Roulette - Probability of losing 10 bets in a row (Martingale)

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?

Your math is right. 16 wipe outs per 1,000 attempts.

Where can you play at $1 min

Quote: rottenluck

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

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).

Quote: evilzebra

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.

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.

Quote: RS

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!!!