DavidsCharger
DavidsCharger
Joined: Sep 6, 2015
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September 5th, 2022 at 8:03:45 PM permalink
I would like to know if anyone can give me the probability of losing 9 times in a row when you bet in Roulette alternating two black then two red back and forth. I know how to calculate the probability of just betting on say black and having a streak of 9 losses in a row.
Having played about 2000 spins I have only seen a loss string as high as 9 and winning on the tenth spin only once.
Thanks for any replies.
unJon
unJon
Joined: Jul 1, 2018
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Thanks for this post from:
DavidsChargercamapl
September 5th, 2022 at 8:15:48 PM permalink
The probability of losing 9 times in a row betting two blacks then two reds, alternating back and forth, is exactly the same as the probability of losing 9 times in a row betting black every time.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
DavidsCharger
DavidsCharger
Joined: Sep 6, 2015
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September 5th, 2022 at 8:54:41 PM permalink
Thank you for clearing that up for me. I often wonder exactly how small a losing streaks probability of losses in a row is to be considered basically zero chance? Say like the probability of losing 30 straight spins in a row.

Have a great Doy UnJon
OnceDear
OnceDear
Joined: Jun 1, 2014
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September 6th, 2022 at 4:49:06 AM permalink
Quote: DavidsCharger

I would like to know if anyone can give me the probability of losing 9 times in a row when you bet in Roulette alternating two black then two red back and forth. I know how to calculate the probability of just betting on say black and having a streak of 9 losses in a row.
Having played about 2000 spins I have only seen a loss string as high as 9 and winning on the tenth spin only once.
Thanks for any replies.
link to original post


Assuming double zero roulette...
For any set of N observations, the probability of losing N even money bets in a row is 0.526315789474 raised to the power of N

1 in a row 0.526315789474
2 in a row 0.526315789474 x 0.526315789474
3 in a row 0.526315789474 x 0.526315789474 x 0.526315789474

9 in a row 0.526315789474^9 = 0.00309897 or 1 in 323

So you won't need to observe many sets of spins before you see a streak of 9 losses. Roughly twice as many to see a streak of ten.
The pattern that you bet is of no relevance, whether it's red, black, red, black or even, even, odd, odd, even, even, odd, odd.

Probability of losing 30 in a row, roughly 1 in 230 million.....
But if you just watched a losing streak of 29, the probability of losing the next one is still 20/38
Beware. The earth is NOT flat. Hit and run is NOT a winning strategy: Pressing into trends is NOT a winning strategy: Progressive betting systems are NOT a winning strategy: Don't Buy It! .Don't even take it for free.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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September 6th, 2022 at 7:36:46 PM permalink
Quote: DavidsCharger

Thank you for clearing that up for me. I often wonder exactly how small a losing streaks probability of losses in a row is to be considered basically zero chance? Say like the probability of losing 30 straight spins in a row.

Have a great Doy UnJon
link to original post


This reminds me of my SUREFIRE GUARANTEED CANNOT MISS ROULETTE SYSTEM:

1. Determine the smallest number N such that it is "impossible" for red (or black, for that matter) to come up N times in a row.
2. Wait for red (or black) to come up N-1 times in a row.
3. Bet pretty everything on black (or red); you should bet some on green "just in case."
4. Oops - did red (or black) come up the Nth time in a row? Don't look at me; you're the one that said that this was impossible.

Besides, if you want to cover the possibility of losing 30 spins in a row and be able to show a profit if you win at any point, your 31st bet would have to be over 1 billion (yes, billion, with a B) times what your first bet was - and the two possibilities are (a) ending up 1 ahead, and (b) ending up over 2 billion behind.
Ace2
Ace2
Joined: Oct 2, 2017
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September 6th, 2022 at 8:19:48 PM permalink
If your next question is:

How many spins should it take until you get nine consecutive losses (betting any permutation of red or black) on a 00 wheel?

Then use the formula ((1/p)^(n+1) -1) / ((1/p) -1 ) - 1, where p is the probability of a loss and n is the length of the streak. So:

((38/20)^10 - 1) / (38/20 - 1) - 1 =~ 679 spins

Not that remote but not something you'll see every day
Last edited by: Ace2 on Sep 6, 2022
Itís all about making that GTA

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