Roulette Odds Black In A Row
Roulette Odds Black In A Row 3,0/5 5927 votes
If you have a computer roulette simulator, try it out. This means that of the 'eight blacks' sets there are 50% eight blacks in a row FOLLOWED BY A RED and 50% eight blacks in a row FOLLOWED BY A BLACK. The same applies to the 'eight reds' sets. If you plan to play online roulette, stick with the bets whose odds almost mirror their payouts. These 'even money' bets include betting on Even, Odd, Low (numbers 1 through 18), High (numbers 19 through 36), Red, or Black. All of these wagers pay out at 1 to 1. Game Odds Articles.
evilzebra
Roulette Odds Black In A Row 4
Let's say you are a strict martingaler. For simplicity sake, you have $1024 to bet with. Here's how your progression would go
Bet Level
1 1
2 2
4 3
8 4
16 5
32 6
64 7
128 8
256 9
512 10
This means that you can afford to lose your bet 9 times in a row, but not 10 times. Assuming you pick a color (red), what is the probability you will get wiped out here. If you lose on the 10th bet, you only have $1 left.
Well, there are 20 colors that aren't red in the roulette wheel.
So that means that the probability of losing a bet is (20/38) which equals 52.6%
Now, the probability of losing 10 bets in a row is probably (20/38) ^ 10
I get .00163
That means that out of 10000 spins, only 16 times will a streak of 10 or more occur.
Is this math right, or am I missing something?
Bet Level
1 1
2 2
4 3
8 4
16 5
32 6
64 7
128 8
256 9
512 10
This means that you can afford to lose your bet 9 times in a row, but not 10 times. Assuming you pick a color (red), what is the probability you will get wiped out here. If you lose on the 10th bet, you only have $1 left.
Well, there are 20 colors that aren't red in the roulette wheel.
So that means that the probability of losing a bet is (20/38) which equals 52.6%
Now, the probability of losing 10 bets in a row is probably (20/38) ^ 10
I get .00163
That means that out of 10000 spins, only 16 times will a streak of 10 or more occur.
Is this math right, or am I missing something?
rottenluck
Your math is right. 16 wipe outs per 1,000 attempts.
GWAE
Where can you play at $1 minExpect the worst and you will never be disappointed. I AM NOT PART OF GWAE RADIO SHOW
evilzebra
Your math is right. 16 wipe outs per 1,000 attempts.
I think you're off by a magnitude of 10. It should be 16/10,000
DJTeddyBear
Yeah, your math is right, but ... What's the point?I mean, let's say you lose 3 times then win, or 7 times then win, or all 9 times then win. When you win and add up all your money, you're up one whole dollar.
Are you really prepared to risk $1,024 to win one buck?
Well, even if you said yes, the casino isn't prepared to let you try. Casinos typically do not have that much of a range between the minimum and maximum bet.
Additionally, you're gonna have to multiply your bankroll by 5 (or maybe 10) as typical casino minimums are $5 (or $10).I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
OnceDear
Administrator
Administrator
That means that out of 10000 spins, only 16 times will a streak of 10 or more occur.
Is this math right, or am I missing something?
Each of those 16 times you will lose $1023, which is $16368 lost
If you won ALL the other 9840 wagers, that would give you $9840 of winnings to offset against those losses. But that would be pretty miraculous.
You'd maybe win about half (18/38) or 4661 of wagers giving you a more realistic $4661 of winnings to offset against your losses.
Academic really as on wipeout number one you are wiped out.
Roulette Black Red Odds
Marty is good for when you have £1000 and NEED $1001 to escape a firing squad or fly out of a war zone. Other than that it's just a fun way to lose your money.Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
RS
18 red, 18 black, 2 green.
20 losers, 18 winners. 38 total.
(20/38) ^ 10 = 0.00163103767 ~ 0.00164 = 0.164% [rounded to 0.164% to give martingaler benefit of the doubt].
For ease of use, let's say you decide to play 100,000 cycles (a cycle is when you first bet $1, until you win or lose 10 in a row...a cycle is NOT 100,000 wagers).
For 100K cycles, you expect to lose 164 of those. The remaining 99,846 cycles, you will win $1. For the 164 losing cycles, you will lose $1,023.
164 * 1,023 = $167,772.
99,846 * 1 = $99,846.
$167,772 - $99,846 = -$67,926.
20 losers, 18 winners. 38 total.
(20/38) ^ 10 = 0.00163103767 ~ 0.00164 = 0.164% [rounded to 0.164% to give martingaler benefit of the doubt].
For ease of use, let's say you decide to play 100,000 cycles (a cycle is when you first bet $1, until you win or lose 10 in a row...a cycle is NOT 100,000 wagers).
For 100K cycles, you expect to lose 164 of those. The remaining 99,846 cycles, you will win $1. For the 164 losing cycles, you will lose $1,023.
164 * 1,023 = $167,772.
99,846 * 1 = $99,846.
$167,772 - $99,846 = -$67,926.
OnceDear
Administrator
Administrator
164 * 1,023 = $167,772.
99,846 * 1 = $99,846.
$167,772 - $99,846 = -$67,926.
Maybe More accurate. Equally devastatingly bad.
Actually, RS, 100,000 cycles would be way more than 100,000 wagers, More like 200,000. so wouldn't likely number of lost cycles be about double too.. Cannot be bothered to do the proper maths right now. Hmmmmm. will give more thought to RS's cycles approach to see if reality is dumb to the tune of 67K or something maybe half or so of that.
Marty is still SSssoooooo dumb if there is a house edge, and as to double Zero Roulette. PaH!!!
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..