), or do we also count the intermediate squares landed on? If we count the intermediate ones, then does being sent to jail "reset" the doubles counter or not? (In other words can you roll a double, roll a second double and get sent to jail by where you landed, then, roll a double, and remain in jail?) If it does not reset, then you go back to jail on the 3rd, 6th, 9th, etc doubles, right? (Your 4th doubles technically is a third double in a row, but I'm pretty sure that's not how I should be reading the instructions.) If we don't count the intermediate squares, do we do all of the special stuff for intermediate squares? (eg If you land on Go To Jail, then go to jail.)

And a minor technical point. If I land on square 36 (CH3), then get told to go back 3 spaces to CC3, do I immediately draw a community card and possibly get sent to Go or Jail? I doubt that my handling of this square will change the answer, but I'd like to get it right.

Thanks.

[daniel.is.fischer]

First a note. I don't think I've played monopoly since the 80s, so I don't remember the rules at all.

It's not that long since my nieces forced me to play it, but they made up their own rules, so I don't remember them well either.

While most can rely on their knowing the game, I can't, and the word description offered was not entirely clear to me.

What are the rules with doubles? I remember something about going again if you have a double.

Yes, check here: "A player who rolls doubles takes another turn after completing the first one (unless he/she was in Jail). If he rolls doubles again, he takes a third turn after completing the second. If, on the third turn, he rolls doubles again, he does not take that turn and goes to jail."

Do we count only the last square you land on after a sequence of doubles (with that possibly being jail), or do we also count the intermediate squares landed on?

Also the intermediates, those are separate turns, they're just not separated by other players having their turn in between.

If we count the intermediate ones, then does being sent to jail "reset" the doubles counter or not? (In other words can you roll a double, roll a second double and get sent to jail by where you landed, then, roll a double, and remain in jail?) If it does not reset, then you go back to jail on the 3rd, 6th, 9th, etc doubles, right? (Your 4th doubles technically is a third double in a row, but I'm pretty sure that's not how I should be reading the instructions.)

AFAIK, if you're sent to jail, that's it, doubles counter reset, next player.

If we don't count the intermediate squares, do we do all of the special stuff for intermediate squares? (eg If you land on Go To Jail, then go to jail.)

And a minor technical point. If I land on square 36 (CH3), then get told to go back 3 spaces to CC3, do I immediately draw a community card and possibly get sent to Go or Jail?

Yes, whenever you land on a CC/CH square you draw a card.

I doubt that my handling of this square will change the answer, but I'd like to get it right.

Thanks.

[btilly]

I finally got around to solving it according to the clarification in viewtopic.php?f=5&t=1130. My solution says that after 2^50 goes around the board for 6-sided dice the probabilities of the 3 most popular squares are 6.21951468197554% for JAIL, 3.18576628665498% for E3 and 3.09612303341043% for GO. Just to verify that those probabilities are stable I calculated the answer after 2^100 goes around and found the answers unchanged. Rounding to 2 decimal places that wold be 6.22%, 3.19% and 3.10% respectively, which is slightly different from the 6.24%, 3.18% and 3.09% listed in the problem statement.

Reading the thread, I used the same strategy as Alvero but got different answers. However my answers are in agreement with http://www.durangobill.com/MnplyDiceRoll.html. The probable cause of the discrepancy is that I (in accord with the rules of the game) reset the doubles counter if you are sent to jail, while I suspect that Alvero does not. This would mean that if you rolled, say, double 1s, got sent to jail, then double 1 and double 1 then I would put you at square C3 while Alvero would put you back in jail. This change reduces the odds of jail slightly, and increases the odds of other squares.

I am posting my (admittedly ugly) code to the thread in case any admins wish to inspect it to investigate my claim.

[RedNine]

I seem to recall from my Monopoly days that if one rolls a double and lands on Community Chest or Chance, one rolls again (or goes to Jail if that was the third double) without stopping / drawing a card, but that if one lands on Go To Jail on a double, one goes to Jail immediately without any further rolls. The problem doesn't seem to say anything about any exceptions to the "roll again after a first or second consecutive double" rule, so I'd just wonder what rule I should assume in my attempt to solve the problem.

Thanks in advance!

[spiketheimpaler]

There's no rule about rolling double and landing on Chance/Community Chest. The only doubles rules are the "3 in a row sends you to jail" rule and the doubles to get out of Jail, both of which are ignored for this problem.

[RedNine]

Not sure what you mean. The problem states that the "double to get out of jail" rule is ignored for this problem, but that the "3 consecutive doubles sends you to jail" rule is still in place. Anyway, my issue was with what happens when I land on Go To Jail with a double. In that scenario, would I roll again (as per the normal rule for doubles) or just go to jail (as per the normal rule for that square)? The thing is just that there're two conflicting rules for what to do, so I'd like to know which prevails in that instance.

[RedNine]

Now that I reread the problem, it doesn't even mention the "go again if you rolled a double" rule. Should I just be ignoring doubles altogether?

[DJohn]

You should ignore all of the rules that you think you know from the real game, and use only the rules stated in the problem description. I think it's all covered there. In particular, ignore all doubles except for the "three doubles go to jail" rule.

The real game does have a "go again after a double" rule, but since that has no effect on the squares you visit (just the order in which different players visit them), it can be ignored. There's nothing about a double landing on Chance or Community Chest. That must have been a house rule.

[RedNine]

Thank you. The problem clearly states that we care about the final square visited on each *roll*, but for some reason I misread it and was trying to find the final square visited on each *turn* - i.e. if I start on GO and roll double 4s, I don't count square 08 as having been landed on because I move off it before my turn ends. So, thanks for your help in seeing what I should have understood already.

[MaJJ]

What are doubles exactly? Only 6 and 6, or both dice the same (1 and 1, 2 and 2, all the way to 6 and 6)?

[daniel.is.fischer]

Both dice showing the same, so there are six doubles.

[dconrad]

Well, I managed to solve this problem, but I cannot get my results to match up to everyone else's. For 6-sided dice my numbers are close enough, but for 4-sided dice I get different results, consistently. Jail is still up there, but the next three are very close: a property that isn't part of the answer, and a railroad and a property that are part of the answer. (I'm being intentionally cagey due to the "don't post any results" rule.)

The second through fourth items are so close together in value that I surmised that perhaps they should be in a slightly different order, and after a guess or two, got the correct answer. But I wish I knew why my numbers are off.

My solution is a Monte Carlo simulation in Java, and I modeled everything in detail, with Card and Deck and Square and Board classes, etc. As a result it's rather longer than most of the solutions, so I don't know if anyone would want to look at it to see what I might have done wrong, but on the other hand it's really explicit and straightforward about the rules, so it's rather hard to see what I could have done wrong.

Well, if anyone would like a look, reply or PM me. Otherwise, I guess I'll just keep scratching my head.

[rayfil]

Let's hope you have been rolling the two dice individually. Their total is not evenly distributed if you should be getting a random number for the total instead of a random number for each die.

[dconrad]

Let's hope you have been rolling the two dice individually. Their total is not evenly distributed if you should be getting a random number for the total instead of a random number for each die.

I do roll the dice individually, but I guess I could check on the probability distribution that java.util.Random's nextInt(n) method is giving me. Thanks, rayfil.

[thedoctar]

I've just finished this problem! 2.5 seconds

However, when debugging my program I found when simulating 300,000 dice rolls (for a six sided dice), I would get the correct answer, Square 10 = 6.27%, Square 24 = 3.18%, Square 0 = 3.11%. But when I simulated through 3,000,000 dice rolls, I would always get the answer, Square 10 = 6.25%, Square 24 = 3.19%, Square 19 = 3.09%

I used normal python when simulating 300,000 rolls, but I used pypy when simulating through 3,000,000. Could the reason for the two different answers because of pypy, or maybe because when generating random integers over a large range, the random integer generator in python stops being as random? E.g. when simulating over 300,000 rolls, the integers seem random, but over 3,000,000 rolls, the integers start to follow a pattern, as they are only psuedo-random.

Also, is it possible to solve this problem by hand? I tried to initially, but I couldn't be bothered.