Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that J

Leila Jennings 2022-07-22 Answered
Geometric series with probability
Archeological evidence indicates that the ancient (pre-Roman) Etruscans played dice using a dodecahedral die having 12 pentagonal faces numbered 1 through 12 (figure above). One could simulate such a die by drawing a random card from a deck of cards numbered 1 through k. for your own personal value of k, begin with the largest digit in the sum of the digits in your student ID number. This is your value of k unless this digit is less than 5, in which case subtract it from 10 to get your own value of k. a.) John and Mary draw alternately from a deck of shuffled k cards. The first one to draw an ace-the card numbered one- wins. Assume that John draws first. Use the formula for the sum of a geometric series to calculate (both a rational number and a four place decimal) the probability J that John wins, and the probability M that Mary wins. Check that M + J = 1. b.) Now John , Mary, Paul draw alternately from a deck of k cards. Calculate separately their respective probabilities of winning, given that John draws first and Mary draw second. Check that J + M + P = 1
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Cael Cox
Answered 2022-07-23 Author has 11 answers
Step 1
We look at the three-person game, since it is more interesting. We also assume that the deck is shuffled between draws, in order to simulate tossing a fair k-sided die. You will find it easy to adapt the idea to the simpler 2-person game, and if you wish, to a d-person game.
Let our players be A, B, and C. We suppose A tosses first, then (if necessary) B, then, if necessary, C, then, if necessary A, and so on.
To make the notation simpler, let p = 1 k .
Player A can win on the first draw, the fourth, the seventh, the tenth, and so on.The probability she wins on the first draw is p.
In order for A to win on the fourth draw, A, B, and C must all fail to win on their first three draws, and then a must win. The probability of this is ( 1 p ) 3 p.
In order to win on the seventh draw, A, B, C must all fail twice, and then A must win. This has probability ( 1 p ) 6 p.
Step 2
Similarly, the probability A achieves her win on the tenth draw is ( 1 p ) 9 p. The probability A achieves her win on the thirteenth draw is ( 1 p ) 12 p. And so on.
So the probability that A wins is p + ( 1 p ) 3 p + ( 1 p ) 6 p + ( 1 p ) 9 p + ( 1 p ) 12 p + .
The above is an infinite geometric series. Recall that the infinite geometric series a + a r + a r 2 + a r 3 + has sum a 1 r (if | r | < 1). In our case, a = p and r = ( 1 p ) 3 , so the probability A wins is p 1 ( 1 p ) 3 ..
This can be "simplified" to the less attractive expression 1 3 3 p + p 2 .
We could go through a very similar calculation for the probability that B wins. However, there is a shortcut. Suppose that A fails to win on her first throw (probability 1 p). Then effectively B is now first, so has probability of winning p 1 ( 1 p ) 3 . So the probability B wins is ( 1 p ) p 1 ( 1 p ) 3 ..
A similar argument shows that the probability C wins is ( 1 p ) 2 p 1 ( 1 p ) 3 ..
Did you like this example?
Subscribe for all access
Emmanuel Pace
Answered 2022-07-24 Author has 6 answers
Step 1
We can avoid summing an infinite series. Let a be the probability that A ultimately wins. As discussed earlier, the way for B to win is for A to fail on her first draw. Then effectively B is the first player. So the probability B is the ultimate winner is ( 1 p ) a. Similarly, the probability C is the ultimate winner is ( 1 p ) 2 a.
Step 2
But it is (almost)clear that someone must ultimately win, the probability the game goes on forever is 0. It follows that a + ( 1 p ) a + ( 1 p ) 2 a = 1 ,, and therefore a = 1 1 + ( 1 p ) + ( 1 p ) 2 ..
A little calculation shows that this is the same answer as the one obtained earlier.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-08-11
The number of Bernoulli trials required to produce exactly 1 success and at least 1 failure.
Let X be the number of Bernoulli (p) trials required to produce exactly 1 success and at least 1 failure. Find the distribution of X.
How can I answer this question if I don't know how many trials have taken place?
For context, the section this comes from was about Poisson distribution.
( n k ) p k ( 1 p ) ( n k ) e μ μ k k !
as  n  and  p 0  with  n p = μ
However I read in a different book that you can use the Geometric probability model for Bernoulli trials: Geom(p)
P ( X = x ) = q X 1 P
where p is prob. of success, X is number of trials, and q is prob. of failure
asked 2022-08-20
If X is a nonnegative σ-subGaussian random variable with P ( X = 0 ) p, what is a good upper bound for P ( X h )?
asked 2022-07-14
Probability X is odd in a geometric distribution
Let X be a Geometric distribution with parameter p = 1 10 .
asked 2022-11-17
Negative Binomial Vs Geometric
So I am trying to get the difference between these two distributions. I think I understand them but the negative binomial has me a bit confused.
The Geometric is the probability of some amount of successes before the first failure.
The Negative binomial is the portability of some amount of successes before a specified number of failures? for example the number of successes before the 8th failure?
asked 2022-08-14
Geometric or binomial distribution?
A monkey is sitting at a simplified keyboard that only includes the keys "a", "b", and "c". The monkey presses the keys at random. Let X be the number of keys pressed until the money has passed all the different keys at least once. For example, if the monkey typed "accaacbcaaac.." then X would equal 7 whereas if the money typed "cbaccaabbcab.." then X would equal 3.
a.) What is the probability X 10?
b.) Prove that for an random variable Z taking values in the range {1,2,3,...}, E ( Z ) = Summation from i = 1 to infinity of P ( Z i ).
c.) What's the expected value of X?
First, is this a binomial distribution or a geometric distribution? I believe it is a binomial but my other friends says that it is geometric. As for the questions above, for a can I just do 1 P ( X = 9 )   or   1 P ( X < 9 ), but I don't know how I will calculate X < 9, I would know how to calculate P ( X = 9 ), I don't know how to do b or c.
asked 2022-10-21
Probability of two successes, possible geometric distribution problem
Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula P ( Y = k ) = ( k 1 ) ( p 2 ) ( 1 p ) k 2 .
Okay I know the outcome space is Y := 2 , 3 , 4 , . . . , the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra ( k 1 ) p in front.
asked 2022-08-10
A problem on geometrical probability
Two points P and Q are taken at random on a straight line OA of length a,show that the chance that P Q > b, where b < a is ( a b a ) 2

New questions