A friend of mine was trying to calculate the odds of loosing it all playing Martingale’s strategy at roulette. There have been a number of threads about why is’s a losing strategy, but I was trying to figure out how to calculate her chances of loosing 5, 6 or 7 times (the number of loses she could cover based on her first bet) in a row if she played 20 times or so. By analogy, I tried to use flipping a coin (though that’s better odds than the roulette wheel), but factorials and stuff are not my strong point.
So, if you flipped a coin 20 times, what are your chances of getting a string of 5, 6 or 7 tails? Or for that matter, what are the odds of 5, 6 or 7 losses if you’re playing red or black on a roulette table?
I just realized that since he’d walk away after his limit was reached, it isn’t necessarily 20 throws, but up to 20 depending on the chance of first getting to the string of tails/losses.
The expected number of “r-runs” of one side of a coin in n tosses is:
n/2**(r+2)
Your odds on a particular spin at roulette is slightly less than 50/50 - the wheel usually has one or two slots, “0” and “00” only are house wins, with 18 red and 18 black ones as well. Longer strings of losses are a lot more likely than in coin flipping because of this.
On thinking it over, if you played a long series of of tries and broke it up into sets of 20 tries each wouldn’t the probability that any particular 20 try set contained 5 in a row be 20/32?
A moment’s thought should convince you: Scenerio 1: you flip a coin five times. Scenerio 2: you flip a coin a hundred times. Isn’t it obvious that the odds of getting five heads in a row somewhere in a sequence of a hundred flips is better than the odds of getting five heads in a row in a seqence of five flips? In five flips, there is only one seqence of five, one chance to see five heads; in a hundred flips, there are 96 sequences of five, 96 chances to see five heads.
Scenerio 3: you flip a coin a million times. The odds of getting five heads in a row somewhere during those million flips (999,996 sequences of five) has gotta be greater than the odds of getting five heads in a row sometime during a hundred flips (96 sequences of five).
The fact that you intend to stop as soon as you hit those five heads in a row is irrelevant to the calculation of what the odds are of five heads in a row sometime in n flips.
Slightly? Try enormously. The house edge is 5.26%, which makes it a sucker bet. The OP’s friend would be well served trying her martingale on the Don’t Pass line in craps for several reasons. First, the house edge is only 1.40%, which is close to the most you would normally want to tolerate. Second, it takes longer in real time, on average, to resolve a point in craps than to spin a roulette wheel once, which means the inevitable loss of all money will take longer. And third, you have complete dominion over your space in craps, whereas in roulette it’s akin to feeding time at the zoo when it’s time to place your bets.
To address the OP, first off the “20 spins” concept is flawed. You don’t care about spins, you care about series. Whether it takes 1 spin, 4, or 20 to finally get a winner, all those spins equate to a single wager, and should therefore be considered as a single entity. “Series” works as well as any other word for this.
So, what’s the chance of a single series failing? Let’s see, using the Martingale, and capping (as the OP did) at 5, 6, or 7, you can figure out the single series chance pretty easily.
Chance to lose roulette is 20/38. Chance to lose 5 times in a row? (20/38)[sup]5[/sup]. nth times? (20/38)[sup]n[/sup]
Let’s go with 6, because it’s in the middle, and also this way she can lose twice the money she would have had she capped losses at 5.
(20/38)[sup]6[/sup] = 2.13%
So let’s say we’re talking about a $5 wager, doubled after every loss, up to six losses max. Lose 5, 10, 20, 40, 80, 160. This is common, as bringing $320 to the casino is about right. You’ve got a 2.13% chance to lose $315, and 97.87% chance to win $5.
How many did you say? 20 spins? I happen to know that 50/50 martingale train(wreck)s are resolved, on average, approximately every 3.5 trials or so. So that 20 spins would involve about 6 series, which would net you $30, but also involve a chance to win (to the 6th power when discussing 6 wins in a row) of 87.91%
So there’s your chances, assuming you meant 20 spins, and not series. 87.91% chance to win $30, 12.09% chance to lose $315.
If you meant 20 series, so that you could win $60, which IMO is much more likely, you’re talking about a 65.07% chance to win 20 martingale series’ in a row. In short, the martingale train will have over a 1 in 3 chance to wreck in 20 resolutions.
Allow me to attempt to pinpoint exactly what the fallacy of the martingale is. I’ll just quote myself from this thread:
I sincerely hope that helps. Nobody should suffer the anguish of having to place their final bet in a martingale train.
