Question

In the New York State "Win 4" game, four digits are randomly drawn, with replacement. The player chooses four 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. There are four kinds of "box" bets. In each, the player wins by matching the digits drawn in any order. "24-Way Box": The player's digits are all different. "12-Way Box": The player names one digit twice. "6-Way Box": The player names two different digits twice each. "4-Way Box": The player names one digit three times and a different digit once.

Find the probability of winning each of these bets.

Answer 1

a) Straight. The total number of possible outcomes with their order are (10)4, as there are 10 choices for each digit. Each outcome is equally likely, and the player only chose 1 of them. Thus,

P(winning in straight) = 1/10000

b)

"24 way box": The player has chosen 4 distinct digits. These can be arranged in 4! many ways. Thus, they can correspond to 24 different outcomes.

P(winning "24 way box") = 24/10000

c)

"12 way box": The player says one digit twice. This halves the number of permutations from above. So the number of outcomes it can correspond to is 12, as we choose 2 out of 4places for the double and arrange the remaining 2 digits in the remaining 2 places in 2! ways.

P(winning "12 way box") = 12/10000

d)

"6 way box": The other 2 digits are now also same. The total number of permutations is reduced to choosing 2 out of 4 places for the 1st pair in 6 differet ways.

P(winning "6 way box") = 6/10000

e)

"4 way box": 3 of the digits are same, so now we just chose the 1 out of 4 place to put the unique digit, in 4 ways. This only corresponds to 4 outcomes. Thus,

P(winning "4 way box") = 4/10000