Probability of landing on Reading Railroad on the first roll in Monopoly?

Question

5th space.

Doubles you roll again.

Chance is the 7th space.

One Chance card says to go to Reading Railroad.

One says to advance to GO.

There are (I think) 14 chance cards.

Chance is also the 22nd space.

FIRST TURN, NOT ROLL.

SORRY
.
.

AND THERE ARE 16 CARDS.

I'LL CHECK BACK TOMORROW.

Answer 1

You could get to Reading Railroad if you rolled a 5.
Possible good rolls: 1+4, 2+3, 3+2, 4+1
Total possible rolls of two dice = 6² = 36
Chance = 4/36 = 1/9

You could also get to Reading Railroad if you rolled a 7 and then drew the Go to Reading Railroad card.
Possible good rolls: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1
Chance = 6/36 = 1/6
But then you need to draw the right card (1/14)
So the chance of rolling 7 and getting the right chance card is:
1/6 * 1/14.

Putting it all together, the chance of getting to Reading Railroad on a direct roll, or an indirect roll via chance is:
1/9 + 1/6 * 1/14
=1/9 + 1/84

Common denominator is: 252
= 28/252 + 3/252 = 31/252
≈ 0.123015873
≈ 12.3%

Edit: Further research shows that there are indeed 16 Chance cards and 16 Community Chest cards. At first I thought the "Advance to Nearest Railroad" card or "Go Back 3" card could be useful, but they aren't. Advance means go forward, so it wouldn't help in this case for Reading Railroad.

Also, you could get doubles and get to either of these spots on the next roll. I forgot to account for that.
1/36 x 2/36 --> double ones, then roll a 3.
1/36 x 4/36 x 1/16 --> double ones, then roll a 5, then draw the right chance card.
1/36 x 2/36 x 1/16 --> double twos, then roll a 3, then draw the right chance card.
1/36 x 3/36 x 1/16 --> double sixes, then roll a 10, then draw the right chance card
1/36 x 1/36 x 1/16 --> double fives, then roll a 12, then draw the right chance card

In addition there is the chance that you roll double sixes, then double fives (to get to 2nd chance), draw the "Advance to go" then either roll a 5, or roll a 7 and draw the "Advance to Reading Railroad" (this time with a 1/15 chance).
Case 1: 1/36 x 1/36 x 1/16 x 4/36
Case 2: 1/36 x 1/36 x 1/16 x 6/36 x 1/15

Or you can do this with a double five, then double six.
Case 3: 1/36 x 1/36 x 1/16 x 4/36
Case 4: 1/36 x 1/36 x 1/16 x 6/36 x 1/15

Or you can do this with a double five, then double four, then a normal 6 (no doubles)
Case 5: 1/36 x 1/36 x 4/36 x 1/16

Or a double five, then double 3, then a normal 8 (no doubles)
Case 6: 1/36 x 1/36 x 4/36 x 1/16

Or a double five, then a double 2, then a normal 10 (no doubles)
Case 6: 1/36 x 1/36 x 2/36 x 1/16

Or a double four, then a double five, then a normal 4
Case 7: 1/36 x 1/36 x 2/36 x 1/16

Or a double four, then a double four, then a normal 6
Case 7: 1/36 x 1/36 x 4/36 x 1/16

Or a double four, then a double three, then a normal 8
Case 8: 1/36 x 1/36 x 4/36 x 1/16

Or a double four, then a double two, then a normal 10
Case 9: 1/36 x 1/36 x 2/36 x 1/16

Or a double three, then a double six, then a normal 4
Case 10: 1/36 x 1/36 x 2/36 x 1/16

Or a double three, then a double five, then a normal 6
Case 11: 1/36 x 1/36 x 4/36 x 1/16

Or a double three, then a double four, then a normal 8
Case 12: 1/36 x 1/36 x 4/36 x 1/16

Or a double three, then a double three, then a normal 10
Case 13: 1/36 x 1/36 x 2/36 x 1/16

Or a double two, then a double six, then a normal 4
Case 14: 1/36 x 1/36 x 2/36 x 1/16

Or a double two, then a double five, then a normal 6
Case 15: 1/36 x 1/36 x 4/36 x 1/16

Or a double two, then a double four, then a normal 8
Case 16: 1/36 x 1/36 x 4/36 x 1/16

Or a double two, then a double three, then a normal 10
Case 16: 1/36 x 1/36 x 2/36 x 1/16

Or a double one, then a double six, then a normal 8
Case 17: 1/36 x 1/36 x 2/36 x 1/16

Or a double one, then a double five, then a normal 10
Case 17: 1/36 x 1/36 x 2/36 x 1/16

But don't count getting double sixes 3 times (to get to the 3rd chance space). That will send you to jail

You've got the numbers, now you can multiply it all out.

Answer 2

Average throw when throwing two dice: 7 (2, 12 same prob. 1/36, each; 3,11 same prob, 2/36 each; 4,10 prob, 3/36, each; 5, 9 prob 4/36, each; 6, 8, prob 5/36 and 7 prob 6/36). Therefore, will land on Boardwalk, on average once every 7 trips. The point at which the odds are > 50 % that the player will have landed on Boardwalk is the 5th trip around the board at which point the odds will be about ~ 54%. (1- (6/7)^5) that the player will have landed at least once. (Note, after 7 trips, which is the consensus, the odds are only 66% that the player will have landed on Boardwalk) As for the probablity of landing on the first trip this is ~1/7 or 0.14286. By the time you have rolled the die 4-6 times, you are going to have a broad, flat distribution curve so that it is essentially random (i.e., 1/7) on whether the player will land on Boardwalk. (You could count up all the ways you could add up to 39 with 4, 5, 6 ... 19 rolls to figure the odds exactly.). This is confirmed by all of the computer simulations. >>>>>>>>>>>>>>>>>>>>> Edit: I really like Dr. D's "Alternate Mathmatical Approach." That is the best easy way to solve for landing on 39 the first time around. (Sure beats using a spreadsheet)

Answer 3

landing directly on reading railroad: 4/36

since chance is on the 7th you would not be rolling a double to get there...

landing on chance: 6/36 then picking up the reading railroad: 1/16

thus landing on reading on first roll:
4/36 + (6/36*1/16)


§
edit: it turns out that there are 16 chances and community chests