Question

Amanda, Becca, and Charise toss a coin in sequence until one person “wins” by tossing the first head. a) If the coin is fair, find the probability that Amanda wins. b) If the coin is fair, find the probability that Becca wins. c) If the coin is fair, find the probability that Charise wins. d) The coin is no longer fair. The probability that the coin comes up heads on an individual toss is p, for 0<p<1. Plot each players probability of winning for 0<p<1 on a single set of axes.

Answer 1

We first solve (d) and then putting p=0.5 (in case of fair coin) we can obtain results corresponding to (a), (b) and (c) easily.

(d)

To Amanda become winner first head should appear in any of 1st, (1+3)=4th, (4+3)=7th, (7+3)=10th, (10+3)=13th draw or so on.

So, required probability is given by

p+ (1 - p)3*p+ (1 - p)**p+(1 - p)**P+ (1 - p)2 *p + .....

=p[1+ (1 - p)3 + (1 - p) + (1 - p) + (1 - p)12 + .....

= p [1+ (1 - p)3 + ((1 – p)3)2 + ((1 – p)3)3 + (1 - p)3)4 + ...

   [Sum of infinite GP series with p<1]

=p* 1-(1 - p)3 = 1-11-p)

To Becca become winner first head should appear in any of 2nd, (2+3)=5th, (5+3)=8th, (8+3)=11th, (11+3)=14th draw or so on.

So, required probability is given by

(1 - p)p+ (1 - p)* * p + (1 - p)'*p+(1 - p)10 *p+ (1 - p)13 *p+ ....

=p(1-P) [1+(1 - p)3 + (1 - p) + (1 - p) + (1 - p)12 + .....

= p(1 – p) [1+ (1 - p)3 + ((1 – p)3)2 + (1 - p)3)3 + ((1 – p)3)4 + .....]

   [Sum of infinite GP series with p<1]

p(1-P) 1- (1 - p)3 "1- (1 - p)3

To Charise become winner first head should appear in any of 3rd, (3+3)=6th, (6+3)=9th, (9+3)=12th, (12+3)=15th draw or so on.

So, required probability is given by

(1 - p)p+(1 - p)5 *p+(1-p8*p+ (1 - p)1*p+ (1 - p) 14 *p+.....

=p(1 - p) [1 + (1 - p)3 + (1 - p) + (1 - p) + (1 - p)2 + .....

=p(1 - p) [1 + (1 - p)3 + (1 - p)3)2 + (1 - p)3)3 + (1 - p)3)4 + .....

   [Sum of infinite GP series with p<1]

p(1 - p) =p(1 - p)2 * "1-(1-p) 3 1 -(1 - p)3

Plotting probabilities of winning we get as follows.

1.0 0.8 0.6 Probability 0.4 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 p value

In the figure red, blue and green lines respectively denote probabilities of win corresponding to Amanda, Becca and Charise for different values of p.

(a)

In case of fair coin, probability to get head in a toss p=0.5.

So, probability that Amanda wins is given by

1 - (1 - p)3 + 1- (1 - 1)3 = 7 = = = 0.5714286

(b)

In case of fair coin, probability to get head in a toss p=0.5.

So, probability that Becca wins is given by

p(1 - p) 1*(1 - 1) 1 - (1 - p)3 1-(1 - 13 2 0.2857143

(c)

In case of fair coin, probability to get head in a toss p=0.5.

So, probability that Charise wins is given by

p(1 - p)_ *(1 – )' 1 1 - (1 - p)3 + 1 - (1 - 1)3= = = 0.1428571