Tuesday, August 20, 2019

Monopoly PR2: Throwing Doubles and Probability of Turn Outcomes.

What is the probability of throwing doubles?

Each turn of monopoly starts with a roll of two dice.   If the player rolls doubles (the same value on each dice) the person rolls again.   If the player rolls three doubles in a row the person goes directly to jail.   The player who rolls three doubles in a row does not get to buy a property or use a community chest or chance card.

The monopoly turn ends after the first roll of the dice if the dice roll is not doubles and after the third roll of the dice regardless of the dice roll outcomes.

Question one:  The Chance Square exists seven squares away from GO.  What is the probability that a player starts as Go and lands on Chance on the first turn of the game?


The person can roll a seven on the first roll or roll a double go again and hope the sum of all rolls adds to seven.   One also can get to chance on three rolls.

As noted in a previous post 7 is the most common outcome from the sum of two dice rolls.   There are six combos of two dice rolls that add to 7 – (1,6), (6, 1), (2,5), (5,2), (3,4), and (4,3).   Each of these outcomes has probability 1/36.   The sum of the six mutually exclusive outcomes is 1/6 or 0.16667.

The person can also reach 7 by throwing a (1,1) or a (2,2) on the first turn and throwing and taking a second turn.   (Note (3,3) even though sum is less than 7 is not an option.  Explain why.)

To get to seven on two throws of the dice after throwing (1,1) one must throw (4,1) (1,4), (2,3) or (3,2).    The probability of this occurring is 4/36 or 0.11111.  The probability of getting to seven after two throws of the dice after throwing (1,1) on the first throw of the dice is probability of throwing (1,1) (1/36) multiplied by probability of getting dice rolls to sum to 5 (4/36).  This product is (1/(9x36)) or 1/324 or 0.00308642.

To get to seven after throwing (2,2) on two throws of the dice one must throw (1,2) or (2,1).   The probability of throwing (2.2) is 1/36.   The probability of throwing dice that sum to 3 is 1/18.   The product of both occurring is the product of the two or (1/(36x18)) or 1/648 or 0.00154321.

There is one other way to get to seven on one turn.  Throw (1,1) on the first turn, (1,1) on the second turn and dice that sum to 3 on the third turn.   The probability of this sequence happening is (1/36) x (1/36) x (1/18) 1/23328  or 0.0000428669.

The probability of getting to seven on the first turn is the sum of the probabilities of all the ways to get to seven, which is 0.171339163

More monopoly probability problems will follow.

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