Roulette Odds 10 In A Row
Odds Of 10 Black In A Row Roulette, door mail slot with basket, lunch casino cosmopol, blackjack casino vilamoura. Game contribution weightings apply to wagering requirements. Maximum bet while playing with bonus Odds Of 10 Black In A Row Roulette is Odds Of 10 Black In A Row Roulette £5. Minimum deposit of £10 + 2.5% fee (min 50p). Bonus is not available when depositing with Neteller or Skrill. The odds that the Roulette ball will land on red is 18/37 = 48.6%. The odds that the ball will land on red two times in a row is 48.6% x 48.6% = 23.6%. The odds that the colour red will hit three times in a row is 48.6% x 48.6% x 48.6% = 11.4%. Etcetera Etcetera. De odds that the Roulette ball will hit a red slot 8x in a row. The record was registered in 1943, when red color came up 32 times in a row! The probability of such event in French Roulette is (18/37)32 = 0.96886885 with the corresponding odds 10,321,314,387:1. Focusing of European Roulette, the odds that your colour will not hit for 10 rounds in a row is 1 to 784. This might seem good, but keep in mind that the odds are like this only at the start of the game.
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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?
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
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).
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.
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.
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.
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!!!
I'm hoping I can win at least 70 times (or 700 times) before a bust.
For Double-zero: ~ 2/89 chance of losing 10 out of 10 'one dozens bets'
or
for Single-zero: ~ 2/101 chance ...
Administrator
What are the odds of losing 10 one Dozens bets in a row? (Roulette)
I'm hoping I can win at least 70 times (or 700 times) before a bust.
(25/37)^10 = 1.32% 1.98%= 1 in 50 approx
or (26/38)^10 = 2.25% = 1 in 45 approx
(25/37)^10 = 1.32% = 1 in 50 approx
or
(26/38)^10 = 2.25% = 1 in 45 approx
1.32% is closer to 1 in 75.
Administrator
1.32% is closer to 1 in 75.
I stand corrected... But it was the 1.32 that was wrong. I'd initially taken 24/37.Roulette Odds 10 In A Row 4
OnceDearAdministrator
I don’t math very well but is there a minimum and maximum variance percentage or something like that? That may be what’s expected over time? Or should I just stop talking lol
for discrete wagers like this, the maximum and minimums are 'Win all wagers' or 'lose all wagers' in the sequence.More broadly the answer is 'no'
There are confidence limits that can be calculated. E.g. I can be x% confident that the max number of wagers that I will lose sequentially will be N.
For a large number of wagers that follow a 'Normal distribution' we can say things like 68% of outcomes will be within 1 standard deviation of the mean and 90% will be within 2 x standard deviations. Lot's of 'ifs' If it's a normal distribution... If you know the standard deviation...
But (near) 100% confidence needs absolute maximum extremes of variance. E.g. Toss a coin a million times and you can say with roughly 100% confidence that you will have between 0 and a million occurrences of 'heads'. In reality it would be closer to 1/2 a million heads most times and 1 million heads in a row would be somewhat unlikely :o)
Casino Roulette Odds
Roulette software likes to make an example of me by losing 80 dozen bets, or 4 sessions of 20 bets of $400, or 4 sessions of $8,000, or $32,000, in one sit down.