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Probability in a Dice Game


Date: 01/04/2000 at 11:18:38
From: Leona
Subject: The Dice Game

Dear Dr. Math,

This problem is on probability and I have no clue as to how to do it. 
Please help.

THE DICE GAME

Three Players A, B and C play a game with a single die. The rules of 
the game are:

Player A ALWAYS goes first.

A rolls the die. If the die lands showing a 1 then A wins the game. 
If A does not throw a 1 then B has a turn.

B rolls a die. If the die landing shows a 2 or 3 then B wins the 
game. If B does not throw a 2 or a 3 then C has a turn.

C rolls the die. If the die lands showing a 4, 5 or 6 then C wins 
the game. If C does not throw a 4 or a 5 or a 6 then A starts again.

This procedure continues until there is a winner.

Investigate any or all of:

1. The probabilities of each of A, B or C winning the game.

2. Who will be the most likely winner?

3. The most likely length of the game in terms of the number of rolls 
of the die to produce the winner 

Leona


Date: 01/13/2000 at 12:53:40
From: Doctor TWE
Subject: Re: The Dice Game

Hi Leona - thanks for writing to Dr. Math.

To show how to solve this problem, let's look at a similar but simpler 
game. Suppose we have a two player game (A and B), where A rolls first 
and wins on a 1 or 2. If A fails to win, B rolls and wins on a 3, 4, 5 
or 6. If neither has won, the game continues with A rolling with the 
same rules.

Let's investigate what happens one event at a time, and look at the 
probabilities. First, A rolls the die. There is a 2/6 = 1/3 
probability that A will win on the roll (a roll of a 1 or 2), and a 
2/3 probability that the game will continue.

     P(A wins) = 1/3
     P(cont.)  = 2/3
     ---------------
     Total     = 3/3 = 1   (that's good - the total should always = 1)

If A didn't win, then B rolls the die. There is a 2/3 chance that B 
will win on that roll (a roll of a 3, 4, 5 or 6) - but this is only 
2/3 of the 2/3 chance that B even gets to roll. There is also a 1/3 of 
2/3 chance that the game will continue.

     P(A wins) = 1/3       = 3/9   (I converted to 9ths for addition)
     P(B wins) = 2/3 * 2/3 = 4/9
     P(cont.)  = 2/3 * 1/3 = 2/9
     ---------------------------
     Total                 = 9/9 = 1

At this point, the game repeats itself, so let's look at what we have 
so far. There is a 7/9 chance that the game will be over, and a 2/9 
chance that the game will continue. A wins 1/3 of the time, which is 
(1/3)/(7/9) of the combinations where the game has ended. Likewise, B 
wins 4/9 of the time, which is (4/9)/(7/9) of the combinations where 
the game has ended. In other words:

     P(A wins|game ended) = (1/3) / (7/9) = 3/7
     P(B wins|game ended) = (4/9) / (7/9) = 4/7
     -------------------------------------------
     Total                                = 7/7 = 1

Note that P(A wins|game ended) is read as "the probability that A wins 
given that the game has ended". Since the game repeats itself at this 
point, the odds of A and B winning don't change. We can think of this 
as a "draw" and start a new game.

Another way to look at this is that A wins in round one 1/3 of the 
time, plus A wins in round two 1/3 of the 2/9 of the time that neither 
player won in round one, plus...

Similarly B wins in round one 2/3 of the 2/3 of the time that A didn't 
win, plus B wins in round two 2/3 of the 2/3 of the 2/9 of the time 
that neither player won in round one, plus...

   P(A) = 1/3 + (1/3)*(2/9) + (1/3)*(2/9)^2 + (1/3)*(2/9)^3 + ...
   P(B) = (2/3)*(2/3) + (2/3)*(2/3)*(2/9) + (2/3)*(2/3)*(2/9)^2 + ...

These are geometric series. The formula for the sum of a geometric 
series is

     S = k[1/(1-r)]

where k is the coefficient of each term and r is the ratio of the 
terms. P(A) has a coefficient of 1/3 and a ratio of 2/9. P(B) has a 
coefficient of (2/3)*(2/3) = 4/9 and a ratio of 2/9. So we get:

     P(A) = (1/3)[1/(1 - 2/9)] = (1/3)(9/7) = 3/7
     P(B) = (4/9)[1/(1 - 2/9)] = (4/9)(9/7) = 4/7

Which is what we got before. In this game, B is more likely to win.

Your last question was what is the most likely length of the game in 
terms of the number of rolls of the dice to produce the winner. In our 
game, there is a 1/3 chance that the game will go exactly one roll, a 
4/9 chance that the game will go exactly two rolls, and a 2/9 chance 
that it will go more than two rolls. Since the chance of it going 3, 
4, 5, etc. rolls is non-zero (and can't be negative), we know that 
each of these probabilities has to be less than 2/9. So the most 
likely number of rolls for our game is 2. Similar logic will work for 
your game.

I hope I have explained it well enough for you to apply to your game. 
If not, or if you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Probability

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