Question

In the New York State "Numbers" game, three digits are drawn, with replacement, each evening on television. The player chooses, in advance of course, three digits and may pick from several types of bets:

a. "Straight": To win the player's digits must match those drawn in the order they were drawn.

b. "Six-Way Box": The player chooses three different digits and wins if they match those drawn in any order.

c. "Three-Way Box": The player announces a digit once and a different digit twice. To win, these three digits must match the three digits drawn in any order.

Find the probability of winning each of these bets.

Answer 1

Solution

Back-up Theory

Probability of an event E, denoted by P(E) = n/N ………………………………….....................…….........………(1)

where

n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Number of ways of arranging n distinct things among themselves (i.e., permutations)

= n! = n(n - 1)(n - 2) …… 3.2.1………………………………………………………………...........................….….(2)

Number of ways of arranging n things taking only r at a time, such that the same thing

may repeat itself a maximum of r times = rⁿ ……........................................................................................….(2a)

Number of ways of arranging n things of which p things are identical, among themselves = (n!)/(p!) ……......(2b)

Now to work out the solution,

Vide (2a),

3 digits out of 10 digits, namely, 0, 1, 2, ......, 9 can be selected in 10³ = 1000.

So, vide (1), N = 1000. ......................................................................................................................................(3)

Part (a)

Out of these 1000 possibilities, there is only one possibility where the player's digits

match those drawn in the order they were drawn. So, vide (1), n= 1 and

hence vide (1) and (3)

Probability of winning "Straight"

= 1/1000 = 0.001 Answer 1

Part (b)

Vide (2), the winning 3 digits can be ordered in31 = 6 ways. Thus, the player's digits

could match those drawn in any order in 6 ways. So, vide (1), n= 6 and

hence vide (1) and (3)

Probability of winning " Six-Way Box"

= 6/1000

= 0.006 Answer 2

Part (c)

In this case, the player announces 3 digits of which 2 are identical and so these 3 digits can be ordered in {(3!)/(2!)} = 3. So, vide (1), n= 3 and

hence vide (1) and (3)

Probability of winning " Three-Way Box"

= 3/1000

= 0.003   Answer 3

DONE