5 Digit/Buckeye 5

Reply #8

RJOh — Sat, 17 Apr 2004 19:23:29 GMT

280,625/435,897 = 64% of the possible combinations doesn't restrict your choice enough to play all the combinations and trap a winner within those parameters.RJOh

Reply #7

johnph77 — Sat, 17 Apr 2004 01:12:28 GMT

If, on the other hand, one chose to restrict one's choices to those combinations where 1 to 10 comprised the first number and 28 through 37 comprised the last number, there are 280,625 possibilities.

Reply #6

RJOh — Fri, 16 Apr 2004 23:55:02 GMT

Thanks john,You're right.RJOh

Reply #5

johnph77 — Fri, 16 Apr 2004 21:59:45 GMT

RJOh -Is it possible that the number of combinations excluding 01 through 10 and 28 through 37 should read 6188 rather than 4368? Basically, what that does is create a 5/17 lottery draw, of which there are 6188 possibilities. It doesn't significantly alter the substance of youir post, though.....gljohn

Reply #4

RJOh — Fri, 16 Apr 2004 17:01:33 GMT

Quote: Originally posted by RJOh on April 15, 2004Mr Steel,Your theory is a fact. In the Buckeye5 (5/37) game, there are 435,897 possible combinations and 355,167 of them start with numbers 1-10, that 81% of the possible combinations. Like wise 81% also end with numbers 28-37 (the highest ten numbers). Only 4,368 combinations(11-27) or 1% of those combinations would not have either of those conditions.RJOhMr Steel there were some mistakes in the original post, those corrections are in blue.

Reply #3

MrSteel01 — Fri, 16 Apr 2004 14:43:34 GMT

Quote: Originally posted by RJOh on April 15, 2004Mr Steel,Your theory is a fact. In the Buckeye5 (5/37) game, there are 435,897 possible combinations and 355,167 of them start with numbers 1-10, that 81% of the possible combinations. Like wise 81% also end with numbers 26-37. Only 4,368 combinations(11-25) or 1% of those combinations would not have either of those conditions.RJOhDon't you mean 19 % Roy

Reply #2

MrSteel01 — Fri, 16 Apr 2004 14:20:42 GMT

Quote: Originally posted by RJOh on April 15, 2004Mr Steel,Your theory is a fact. In the Buckeye5 (5/37) game, there are 435,897 possible combinations and 355,167 of them start with numbers 1-10, that 81% of the possible combinations. Like wise 81% also end with numbers 26-37. Only 4,368 combinations(11-25) or 1% of those combinations would not have either of those conditions.RJOhTHANK YOU MR ROY

Reply #1

RJOh — Thu, 15 Apr 2004 19:16:07 GMT

Mr Steel,Your theory is a fact. In the Buckeye5 (5/37) game, there are 435,897 possible combinations and 355,167 of them start with numbers 1-10, that 81% of the possible combinations. Like wise 81% also end with numbers 26-37. Only 4,368 combinations(11-25) or 1% of those combinations would not have either of those conditions.RJOh

5 Digit/Buckeye 5

MrSteel01 — Thu, 15 Apr 2004 18:35:12 GMT

www.ohiolottery.comWell it happened 5/6 times since 5/5/91? 2,500 draws+-08-22-23-30-37=120 on 8/14/03 10/3/9706-20-21-24-36=107 on 10/18/99 6/22/9304-10-14-21-26=75 on 12/22/98 10/3/9609-11-18-25-36=99 on 8/21/97 2/14/97?04-07-11-27-30=79 on 7/4/97 6/17/9601-20-22-26-29=87 on 12/01/94 4/13/93Numbers 1 thru 9sums 75 79 87 99 107