Gambling with the Martingale Strategy - Numberphile
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- čas přidán 15. 05. 2021
- Tom Crawford discussing roulette and gambling with a famed strategy.
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Me not having an infinite amount of money does usually come up as a problem
if you have infinite money then why gamble your money in the first place 🤔
I find that it's a bit of a nuisance, most of the times
@@colarboy1720 A better question would be "why not?".
@@colarboy1720 If I had infinite money, I'd gamble it away every which way
Many politicians don't seem to understand this, and think they can solve it by printing an infinite amount of money.
My calculus teacher used to say that gambling is a tax on people who don't understand mathematics
we straight up call it the stupidity tax
National lottery is a form of tax. But hey! it' s the only tax whereby I earned half a million euros.
Gambling is not a tax but a business. The State taxes their benefits, but the major part doesn't serve public goals.
So are lotteries
@@MarcusCactus My dude, it's just an expression meant to drive home the risks of gambling.
Alcohol, weed, gambling, cigarettes. All hail the cash cows of Canadian provinces
Everybody gangsta until it goes green
Man, if only I had seen this video last time I only had $100 and wanted to buy a $101 item at the store next to the casino.
I thought about this problem, usually starting with a real casino game that moderately favours the house (e.g. roulette or craps). The Law of Large Numbers works against you if you make many bets, so I think you should get the gambling session over as quickly as possible. I assume that you can afford to spend more than $101 on the item but have some temporary problem with your bank or credit card. Part of the strategy is to leave the casino with either $0 or just enough to buy the item.
In your example, I follow the Martingale strategy as long as I can. If I lose 6 times then I have $38 left, and I bet all of it on the next play, and if I lose again then it's over. If I win then I have $76, so I bet $25 and hope to reach $101. If I then lose, I bet $50, either winning what I need or coming down to $1. If I really want the item then I keep betting everything I have until I either lose or come back to $64. In that case I then bet $37, and so on. There's no guaranteed end to the session, but the expected number of plays is about 2 (because each play has the potential to end the session in one way or the other). I think that the fast strategy gives the greatest probability of success (at bit less than 100/101 in your example), but I haven't proved it.
A better strategy would be to ask someone for $1 and explain why you need it
@@Flechashe that's begging and that's illegal.
@@rolandasgrigaitis708 Did I stutter
@@rolandasgrigaitis708asking for money is not illegal. Otherwise all charities would be shut down.
"If only I had unlimited money so I could do this, then I'd be rich!"
your expected winning will still be 0
You would already be rich
@@stemacademy2182 That's... the joke
@@SodiumInteresting
No it wouldn't
It's impossible to have an infinite losing streak
Therefore someone with infinite money will ALWAYS win back their money + the original bet.
@@dustinjames1268 no, your expected gain after any amount of bets would be 0 (note: AVERAGE, assuming 50/50 odds).
As someone who once lost $1000 playing roulette at a casino in Las Vegas, believe me when I say “don’t gamble kids”.
Should’ve covered the losses with a £2048 bet
how do i gamble kids?
Lost 4k gambling online at the age of 16, whoops!
Boy, wait until you hear my time losing a game of Russian roulette
I think gambling kids is something most of us would never think of doing. Thanks for the tip tho
Remember everyone, you're only 36 straight wins of roulette away from being a billionaire.
My father in law came up with a slightly tweaked version of the martingale, I can’t remember the differences but they were significant. I downloaded a roulette app and started with a fictional float and played for a whole weekend. Come Sunday afternoon I was up by so much I was considering setting up a live account. Then I lost it all in about 10 minutes.
statistical variance and law of large numbers...in your case, it was nearly a universe sign to save you...
What was the point of this comment. It essentially says, there is this secret martingale variant that I essentially made up to tell a story about something insignificant.
@@TheAcidicMolotov and yet you came here and wasted your time replying to it. Well done you.
Thanks for your story, I enjoyed it, and I hope you enjoyed that weekend.
Ah yes sounds familiar...
I actually tried this once a long time ago at the Blackjack tables in Vegas and found out very quickly why there are table betting limits.
This isn't the reason for the limits. Just a bi-product. They limit bettering for other reasons.
@@TuberTugger interesting, like what?
@@gaeb-hd4lf The fact they don't keep a gajillion dollars on hand to honor your absurdly high bet if you were to win it.
So they'll generally adjust the bet to what they can reasonably give in winnnings for a night.
They should theoretically earn more on average but chance is a cruel mistress so some nights might dig into their margins.
Also the house always wins, you dont get poor Vegas casino bosses, unless they offer 50:50 roulette wheels 🤔😉😆
@@jeffcarr392 The house does not always win, or they'd let you bet however much you like.
'the richer I am, the less likely to lose all my money' ... quite the life lesson here!
You would think that's the case, but most rich people end up squandering their wealth unless they are also members of the government.
The less likely you are to bet everything on a wheel
@@HenriFaust Thanks for telling us that you have never met a wealthy person.
Also the richer you are, the more money you can lose
@@CK-nh7sv what's the point of this "gotchya!"? you added nothing of value except for just being like "erm... no."
I played this game in the university library computers with my mate one afternoon. We both started with £128 which gave us 7 spins before losing. Vividly remember we both had a balance over £200 after a couple hours, and both lost everything within 2 minutes of each other. Some adrenaline rush when we were on the 5th and 6th spins quite a few times, but we learned an important lesson about gambling probability that day.
How many times you lost in a row?
@@samiulhaquerounok5787 well he lost it all so that would be 7
Wrong.
The real lesson is, the richer you are, the more likely you are to win overall because your pockets are deep enough to cover a bad run of bets, until you finally win it all back and start again.
@@allancouceiro9255not true at all
blahem
The beauty of a casino is that you could make this video required viewing before entry and you’d still have a dozen people every day walking in convinced they cannot lose
I encountered this strategy a few years ago and I just wrote a short code simulating the whole thing. And I was quite amazed at how every now and then there'd be a ridiculously long losing streak that basically destroys you.
yeah
i worked out an excel sheet that just throws a coin 200 times and counts the repeats and highlighting the highest one.
the number is almost never below 5 and quite often over 7
I mean, if you play infinite many games, this is bound to happen. But many people don't realize that.
I have seen streaks if 16, so ridiculous
I did the same years ago, though I wasn't ever sure if the random num generator was significant variable, I wasn't working with any sophisticated programming. Even graphed it over a few hundred thousand rounds. It would appear to work over many short stretches of a few hundred or few thousand, but like you noticed there would be those few outliers where the deficit would hit tens of millions.
Don't we need to also look at winning streaks. £1 is the minimum win, but we are waiting for say 2 wins in a row, continuing the double strategy.
In order to make a small fortune at the casino, one must first start with a large fortune.
Even better. To make a fortune at the casino one must OWN the casino. Don't bet against the house, BE the house.
I would say infinite fortune
@@androsida8704 not necessarily.
You start with any one fortune of let us say $k0. You enter a casino. You play. You leave the casino with a fortune of $k1.
In all but finitely many cases $k1
It's a probability thing, most people lose, some people win, casino always wins. Still better than national lotteries, they pay out even less than casinos - in my country ~40% of bets, roulette pays out slightly more.
@@saimon174666 what do you mean by "payout less than casinos"?
Is the payout percentage vs the lottery rake relevant at such scales and do any of them have anything to do with your probability of winning?
The odds are just as terrible aren't they?
I always thought it was funny, when you would see hysteria over then fact that the lottery was 200M.
And you'd always hear people say "well, I don't normally play the lotto, but at 200mil? Yeah, I'm in".
As if the odds of hitting all the numbers materially changed from 50mil...
Nnnooowww it makes "sense" to play?... no.
There was very little in this 20 minute video that was not self explanatory once the Martingale Strategy was defined
I remember independently "discovering" this strategy when I was about 18 and for a brief period of feverish joy I thought I was gonna be a millionaire if I could just lay my hands on some odds and then had my hopes crushed by betting limits
haha I remember the same when I was 13, I used my parents account, after two hours of profiting I experienced 15 lost rounds in a row and lost it all (I had 0.01 base bet and I lost about 300usd :D)
always make sure the starting bet
you guys might have forget to jump house to house
@@thureintun1687 - Even if you start at 1%, the system doesn't work due to table limits. The casino already has this factored in.
Random question but what do you do for a living now?
As far as gambling strategies go, the Kelly formula is quite valuable. In the case of 50/50 it essentially says bet nothing. In a case like roulette with the green zeros, it says "be the casino"
Hi! Sorry to disappoint you, but the Kelly formula is demonstrably too theoretical to be used as a strategy basis. For example, it assumes that you are going to play an infinite number of times, each time knowing the exact probabilities, and that these probabilities are in your favour. And, that you want to grow your budget on a constant rate. All this is rarely encountered in real life. It has supposedly been used by card counters in BJ, but never been proved that it was "better" than some other rule.
When you have a limited span (an evening or even a week), the formula says to bet your whole bankroll in one bet.
The only sensible comment here.
@@MarcusCactus You are more or less correct. You would be correct with something like horse racing where you can never be completely sure what your expectation is going to be. In that case you can use a fractional Kelly. You won't make as much profit as a 100% Kelly but you are not going to run the risk of accidentally over-betting your incorrect expectation. In the case of card counting or other games where expectation is known then Kelly works. However, you have the additional risk of the casino shutting you down before you have made a profit.
@@betfairprotrader7159 What you mean with "Kelly works", I don't know. First, in a casino setting, the house has an edge so Kelly says Do Not Bet. Second, in a positive expectation game, any formula "works". Nothing has ever proved that one is "better" than the other. What is "better" anyway, in a probabilistic environment?
@@MarcusCactus If you are counting cards at blackjack then you have an advantage over the house. The same with a roulette computer. In any game where you have complete information and know what the winning bet is going to be then you have 100% expectation; Kelly says, "Bet 100%" of your bank, other systems might say bet 2%. Which is going to make more money in the long run? Anyway, I'll stick to Kelly over fractions.
The expected value is the same whether you use the martingale strategy or just bet it all right away. The 36.8% chance to double neglects the fact that you've been saving your winnings in a separate pot. In most of the 63.2% of cases where you go bust, you'll still have some saved from previous wins, and this will exactly balance the reduced chance to double, so the expected value is the same.
oh, okay yes I was wondering about that. You can't beat the house using this strategy but you also can't lose to the house (on average from a large number of plays) if each of the rolls are 50/50.
@@colorado841that’s assuming the loss didn’t come early
Technically, your expected loss is smaller if you use the martingale system, because you start at a smaller bet.
Unfortunately it is not 50/50 the house has an advantage.@@colorado841
Once you win doesn’t the strategy reset so ther aren’t any stored gains?
Gambling rule number one: The house always wins.
It is the only rule one needs to know 😄
the house always wins until you get the skills to beat their game (advantage play) and then they ban you from playing so they still win.
"the richer I am, the less probability that I lose all my money" also works in regular life. Yay!
This is essentially true. There is serious literature about how being poor contrains ur ability to make profitable gambles/investments. Likewise being able to afford losing money is about the biggest edge normal people can count on in any given investment scenario.
Tom's enthusiasm for all things mathematical is infectious.
Like Covid
@@sillysausage4549 Nice reply, fellow human.
Why did I have to be so mean? I'M SO SORRY, SNOWFLAKES!
@@sillysausage4549 Underrated apologize.
I’ve been thinking about this strategy my whole life, I always knew it must not work because otherwise people would have done it. But it’s really nice to understand _why_ it doesn’t work. Thank you for this, both of you.
I've always thought about what to tell people who think it is foolproof without going deep into the maths behind it or considering very long (and unlikely) loss streaks.
Let's say you've sat on that roulette table for an hour or two. You lost some and won some more, you got lucky a couple times and have now about 31 bucks more than what you started with. Now you lose five times in a row, which really isn't an unlikely thing to happen. So, what's your situation now?
Following your strategy, you would have to bet 32 now, since you just lost a total of 31 in the current streak. But honestly, why? Your balance is now exactly at the point where it was 1-2 hours ago when you started betting only 1. The probability of winning the next round also is exactly the same as it was back then. If you're willing to keep going and bet 32 now, you know you could have done this from the start and not waste a couple hours getting to the point where "the strategy allows it".
I get that you think you're not gonna lose like ten times in a row and all your money is gone, but five losses in a row is nothing. That happens all the time, and that's the only thing standing between that strategy and just starting with big bids.
There's a math theorem called "Optional Stopping Theorem" that states how any strategy that finishes on average in a finite time (so every strategy for a human) can't buy you anything in terms of expected value in a martingale, so either you discover more information that pushes the odds in your favour or you are bound to lose.
@@mileyardgigahertz or you know, just use logic
Same. The only answer I’ve seen elsewhere is that roulette wheels have 0s so the odds aren’t actually 50/50, but that really isn’t the main problem here because the payout is still double your stake. It’s nice to have an actual in depth answer that identifies the real problems.
@@stechuskaktus8318 Is the £1 win the limit of max expected win value ? Because we know we can double our money, about 1/3 of the time, presumably the flip is total loss 1/3 of time and the 1/3 in the middle the limit is something like 1 or 0. Can anyone want to sort the maths numbers for me.
This is a classic exercise in university. Was waiting for you guys to go over it! :) Would've been great with some actual Martingale equations in here as well as the technical proof that we need funds = infinity. Nice video regardless!
I love the videos with Tom Crawford, such a pleasure to watch!
As a long-time Runescape player I've had to explain to many people why the Martingale Strategy doesn't work
Doubling money!
It does work sometimes. In fact, it's 50/50, it either works or it doesn't.
Man, martingale Works with wee use It with a lot of care, only using tree doubling steps
It does if you only do it once. They'll usually double the first 100 gold or whatever to "prove" that it works. So just run away after the first doubling.
Sand casino bad
16:30 It's worth mentioning that the probability of losing all your money is NOT (1-1/e). This is because in the event that you go on a losing streak and lose £100 you get to keep all of the 1£ winnings from your previous successes. If you instead reinvest all of your winnings back into the strategy, the probability that you double your money will be exactly 1/2. In fact, due to the optional stopping theorem, ANY strategy which involves starting with a fixed sum of money, leaving when you have doubled your money, and not overbetting (so that you never have over double your initial money, even if you win) will have exactly a 1/2 chance of doubling your money.
Thanks. I was hoping someone would get into this
Your bets also don’t evenly go into 100. If you start with 100, you are effectively starting with £63. No matter what you are keeping the remaining £37. This also plays into the probability making it a lot less likely to double your 100 than what they claimed in the video
Thank you. The assumption that all the winnings were put aside was not stated. It would be interesting to look at the expected return using different starting sums with this method though.
@@colbygann4674 It was stated that the bets could be easily calibrated to take account of this. However your right that you'd only be able to split your bets down into certain denominations.
Ah this makes a lot of sense, I was wondering if you reversed the strategy and played as the casino (double your bet every time you win, and start at 1 again after you lose) then you'd theoretically have a 63% chance of winning based on what we learned in the video which makes no sense.
Best advice I ever heard is "bet with the streak, or not at all. I have played the Martingale strategy, but with no loyalty to any color.. whatever color came up last, is the color I bet next, with the amount of my bet returning to "table minimum" after any win.
… What? Why would the previous color have any bearing on the likelihood of the next color?
In French, we have a saying, gambling is a way of separating idiots from their money
I did this on Runescape at the Duel Arena back in the day, it's a great way to clear your bank trying to double a bunch of lost bets.
As did I... Sigh...
Should have just DDS spec’d harder
Same, even though I was the casino (had better stats). If you don't have an infinite bank you're bound to get stuck. I did manage though to go from a 500k bank to like a 10mill bank before I got cleaned.
LOL RIGHT? I KEEP GOING ON -12 STREAKS WITH THIS STRATEGY
@@ZaneZephyr no arm, whip only bruv
There's a math theorem called "Optional Stopping Theorem" that states how any strategy that finishes on average in a finite time (so every strategy for a human) can't buy you anything in terms of expected value in a martingale, so either you discover more information that pushes the odds in your favour or you are bound to lose.
omg hi
you dont need a complicated maths theory to know that if you have a less than 50% chance of winning your more likely to lose the more you play.
Great video. He ends the video by answering a question he poses, which is why one would employ the strategy at all when one wants to double their money. The answer is that you might not want to double your money. The odds of taking your million and winning a small fraction of it are huge, which is the key part to him disregarding al the games you win before losing. This means a loss likely actually loses less than if you just put it all on red in one go. It doesn’t make the method more sound, but just explains why you lose about 14% of your “luck”
I've both theorized and tried out variants of this. I tried tripling instead of doubling, and instead of keep going up the ladder I kept betting on the 4th/5th/6th until I won more times than I lost, then starting again at the initial bet. My experience exactly reflects the odds - it trickles along until you wash out. I suspect you can't game these odds in your favor, but haven't done the math on all my variants.
The expected value still is 0. This strategy just maximizes the chances of a small win by amplifying the negative value of a loss. It's the exact opposite of a fair lottery where your chances of winning are slim but the value of that win is greatly amplified.
The expected value is always 0.
If you bet k pounds and you win, you win k pounds.
If you bet k pounds and you lose, you lose k pounds, so you win -k pounds.
Averaging the two to get the expected value is 0.
If probability to double your money is 38,6 percent and probability to lose all your money is 61,4 then the expected value is negative.
@@themasterofthemansion3809 No, you are missing a point here. With the martingale as the guy explained it in the video you are not counting the extra profits you get every time an iteration wins a pound, this is not added to your bank. He did that for simplicity reasons. So this extra money you will end up having everytime you actually lose all your bank will on average account for the extra 11.4% to get you to 50-50. With a fair 50-50 game, whatever strategy you follow your value is zero so doubling your bet is 50-50 whatever you do. The strategy is only affecting the variance not the expected value.
@@themasterofthemansion3809 But it's not what happens here. When you hit the losing stop condition the expected amount of money you won beforehand exactly compensates this negative value.
So you have 38.6% chance of doubling your pot, and 61.4% chance of not doubling it and leave with anywhere from 0 pound to your initial pot.
I would have really liked if he had explained that or even calculated the expected value to show it's 0. Our prof. used a seesaw as an analogon with weights on both ends, where one weight corresponds to winning and one to loosing. The size of the weights equal the amounts you get/lose and the distance from the anchor point the probability to win/lose. In a fair game like the one in the video the seesaw is balanced and different strategies are using different weights at different distances to keep the balance. Like @HammerTh said the shown strategy is having a small winning weight close to the center and a very big losing weight far out, where classic lottery is the exact opposite.
A friend of mine was obsessed with this strategy and told me about it. This strategy got me so curious, that I wrote myself a simple simulation, that would run play this strategy... The result was, that it worked in the short term, but after a while, there would always be a case, where the same color would keep coming over and over again, up to the point, where all money is lost (at the point, where the amount of betting exceeds the balance, the simulation would stop). Then I increased the start balance, but it didn't matter, even when starting with millions, billions or trillions, despite the initial bet was still 1. At some point, everything is certainly lost.
The odds of 26 blacks in a row is 66.6 million to 1.
The odds on 1000 blacks in a row, 64.8 quintillion to 1.
@@lordfarquard9902 A low probability doesn't mean it won't happen... Let the game run for long enough and it's almost certain, that it will happen at some point.
With a software simulation, which executes millions of iterations pet second, it almost happens instantaneously.
@@lars7898 1000 in a row?
Maybe over a few billion years it would occur, but highly highly doubtful tbh.
@@lordfarquard9902 It basically depends on how fast you play and on how big your start balance is. A simulation sacrifices all the money instantly, whereas playing in real life might take ages...
It's probable, that it works in the short term period. The higher the start balance, the longer the initial working period.
But when running the simulations, I noticed that the balance usually maxes out at 1.3-1.5 times the start balance. After that, there is usually a chain of blacks or reds, which makes the player go bankrupt.
@@lars7898 I have a system of starting with 635 and quitting at 800 using 5 as a stake. Basically trying to get 33 blacks before a chain of 7 reds/0s in a row occurs. It’s been successful so far but I’m not delusional.
It will get the full 635 eventually. Just waiting for it to happen. 12 days in a row lucky so far haha
I just nearly failed Calc 2 because I was having trouble with series. This makes me feel so much better about it! Even if I don't truly understand, at least applied versions of it can make me smile for 19 mins
It is this Beauty of mathematics which I love : " to be rich you must have infinite money to start with"
I love the fact that the moral of the story is that you have better odds of doubling your money if you just bet it all on the first bet than you do using this method.
Truth!
The thing is if you bet all of yor money 1 time and you lose you'll have nothing. If you'll keep betting your money with the Martingale strategy and you lose you'll still have some money that you've won previously. So it's evens out
@@Slyzor1 No, you will not. Do the math.
@@NeverTalkToCops1 The strategy described includes not adding your winnings to your stake, so you can only lose all of your money if it happens on the first attempt. If it happens on the nth attempt, then you've already won n-1 times and so still have all of those winnings.
@@NeverTalkToCops1 Yes, you will. If you go through 63 iterations before losing you'll end up with $63 at the end. If you bet all your money on the first round and lose you end up with $0 at the end.
I remember wondering about this as a kid, love that in ten minutes they described what took me years to understand.
You mean 18 minutes.
My analysis of the martingale (or any other) strategy is that yes you can alter the probability of winning, but you can't alter the expected gain. Like if there is a roulette without a zero, then the expected gain is 18/36 - 18/36 = 0% per play, i.e. in the long run you neither lose nor win any money, and that will be so regardless of whether you use a system like martingale or not. And if there is a zero in the roulette, then the expected gain will be 18/37 - 19/37 = -2.70%, i.e. in the long run you have an expected loss of 2.7% of your betted money in each play, regardless of strategy. So martingale changes the probabilities in the sense that you have a big probability of winning a little, and at the same time a small probability of losing a lot. But the expected gain (or loss) remains the same; 0% gain or loss if the roulette has no zero, and 2.7% loss (per play) if the roulette has one zero, just as it would be if you did not use the martingale strategy.
Pretty sure thats not true, betting the same on a 50/50 over and over is guaranteed to win eventually. The strategy, if no maximum bet is in place, does work, and I have personally used it at a 100% success rate of coming out with more than I started.
@@GoddaryuTUBE No. You need to check more thoroughly how these things work.
@@BigParadox Nah it's most definitely a thing in probability, that if you bet the same thing over and over in a 50/50, even though each toss is individual, the odds of getting it correct increases with each attempt. So maybe it's you that doesn't truly understand how things work.
@@GoddaryuTUBE lol, make this thought experiment: A and B toss a coin (50/50) repeatedly. They bet money on each toss. According to your belief both are going to be winners in the long run. That is clearly not possible.
quite literally if you have unlimited money and no bet limits and you double your bet every loss there is absolutely no way you dont come out a winner how you can't comprehend that eventually you will win and a single win even after 100 loses is still coming out an overall winner@@BigParadox
I remember reading this strategy online as a young'un and thinking it was pretty great. Never had a chance to use it.
Thankyou for the breakdown. Isnt maths fun :3
5:32: the zero does not change anything. In fact, the strategy with unlimited money does not depend on the probability of winning at all (as long as it is >0). All that matters is that you win twice your money.
In the real world, a probability below 50% increases the chances that you hit your limit of course.
Thank you! I was hoping someone would notice this.
A Numberphile video about Scott Steiner maths would absolutely make my whole week.
There is a 144.3% chance of that not happening.
@@TheJohnmmullin A video about that would spell DISASTER for YOU
@@lenmetallica what a sacrifice
I feel kind of proud that I came up with this on my own without knowing it was a named mathematical strategy
I've tried this in Fallout New Vegas and it works pretty well. It helps that I can reload an old save whenever I inevitably end up having so long a losing streak I run out of money, but this happens so rarely I'm not slowed down by the built-in anti cheating thing in the game.
I laughed at the idea of this working if you had an infinite amount of money. Because there would be no reason to enter into the casino in the first place.
😂
Problem is you can double your money or halve it anytime, casino or not (:
"Yay! My amount of money got infiniter!"
No sane person goes into a casino expecting to come out richer. So I would say infinitely rich people have at least as much reason to enter as regular folk.
Fun?
To make it very slightly less silly, I think it would also work if the casino was willing to give you infinite credit to gamble with.
If you always use martingale strategy, you will eventually lose all your money no matter how much money you have. Even if you use your winnings to bet.
Thats why you reset to 1 everytime you win
@Oyvind Lie Did you even watch the video?
@@Emetris No. Play long enough, and you will eventually have a streak of bad luck sufficient to either bankrupt you or take you past the maximum allowed bet.
No, that's not the take away. You're odds are 50/50. No matter the strategy. Some people will use this strategy and be fine. And some will lose it all. Regardless of how long you play for.
No, you won't live forever. Your odds of winning are equal to losing if you have infinite money and finite time.
Great video, smart conclusion! Keep them coming ;)
thanks tom great strategy, can't wait to make some dosh from my student loan in april!
This reminds me of the St. Petersburg paradox in economics. I would have love to see a video on that. Or videos related to (mathematical) economics. There are many interesting problems and paradoxes in that field that would be in line with numberphile’s scope of interest I think.
As soon as I saw the 1/N and raised to the N power, I knew e was involved.
Yeah - my suspicions were raised as soon as he said 0.366 (i.e. approx 37%) for the first case. I was sure 1/e was coming. It comes up all over the place in these probability/strategy problems.
For example, I think it's the probability (limit) of no-one choosing themselves at Secret Santa. Also, it comes up in the "Secretary Problem".
I first remember encountering them during nuclear decay and stuff. Nowadays the 1/e and 1-1/e appears everywhere.
@@matthewcooke4011 That (1+1/n)^n converges to 1/e is actually the reason you see the e^(-m) in the probability function for a Poisson distribution e^(-m)*(m^x)/x!. Plug in m=1 and x=0 and you get that the probability that nothing happens over a time period where it on average ought to have happened once in that time period is.... wait for it... 1/e :)
I just want to say that I appreciated the brilliant thumbnail art.
Really like how this is relatable to, not just abstract maths.
it would be an interesting kind of mini series talking about the math behind various casino games :)
Slot machine: the computer choses what it gives you. They can even be reprogrammed to give better or worse average output.
@@Lulink013 also, the operator can and will reprogram the machine accordingly
Check out minding the data, he makes videos like that
It's like picking up a dime in front of a steamroller.
This brings me back lol, I remember back in hs when someone brought up with strategy and I proved that the expected value is still the same.
I had thought up this same exact roulette strategy on my own some years ago but reasoned in the long run it would be untenable . I knew that almost certainly I wasn't the first to think of it, but it's cool to finally have a proper name I can refer to it by.
Wow, its really awesome how detailed you think about this strategie, thanks for the vid!
I've tried this strategy at the duel arena on OSRS when I was younger. Thought I was outsmarting everyone. I lost nearly everything, it was very sad times
This video was fantastically clear to understand and very entertaining. Well done! :)
If you do consider the pool of winnings that you put to the side, I believe the expected return tends to 0, not a loss (in the 50-50 casino). Otherwise you could just reverse the strategy and give the casino unfairly bad odds (doubling if you win, resetting if you lose).
I'm awful at math, literally one of the worst things I'm at but since starting to watch this channel ice started to atleast get interested in it.
Love this guy and his enthusiasm, keep it up Tom!
@@TomRocksMaths you are more precious than knowing negative number
Amazing video! The flow is just superbe
I really wasn't gonna watch this video to the end, once it got started I couldn't stop😳 This was super fun to watch 🙌🏾
Finally a video on this to show to my brother who keeps insisting "it just works"
if he didn't figure it out by himself i doubt this video will help him
That kind of person is a rarely convinced by a reasoned argument.
Either your brother is really lucky and it does seem to work for him
Or you could just offer to play against him.
Is he rich?
Imagine taking exams with mathematical formula tattooed on your body? 😎
I mean he has his PHD already, so he probably doesnt really do exams anymore ;)
Also, e is preprogrammed in most calculators. The navier stokes equation he has could be of more use but he literally wrote his thesis on that so I guess hes got it down already.
Could come in handy for some students taking exams with him watching though
@@kcbsuiejd jajaja
He's not allowed to play Pokémon Crystal anymore too
we did imagine it at our university and now it's in the regs: all tattoos must be covered for the duration of the exam. We even (allegedly) provide the exam invigilators with a box of plasters to stick over tattoos. You can never be too prepared or too cunning when it comes to university exams.
Great video, i remember reading about this strategy. Nicely explained here.
The one thing i don't understand is if you took £N with you, after a few losses you would no longer have £N available to bet.
Does it just mean that £N is the single-bet limit imposed by the casino?
I got the impression they were switching between those two ideas of what £N represents (i might need to rewatch!).
Hi @numberphile - do you think you could do a compilation video covering some of the most mysterious and wonderful parts of maths, the interesting oddities and features of maths? Specifically for kids/teens etc, to show them that maths is actually really interesting beyond the basics?
This is the best explaination of the subject I could have ever wished for. Thanks alot.
I love the animations! And the maths, of course :D
it doesn't matter whether the odds are 50/50 or anything else (e.g., 20/80), it might only make longer the time you might have to wait to win. But in any case you will win the initial bet, once the color you bet on will come out.
I thought it was unrealistic to have the rule that you can't gamble with your winnings, so I figured what the probability was to double your 100 pounds assuming you could touch your winnings and... ...yeah it's 49.75%. Still better just going all in on your first game.
Isn't it 50%?
Thank you for the informative video❤️❤️
The maximum odds one can have playing roulette would be one's first game of roulette. Therefore. I plan to play roulette once in my life.
When are your plans...I'll join
and before we do that, we want to earn as much capital as we can!
@@ethancheung1676 up to the highest possible maximum bet... though I also plan to use a casino promotion for free chips because they often do those and it can "hedge" the bet a bit
It doesn't have to be a 50% game. As long as there is a nonzero chance of winning doubling losses will always come out ahead whenever a win occurs. It will take longer on average to get to a win but you will come out ahead.
I love Tom enthusiasm.
This is just like the many supposed "paradoxes" that basically amount to not understanding calculus. Yes, you can get arbitrarily close to 100% chance of winning 1, but at any point, whatever probability remains is equally close to losing infinite money. These particular infinities exactly cancel out, to make your coin flip the same as any coin flip.
It's not a paradox. If you have infinite money, you will keep winning fresh dollar bills. The only issue is that its not physically meaningful to have infinite money.
Jeremy,Please change ur profile pic.
I’ve always been fascinated with this strategy at the sand casino in RuneScape
It should also be mentioned with this strategy, that once you do hit the win you should restart the sequence with your initial bet - rather than continuing where you were
Any one that didn't get that deserves to lose all of their money
Ooh, cool, I recently learned about this. Doob's optional stopping theorem forces one's expected winnings in a fair game to be 0. It does not prevent some (mis)fortune, though. Simply put, if you play a fair game long enough, no matter how you choose your betting strategy, your expected winnings are 0. In other words, the house always wins.
I would love to see the math for reverse Martingale, ie, doubling everytime you win instead of on a loss.
it's exactly the same math as the regular Martingale, except you change the signs: winning becomes losing, and losing becomes winning.
The math is identical. It's just that now you're asking, "If I'm a casino, what are my chancing of making $X if a gambler employs reverse Martingale?"
Keep in mind that with 50% odds, every bet is symmetric: When the player bets $1, the casino is also betting $1. You could model this by having both the casino and the player place $1 on the table, with the winner collecting the $2.
Funnily enough, even if the odds are 60/40 in your favor in this game and you double your bet everytime you win, the expected value of each bet is always greater than zero which means that to reach max profit you should never stop playing, but if you never stop playing you will eventually lose all your money with a 100 % certainty.
That seems like a really bad strategy given that you won't know how many times you lose between winning. With the normal Martingale, the increase on loss means your wins will always put you back in the positive. Reversing that means you could lose enough times in a row that winning, doubling your bet, and winning again may not be enough to even get you out of the hole.
For instance, say you start with a win streak. You bet 1 and win, so your current profit is 1. You double to 2 and win, so your current profit is 3. Say you get to betting 16 before you finally lose. Thanks to the math of the Martingale strategy, we know you'll now be 1 in the negative, so all your winning was for nothing. Now say you lose a few times in a row; the losses will continue digging the hole without increasing the bet, so you get down to negative 4. A win from here will only dig you out by 1, so even winning again the very next round (betting 2) will just get you back to being down by 1.
Essentially, the Martingale strategy isn't a bad strategy for small gain (you'd need a bad losing streak to wind up with nothing), it would just be difficult to go so far as doubling your money with it. Trying to reverse it, however, would be a fast way to lose all your money, even while you're winning half the time.
@@granite_planet Thanks for the broken brain.
There's a small context error in this video, about the actual number of losses afforded to the gambler. In each case that was described in the video, the gambler has a total amount of money that they are bringing to gamble with, at 100, 1000, etc., while the number of losses is simply the number of doubling iterations it takes to get from 1 to n, where n is the largest integer under the previous stated amount of money. But after each doubling, your pocket of money should be decreasing, as you are continuously losing the previous bet in order to get to the next doubling integer.
In the case of 100 pounds, the actual number of losses afforded for this strategy is 6, not 7, because after 6 losses, you would be down 63 pounds, and not have enough funds to make the 64 pound gamble. This also means that effectively, having that extra 37 pounds won't make a difference towards your betting strategy, so n would actually be 63, meaning that your overall chances of success the first time around to win 1 pound would be slightly lower at 98.4%.
What is interesting is that I would have thought that the constant that the odds of doubling was going to trend towards was .5, but instead it's e. I think an explanation for why this is the case would be an awesome follow up video. Tom does talk about how e tends to show up all the time when it comes to gambling, or interest growth.
@@xIPatchy Yeah, it seems weird to me as well - I would have expected the chances of doubling your money to always be 50 % no matter what kind of tricks you do with your betting. I honestly don't understand how just betting with a changing amount of money manages to lower the expected value of total winnings for the night. My math intuition says that everything should cancel out and just approach 50 %.
@@xIPatchy I think it comes from "how many times would you need to play how many times to ..."
@@granite_planet I had a doubt
If 2 players with equal initial amounts bet on opposite colours each time, then each of them should have equal probability of doubling which should be 0.5
But after some thought I realised that not every outcome results in a player losing all money which goes in doubling the opponents pocket. So even if the player loses 1 bet, that's not considered in this probability so we exclude those cases.
I think u had the same doubt. Hope its resolved
@@granite_planet Maybe it has to do with the fact that with this betting strategy, you are always angling towards winning an extra pound, vs simply betting the same amount every time would average out to being 0 sum. I don't understand it either.
There is a flaw in the "double your money" reasoning. Even though the probability of doubling your £100 is only 36% it doesn't mean that the probability of losing £100 is 64% because you are squirrelling away your £1 wins each time you win a bet so your total losses could be anywhere between £1 and £100 depending on when your losing streak hits. If you calculate the expected value of your winnings on a fair even chance game using Martingale, it still turns out to be 0. (Calculating the expected value if there is a house edge is a different matter).
Thats really interesting. I've actually used this method not knowing about it. In online gambling on an MMO, staking on runescape, made a decent bit. Also wasn't a pure 50-50, used methods to gain 4-5% odds.
I guess I wasn't the only one screaming e in my mind when I saw (1-1/n)^n.
That's not even the formula for E. The correct formula is the limit as n approaches infinity, of the quantity (1 + 1/n)^n
@@NeverTalkToCops1 yup I know. But it's clear that it will be related to e.
Great video and Tom is really amusing to watch. Funny enough this strategy is often advertised for actual financial trading by scammers, specially in binary options and forex world where there's even transaction costs on top of that lol.
I think there is a tiny math mistake here: At 10:57 the probability cannot be 1/(N).
If N equals 1, you lose all your money if you lose k=1 times in a row. This means the probability has to be 0.5
In my opinion the mistake originates at 8:21 - You can already lose everything earlier, because you already paid £ 2^k-1 after k steps. This means you can only bet £ N-(2^k-1).
If you take this into account things get quite messy because N can be an arbitrary number. To make things easier, i assume that N is of the form N=2^x-1.
Then, your game is over when the money you already paid is equal to your amount of money N which leads to the following inequation 2^k-1 = N.
If this equation is used to calculate the probability for k losses in a row, you get P=1/(N+1) which gives you 0.5 for N=1
but the k is counting.. the losses.. so if k=1.. he already lost (his 1 coin)
You are saying: If I've lost k=1 times in a row, i've lost all my money (1 coin) with the probability = 1.
That's totally correct, but the formula at 10:57 (P(Lose k times in a row) = 1 /N) says something different:
It says the probability to get to this point (lose k=1 times in a row) equals 1 / N = 1 which is not correct.
It’s been a year since u posted this but you’re definitely correct. It definitely should be N=2^k-1
They glossed over smaller details in the interest of time. For the formula to be exact, N would have to be a power of 2
But their main point was to show what happens for large N, and what the pattern looks like
玉林荔枝狗肉节
Not having infinite money; my only weakness
Great video!
However, in practice you wouldn't lose all £100 when you don't double your money. You would also have between 0-£99 left in your pocket from the previous wins.
Hence the 38%+ number. The remaining 12%- number accounts for the money you put aside to bring the true odds back to the starting 50/50. Ultimately what this strategy will do is allow you to win a smaller amount more often and lose a bigger amount less often. In the end, it's always 50/50.
Which of Tom's tattoos don't have a Numberphile video yet? We want 'em all!
...sure? ;-)
Or at least the mathematician ones.
you can multiply by 3 to get higher profit the longer the round gets.
also when you DCA the stock market (or a single stock) you can use the martingle strategy without actually locking a loss when it's not going your way.. just lowering the averge entry price.
This is a video that truly deserves two thumbs up from me.
I found this really interesting. Can you do more around gambling, especially poker. I know it’s not a perfect knowledge game, but it would be cool to understand stuff like, what’s my chances of hitting a flush on the turn or river if I already have 4 of the same suit?
there’s a lot of great poker math videos out there to cover basics like that- it would be interesting to hear them explain nash equilibrium
7:21
"If at any moment we need to bet more money than we have..."
That one is easy, you bet the house.
I was a specialist in the Casino Industry for 13 years, specializing in Tables, Slots and Surveillance.
Table limits render this strategy useless. We actually tested this on numerous occasions, and in the end...the house always won.
I developed this in high school and was obsessed with it for about a week before I determined it's downfalls (you'll always lose it all unless you have infinite money)
Later I found out it was a legit strategy and was proud of myself
Had a losing streak of 13 losses while martingaling once. It was a tiny amount of money, but it definitely made sure I never gambled tangible amounts of money ever after.
I've lost at least 13 times in a row betting on 25 cent video roulette. I was winning for a while until I went on a huge losing streak.
At the end, when you don’t have enough money to make the bet, you might still have some money. If we change the strategy to bet everything you have left at that point, your probability of doubling your initial money goes up to 50% (from 37%). The reason why it’s 37% is because you stop playing when you still have some money left.
it's the 37$ you have left.... 1+2+4+8+16+32 = 63 100-63 = 37
I've done this a few times in Vegas. Basically, every time you win, you gain whatever your original bet was (I usually start with $3 since most Vegas roulettes have that as the lowest minimum). The problem is, most of those $3 min tables have a max of, say, $100. If you lose 6 times in a row (so a $96 bet), you've hit the max and you lose money even if you won the 7th spin (you'd win $100 on that roll, but you lost $189 on the previous 6).
In summary, if you had unlimited cash and a no-limit table, you always win money (it's just a slow process).
I’d liked to see this for other bets. Like where you bet the column for a 2:1 payout instead of 1:1, but where you only have a 1/3rd of a chance of winning.