2006-12-19 02:53:21 · 7 answers · asked by Anonymous in Games & Recreation ➔ Gambling
the numbers can only come out once and must be in the same order as drawn.
2006-12-20 19:12:02 · update #1
63 :-D
123 -
234 |
345 |
456 | = 7
567 |
678 |
789 -
9 numbers x 7 outcomes = 63
2006-12-19 03:06:05 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Assuming 3 sets of balls numbered 1-9.
Also order needs to be correct (if the numbers drawn are 1-2-3, then a ticket with 2-1-3 is a loser)
9x9x9
odds of winning 1 in 729.
2006-12-19 11:30:50 · answer #2 · answered by H_A_V_0_C 5 · 0⤊ 0⤋
Lets assume that the numbers cannot be duplicated (i.e. like a ping-pong ball lottery). The three numbers will come out in some order. Call it A-B-C. And you could position the 3 winning numbers in any one of six different ways. ABC, ACB, BAC, BCA, CAB, CBA.
Only one of those six will be in the correct numerical sequence (which I think is your question). So ... no matter whether the ping pong balls are numbered 1-9 or 1-9999 or whatever... the odds are one in six that they will be drawn in numerical order.
2006-12-19 11:14:03 · answer #3 · answered by ya_jerry 1 · 0⤊ 0⤋
total number of possible outcomes in the correct order is:
9*8*7= 504
therefore the answer is 503/1
2006-12-21 15:16:14 · answer #4 · answered by ? 3 · 0⤊ 0⤋
1-9
2006-12-20 15:03:35 · answer #5 · answered by munchie 6 · 0⤊ 1⤋
Chances of getting 1st one: 1/9
Chances of getting 2nd one: 1/8
Chances of getting 3rd one: 1/7
1/504
2006-12-19 11:39:10 · answer #6 · answered by rachfan1991 1 · 0⤊ 0⤋
I don't know it's all a bit of a lottery to me
2006-12-19 11:02:53 · answer #7 · answered by Anonymous · 0⤊ 1⤋