Monopoly RulesQuote:WizardDo they publish a tally of how many of each piece was made? It would make for a good math problem on the expected number of tickets you need to buy to complete any given set.

If you scroll down to #6, it states that they plan to make approximately 513,591,720 total game pieces. Further down, it lists how many of each prize are available.

There are 513,591,720 pieces

Odds of winning Boardwalk

One (1) prize is available in the Game. 1 in 513,591,720

Odds of the 5k through 50k There are 5 of each winning piece

Five (5) prizes are available in the Game

the approximate odds of collecting the Winning Combination (Pacific Avenue, North Carolina Avenue and Pennsylvania Avenue) are 1 in 513,159,085

Odds can't be near the same as the boardwalk, can they?

So, to help simplify this I created a "weight" column that shows how many of each property/food I believe to have been printed.

Piece | Inverse | Weight |
---|---|---|

Mediterranean | 11 | 46,690,156 |

Baltic | 114,132 | 4,500 |

Oriental | 11 | 46,690,156 |

Vermont | 51,359,172 | 10 |

Connecticut | 11 | 46,690,156 |

St. Charles Place | 11 | 46,690,156 |

States | 873,498,405 | 1 |

Virginia | 11 | 46,690,156 |

St. James Place | 10 | 51,359,172 |

Tennessee | 102,718,344 | 5 |

New York | 10 | 51,359,172 |

Kentucky | 102,718,344 | 5 |

Indiana | 10 | 51,359,172 |

Illinois | 10 | 51,359,172 |

Atlantic | 10 | 51,359,172 |

Ventnor | 102,718,344 | 5 |

Marvin Gardens | 10 | 51,359,172 |

Pacific | 10 | 51,359,172 |

North Carolina | 10 | 51,359,172 |

Pennsylvania | 513,591,720 | 1 |

Park Place | 11 | 46,690,156 |

Boardwalk | 513,591,720 | 1 |

Reading | 11 | 46,690,156 |

Pennsylvania Railroad | 11 | 46,690,156 |

B&O | 11 | 46,690,156 |

Short Line | 2,567,959 | 200 |

Food | 2.619110761 | 196,093,929 |

Total | 1,027,183,437 |

Here are the prizes for each set, listed by the most expensive property in each set.

Set | Prize |
---|---|

Baltic | $50 |

Connecticut | $1,000 |

Virginia | $2,000 |

New York | $5,000 |

Illinois | $10,000 |

Marvin Gardens | $25,000 |

Pennsylvania Ave. | $50,000 |

Boardwalk | $1,000,000 |

Railroads | $500 |

I show the expected value of 1,000 cards (or 2,000 properties) to be $3.09. In other words each card (of two properties) is worth 0.3 cents.

Quote:GWAEI am confused. Maybe I am dumb or they just screwed up.

There are 513,591,720 pieces

Odds of winning Boardwalk

One (1) prize is available in the Game. 1 in 513,591,720

Odds of the 5k through 50k There are 5 of each winning piece

Five (5) prizes are available in the Game

the approximate odds of collecting the Winning Combination (Pacific Avenue, North Carolina Avenue and Pennsylvania Avenue) are 1 in 513,159,085

Odds can't be near the same as the boardwalk, can they?

The odds of getting the set of pieces can.

Note that each "game piece" has 2 stamps (a property or one of the "instant win" prizes ($25, $50, or food)).

The probability of getting Boardwalk is 1 in 513,591,720.

The probability of getting both Boardwalk and Park Place (and thus winning the million dollars) is also 1 in 513,591,720, which I interpret as being possible only if the game piece with the Boardwalk stamp also has a Park Place stamp.

On the other hand, the probability of getting one of the five Pennsylvania Avenue stamps is 1 in 102,631,817. Apparently, somehow they figure the probability of getting both North Carolina Avenue and Pacific Avenue is 1 in 5.

Now there's a problem: if 1/10 of the game pieces have Pacific Avenue, 1/10 have North Carolina Avenue, and no game pieces have both, how many game pieces do you need to have a 1/5 chance of having them both?

Assume you have a bowl with 10 balls - one says Pacific, one says North Carolina, and the other eight are blank. The question is, for what value N does the probability that N draws with replacement will result in at least one Pacific and one North Carolina = 1/5?

If there are N draws, there are 10

^{N}possible sets of draws. 9

^{N}have no Pacifics drawn, and 9

^{N}have no North Carolinas drawn, but this counts the 8

^{N}draws that have neither Pacific nor North Carolina twice.

The total number of draws that do not have both = 9

^{N}+ 9

^{N}- 8

^{N}

The total number that do have both = 10

^{N}- (9

^{N}+ 9

^{N}- 8

^{N}) = 10

^{N}- 2 x 9

^{N}+ 8

^{N}

The probability of getting a draw with both = (10

^{N}- 2 x 9

^{N}+ 8

^{N}) / 10

^{N}

If this is 1/5, then 10

^{N}- 2 x 9

^{N}+ 8

^{N}=10

^{N}/ 5 = 2 x 10

^{N-1}

2 x 9

^{N}- 8

^{N}= 8 x 10

^{N-1}

Solve for N: 2 x 9

^{N}- 8

^{N}- 8 x 10

^{N-1}= 0

Using Newton-Raphson, I get N = just over 6 as the only positive solution

Quote:ThatDonGuyNow there's a problem: if 1/10 of the game pieces have Pacific Avenue, 1/10 have North Carolina Avenue, and no game pieces have both, how many game pieces do you need to have a 1/5 chance of having them both?

Note: Answer corrected.

Quote:WizardI show with 4 pieces will have a 14.67% of having both. 5 pieces is 19.93%.

Are you sure?

I get 5 pieces having 14.67% and 6 having 19.9262%

What do you get with 2 pieces? It should be 2% (either Pacific and then North Carolina, or the other way around)

Quote:ThatDonGuyAre you sure?

Well, unlike Donald Trump, I've been known to be wrong before. Here is a table showing my probability of having a set by number of cards purchased. Maybe that will help us find our point of departure.

Pieces | Probability Set |

2 | 0.020000 |

3 | 0.054000 |

4 | 0.097400 |

5 | 0.146700 |

6 | 0.199262 |

7 | 0.253121 |

8 | 0.306838 |

9 | 0.359377 |

10 | 0.410017 |

11 | 0.458278 |

12 | 0.503860 |

13 | 0.546602 |

14 | 0.586445 |

15 | 0.623402 |

16 | 0.657543 |

17 | 0.688974 |

18 | 0.717825 |

19 | 0.744241 |

20 | 0.768376 |

Quote:WizardWell, unlike Donald Trump, I've been known to be wrong before. Here is a table showing my probability of having a set by number of cards purchased. Maybe that will help us find our point of departure.

Pieces Probability Set 2 0.020000 3 0.054000 4 0.097400 5 0.146700 6 0.199262 7 0.253121 8 0.306838 9 0.359377 10 0.410017 11 0.458278 12 0.503860 13 0.546602 14 0.586445 15 0.623402 16 0.657543 17 0.688974 18 0.717825 19 0.744241 20 0.768376

Your table shows 19.9262% for six, which is what I said, whereas you said five - unless by "five", you meant how many you needed after the first piece