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

Leila Jennings

Answered question

2022-07-22

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

Answer & Explanation

Cael Cox

Cael Cox

Beginner2022-07-23Added 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 ..
Emmanuel Pace

Emmanuel Pace

Beginner2022-07-24Added 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.

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