Somebody please correct me if I'm wrong. I don't want to mislead, but it seems I have found an edge. I've been messing around with parlay bets lately, running a few test. Again, if I'm wrong in my math or assumptions here please point it out because I don't see it.
Let's use dozens as an example. The odds of winning 3 dozen bets in a row are 1 in 27 (3×3×3). So, it's safe to say in the long run we will see 1 hit in 27 on average, right? Or 10 hits in 270, and so on.
Ok, let's say I bet 27 spins in a row....one unit for each spin. The most I can loose is -$27. However, statistically I should win 3 times in a row once (remember the odds this should happen is 1 in 27).
Now if I have a rule to parlay my winning after a win, and have 3 wins straight then I win $39.
1. Spin 1 - win $3 (parlay winnings plus $1 for the next spin -total bet $4)
2. Spin 2 - win $12 (parlay winnings plus $1 for the next spin -total bet $13)
3. Spin 3 - win $39
Remember, if you loose the parlay bet you're only loosing your original $1 of your money. So, in 27 spins you can only loose $27 of your money no matter how many parlays don't reach 3 wins.
To summarize, 3 wins in a row should happen for dozens 1 in 27 times on average....BUT you will win $39 each time, averaging $12 for every 27 spins....or $120 for every 270 spins (all depends on the unit size you are playing with).
Ok guys, if I'm wrong on any of these statement let me know. I've gone over it a few times and can't find a mistake...but I'm still skeptical it's that easy
Quote from: Scarface on Apr 27, 11:56 PM 2017The odds of winning 3 dozen bets in a row are 1 in 27 (3×3×3).
Where did you read that? No promised winnings in roulette...
Guess I should have ran some test before I posted this. Just ran several test for 270 spins...I thought that it should average 10 wins out of 270. But tests ranged from 4 to 8 wins...probably averaging around 5. So guess this isn't going to work. Still not sure where I messed up on the math.
The odds of winning 3 times in a row is a little less than (1/3)^3 = 3.7%. as we also include 0/00 i.e. in every 100 spins less than 3.7 spins. it can happen anytime during that 100 spins, best case scenario would be your first 27 spins. Sorry, odds are really against you with this strategy.
Hello
8 times out of 9 you will lose $ 3 ($ 24)
1 time out of 9 you win $ 10.
So you lose $ 14 every 9 spins.
Quote from: plolp on Apr 28, 02:44 AM 2017
Hello
8 times out of 9 you will lose $ 3 ($ 24)
1 time out of 9 you win $ 10.
So you lose $ 14 every 9 spins.
excuse me :
you lose 14$ every 18 spins ( 9 try )
Sorry it's not that yet
Sum invested for the first spin: $ 1
For the second: $ 4
so for 8 try (lost) -40
and 1 try (win) +15
so -25$ for 9 try .
Quote from: Scarface on Apr 28, 12:07 AM 2017But tests ranged from 4 to 8 wins...probably averaging around 5
Thanks God, Good Morning, Thanks Scarface.
1) Did you tested more of it?
2) How about playing against it?
regards, SugTips
I see where my assumptions were wrong now. There are 27 different combinations of 3 dozens if you write them all out: 111, 112, 113, 121, 122, 123, etc...
I said that in 1 in 27 spins I should have 3 consecutive wins in a row...this is where I went wrong. It should be 1 win in 27 series of 3, not spins (1×3×27). So, the odds are 1 in 81 spins I would win.
In summary, this is not a good bet
Quote from: Scarface on Apr 28, 07:32 AM 2017
I see where my assumptions were wrong now. There are 27 different combinations of 3 dozens if you write them all out: 111, 112, 113, 121, 122, 123,
A question. Let's say you are looking for 111. If the first spin is 2, then do you restart betting from fourth spin or 2nd spin. If you do from 2nd spin, then the above argument Is not valid.
Quote from: Priyanka on Apr 28, 09:22 AM 2017
A question. Let's say you are looking for 111. If the first spin is 2, then do you restart betting from fourth spin or 2nd spin. If you do from 2nd spin, then the above argument Is not valid.
Actually, preferred method is follow the last spin. So, I guess there would really need to be 4 of the same dozen in a row to get 3 consecutive hits...since playing this way, I miss the first hit.
Going to look into these bets a little more. I still haven't given up on a good parlay progression. Maybe parley series instead of individual spins
Pri, if I missed the first bet and immediately started betting on the second bet do you know what the odds are? The math is really screwing me up here lol. If I were betting for dozen 3 to hit for 3 consecutive times, without skipping bets, what are the odds of this happening? Anyone?
Quote from: Scarface on Apr 28, 10:15 AM 2017
If I were betting for dozen 3 to hit for 3 consecutive times, without skipping bets, what are the odds of this happening? Anyone?
Scarface,
Two answers for the probability that dozen 3 would hit for 3 consecutive spins:
1. If you exclude the 0/00, then for dozen 3 to hit 3 consecutive times is, the probability is:
1/3 x 1/3 x 1/3 = 1/27 (=.037 approximately).
2. If you include the 0/00 for the double zero wheel, the probability is:
12/38 x 12/38 x 12/38 = .0314 approximately.
Obviously, the second answer is slightly more accurate since you are including the zeroes.
Quote from: DoctorSudoku on Apr 28, 11:07 AM 2017
Scarface,
Two answers for the probability that dozen 3 would hit for 3 consecutive spins:
1. If you exclude the 0/00, then for dozen 3 to hit 3 consecutive times is, the probability is:
1/3 x 1/3 x 1/3 = 1/27 (=.037 approximately).
But if this happens 1 in 27 spins (over the long term on average), then why is the payout $39 if I'm only risking $27 of my own money? This is very confusing
Quote from: Scarface on Apr 28, 02:58 PM 2017
But if this happens 1 in 27 spins (over the long term on average), then why is the payout $39 if I'm only risking $27 of my own money? This is very confusing
Scarface,
The 1/27 ratio does NOT refer to 1 in 27 spins.
What the 1/27 ratio refers to is that in
27 sets of 3 consecutive spins, you will have -- on average -- ONE instance of dozen no. 3 repeating in 3 consecutive spins.
Quote from: Scarface on Apr 28, 02:58 PM 2017This is very confusing
Hi Scarface - it is a bit consuming to get ones head around it. I agree. That's the reason I asked that question to you.
There are two ways to look at it.
1. The probability that you got 1/3x1/3x1/3 is correct. What it means is "once in 27 sets of three spins". It does not mean 27 spins. Because what you calculated is the probability of a dozen to appear the same in three spins.
2. Think a bit about how many fours and fives of dozens you got. If you are doing one after the other, rather than starting a fresh set of three spins, then a four will count as two sets of three Ina row. A fiver will count as three sets of three in a row and so on.
Quote from: DoctorSudoku on Apr 28, 04:00 PM 2017
Scarface,
The 1/27 ratio does NOT refer to 1 in 27 spins.
What the 1/27 ratio refers to is that in 27 sets of 3 consecutive spins, you will have -- on average -- ONE instance of dozen no. 3 repeating in 3 consecutive spins.
Scarface,
This is an expansion of my above post.
When you consider 3 consecutive spins, there are
27 separate sets of 3 consecutive spins that are possible for the three dozens. They are as follows (1, 2, and 3 refer to the three dozens):
111
112
113
121
122
123
131
132
133
222
221
223
212
211
213
232
231
233
333332
331
323
322
321
313
312
311
Of these 27 separate sets of 3 consecutive spins,
only one is 333 -- and that is what the 1/27 is referring to.
Thank you DoctorSoduku and Priyanka for clearing this up for me! So, 1 in 27 spins was definitely wrong! So statistically, cycle 333 will appear once in 27 cycles of 3...which is 81 spins. Unless, of course there are two cycles that join to cause it like 123 and 332. Anyways, it's still far off from 1 in 27. Thanks for your help!