Riddle me this, Batman...
Here’s a simple 6 step pos/neg progression I believe I found somewhere on this site.
It takes 2 wins (repeat or parlay) to reset the progression cycle to step 1. It also takes a single loss (parlay) or single NET loss (repeat) to advance to the next step.
. . . . Bet . . . . Loss . . . Win
1) 5 repeat . . . .-5 . . . +10
2) 5 parlay . . . -10 . . . +10
3) 10 repeat . . -20 . . . +10
4) 10 parlay . . -30 . . . +10
5) 20 repeat . . -50 . . . +10
6) 20 parlay . . -70 . . . +10
Now, the Wizard Of Odds says, ignoring pushes, you have a 52.5% chance of losing any hand. You’ve also got a 47.5% chance of winning. The odds of winning 2 hands in a row are about 1 in 4.5 tries. Or twice in 9 tries.
Looking at the chart of the progression, you can say there is a 2/3 chance of any progression cycle ending with a win in the first 3 steps. That’s because 3 tries (steps) times 1 in 4.5 tries, = 3/4.5 = 2/3.
Another way to say this is that on average, of every 3 cycles played, 2 will end in steps 1-3 (with a win) and the 3rd will continue on to steps 4-6. So for 9 cycles played, you’ll have 6 wins in steps 1-3 and 3 cycles continue on to steps 4-6.
Same thing goes for the 3 of 9 cycles that reach steps 4-6. You’ll have 2 wins in steps 4-6, and one time you’ll lose step 6 and the progression cycle ends with a loss.
Totaled up, for each 9 cycles you play, you’ll have 8 wins (+$80) and 1 loss (-$70) for a net gain of $10.
What happened to the “house edge”?
The odds of losing 6 hands in a row are about 1 in 48 tries. The odds of winning 2 hands in a row are 1 in about 4.5 tries. 48/4.5 = just over 10. So the odds of losing 6 hands in a row before winning 2 in a row are about 1 in 10. Since each step pays $10 for a win, on average you should win $90 and lose $70 for every 10 cycles you play. That’s a net gain of $20.
Again, what happened to the “house edge”?
-darkstar-
Here’s a simple 6 step pos/neg progression I believe I found somewhere on this site.
It takes 2 wins (repeat or parlay) to reset the progression cycle to step 1. It also takes a single loss (parlay) or single NET loss (repeat) to advance to the next step.
. . . . Bet . . . . Loss . . . Win
1) 5 repeat . . . .-5 . . . +10
2) 5 parlay . . . -10 . . . +10
3) 10 repeat . . -20 . . . +10
4) 10 parlay . . -30 . . . +10
5) 20 repeat . . -50 . . . +10
6) 20 parlay . . -70 . . . +10
Now, the Wizard Of Odds says, ignoring pushes, you have a 52.5% chance of losing any hand. You’ve also got a 47.5% chance of winning. The odds of winning 2 hands in a row are about 1 in 4.5 tries. Or twice in 9 tries.
Looking at the chart of the progression, you can say there is a 2/3 chance of any progression cycle ending with a win in the first 3 steps. That’s because 3 tries (steps) times 1 in 4.5 tries, = 3/4.5 = 2/3.
Another way to say this is that on average, of every 3 cycles played, 2 will end in steps 1-3 (with a win) and the 3rd will continue on to steps 4-6. So for 9 cycles played, you’ll have 6 wins in steps 1-3 and 3 cycles continue on to steps 4-6.
Same thing goes for the 3 of 9 cycles that reach steps 4-6. You’ll have 2 wins in steps 4-6, and one time you’ll lose step 6 and the progression cycle ends with a loss.
Totaled up, for each 9 cycles you play, you’ll have 8 wins (+$80) and 1 loss (-$70) for a net gain of $10.
What happened to the “house edge”?
The odds of losing 6 hands in a row are about 1 in 48 tries. The odds of winning 2 hands in a row are 1 in about 4.5 tries. 48/4.5 = just over 10. So the odds of losing 6 hands in a row before winning 2 in a row are about 1 in 10. Since each step pays $10 for a win, on average you should win $90 and lose $70 for every 10 cycles you play. That’s a net gain of $20.
Again, what happened to the “house edge”?
-darkstar-
I await enlightenment. hmmm, but I gotta run to the casino first.
