"What is unusual number of losses in row, same game [9/6DDB]? Would 8, 10 or 12 losing games in row be unusual?"
The chance of getting "nothing" in the 9/6 DDB pay table is 55.28%.
The chance of getting nothing twice in a row is 55.28% x 55.28% = 30.56%
The chance of getting nothing 3 times in a row is 55.28% x 55.28% x 55.28% = 16.89%
and so on...
Stick =55.28%^8 into any spreadsheet, and you'll find that the chance of hitting blanks on your next 8 hands is 0.87%, about one in 115.
Here are some values (that'll probably format miserably):
| streak | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| chance | 55.28% | 30.56% | 16.89% | 9.34% | 5.16% | 2.85% | 1.58% | 0.87% | 0.48% | 0.27% | 0.15% | 0.08% |
| 1 in … | 1.8 | 3.3 | 6 | 11 | 19 | 35 | 63 | 115 | 207 | 375 | 679 | 1228 |
BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands. If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.
--Dunbar
---In vpFREE@yahoogroups.com, <whitejeeps@...> wrote :
Good afternoon folks.
Say we play 1000 hands, what is the theoretical loss %? 9/6 DDB. I'm thinking if average is ,say, 55% loss then 550 of 1000 would be losers. Thats assuming I'm thinking right.
What is unusual number of losses in row, same game? Would 8, 10 or 12 losing games in row be unusual?
Hoping someone has the info at their finger tips?
Cheers.....Jeep
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Posted by: h_dunbar@hotmail.com
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