We all know that it doesn’t pay to gamble,

but a week ago, I heard a paradox that seems to show the opposite- that you can earn money

whenever you gamble. Let me show you how it works for a particular

game- but after I’ll show you it generalises to any. The game I heard it in the context of is Roulette. Apparently there’s a way to play the game

where you just need to guess either black or red- both have an equal probability. You bet a certain amount of money and if you’re

right, they give you the amount you betted. If you lose, you pay them. Simple. So convince yourself that this is a game of

pure chance, if you played this game a lot you wouldn’t expect to win or lose anything. By the way, if you played this in a real casino

the odds are slightly against you so that they make money- but that doesn’t make a big

difference to this paradox so let’s look at the fair game for now. It seems like there is nothing you can do

to improve the chance you make money. But let’s think about this strategy. What if you said, I’ll always bet red. First time I’ll bet 2. If I win, I’ll stop. Otherwise I’ll play again but this time I’ll

bet 4. And if I lose again I’ll bet 8, etc, etc. Then say you win on the 3rd round. You win 8, which covers all your previous

loses. The beauty of this is, you will win eventually,

and whenever that is, you win back everything you lost and then some. You might say, yeah but the amounts of money

you’re betting every time get huge quickly. And that’s true. But it’s pretty unlikely you’ll even lose

many times in a row. The probability you’ll lose 3 rounds is 1

in 8. The idea behind this strategy is to chase

your losses- you know, that thing that people tell you never to do. If you lost 3 rounds then you’ve lost 2+4+8=14. So you bet 2 higher than that total loss. That means that if you win this round, you

get back everything you lose, and 2 more. You can check this yourself but this works

for what ever round you finish up on, you bet your total losses so far plus 2. So whenever you play this game, as long as

you go through until you eventually win a round, you always make 2. But this seems like a contradiction- you aren’t

supposed to consistently make money playing this game. You’re supposed to break even. It gets worse than that, because you can apply

this strategy to any betting situation, now matter the odds. You just bet a little bit more than all your

total losses so far… But wait, before you bet on your house, you

can see that this is clearly stupid for most betting situations. Most of the time, bets are stacked well against

you. Let’s say you’re playing a game where you

have a 1/10 chance of winning each bet -which is not at all unrealistic. Then the chance you’ll lose even 12 games

in a row is still very high, about 30%. But at that point you’ll have to spend about

$8000 and still are likely to lose more. So yeah, don’t actually go do this in real

life please. This stratergy doesn’t actually work- it’s

obvious why not when the odds are like this, but when the odds are 50/50, like in the roulette

game, it’s hard to see what the problem is. That’s where the paradox is. I’m going to give you a week to see if you

can figure it out. It took me a lot of frustrated hours with

a friend until we cracked it, but it was good fun to think about which is why I wanted to

tell you guys. It’s a very interesting paradox. And if you do figure it out, here are some

other variations you might try. What if you decide not to be too greedy, and

you only bet your total loss plus 1 each time, instead of 2? Or what if you get super greedy, and you bet

your loss plus some growing amount? In those two scenarios, see how quickly the

amount you bet grows, and what your expected winnings are. If you need some hints, I have some small

ones in the description. Good luck. Try it with a friend so you can discuss it-

I highly recommend it. Oh and, promise me one thing- that you won’t

go gamble this week. Thank you.

This works but if there's a gambling limit then you can't do it. I did it at a small fare with the lowest denomination of money at the time, a penny, and it worked, but it was such hard work and always the threat of they not accepting the bet as they see what you're up to. Exciting though.

1st you dont have infinite amount of money that is needed to play the game and this is REAL problem. You will be winning very small amounts of money (2) and risking more and more, even slightest chance that u will loose all (due to shortage of money) will eventually come true and additional money that u won wont offset the loss.

total loses plus one just means that you started by betting 1 and not 2, you'll make the same bets 2,4,8,16 ,. so on , only difference is that you will win 1 by 1

No paradox here! I'll see you guys at the casino!

When taken as a set, it seems that your chance of winning tends toward 100% each time you lose. However, each round is its own entity consisting of it's own probabilities. ie. a 50/50 game will always be a 50/50 game even after 99 losses.

Now it may be possible to make a small amount on online gambling where each round is part of a single psudo-random seed, assuming you have lots of time and money to use, you also have to be sure that the seed wont change in between games.

Expected gain in N games is 0 and is obtained as the sum

0=P(win)x2+P(always lose)x(-2^(N-1)).

Since P(always lose)=1/2^(N) we think it is negligible but the amount of money we would lose is so huge (2^(N-1)) that the product P(always lose)*(-2^N) is almost -2.

This is called the martingale method. This would work assuming you have an unlimited amount of funds and no maximum bet limits… conditions which unfortunately exists in all casinos.

Here are the problems with this technique as I see it:

– It's only guaranteed you will eventually win if you (and the casino) both have an infinite amount of money, and there is no betting limit. Otherwise…

– You are basically betting all the money you do have for a chance to win just $2. There's a high chance you will win the $2 if you have enough money to go more than just a few rounds, but if you do lose, you lose everything.

If you want to blow some minds, do an episode on the "two envelopes problem"

https://en.wikipedia.org/wiki/Two_envelopes_problem

I just worked it out myself and I didn't expect to see that the expected winnings would always be negative if there any maximum betting limits. For this one does have to assume that your probability of winning one game is less than of equal to 1/2.

The casino has limit and you dont have infinity money, so sooner or later you will run out of money and every round of rullete is independend on the other one, so you could possibly lose 20 times in a row, because the event that happend before doesn't determine event that happends after even if the probability that you lose 20 times in a row is small. Every game of roullete(spinning the wheel) is independent event.

[Bonus Challenge spoiler]

"Loss plus one" strategy

How does one's winnings increase if one bets with the following strategy:

Bet 2

Win: Leave with net win of 2

Lose: Bet 3 (loss plus one)

Win: Leave with net win of 1

Lose: Bet 6 (total loss plus one)

Win: Leave with net win of 1

Lose: Bet 12 (total loss plus one)

And so on…

As one can see, the net win is 2 or 1.

"loss plus increasing amount" strategy

Bet 2

Win: Leave with 2

Lose: Bet 3 (Loss plus one)

Win: Leave with 1 (net win)

Lose: Bet 7 (Total Loss plus two)

Win: Leave with 2

Lose: Bet 15 (Total loss plus three)

And so on…

Interestingly, the more one loses, the more one wins with this strategy when they win, assuming the chance of winning is nonzero.

I dont see a paradox because this does work.

Rulete isnt 50-50 there are 1 or 2 green spots (0) and (00) depends where u play. So at best your ods are 18:37.

second thing is that casino tables have limits lower and upper for betting. u cant bet infinite amount of time. so when u hit upper limit u lose everything.

that's why you cannot bet more than 500

You may have high probability of winning but if you lose, the amount you lose will be significantly bigger than the amount you wanted to win.

If you try to make a lot of money by resetting the amount you bet every time you win (10:lose->20:lose->30:win->10:lose->…) to hack the game, you will eventually lose once (say you can't bet more because you run out of money) and it will cancel all your gains.

Love your videos! What would you say to someone who was thinking of going back to uni to do a degree in Physics/ Aeronautical Engineering after a 8 year hiatus? Need advice. Can't do everyday jobs… The game of life is fun, no mater how it turns out I guess ðŸ™‚

often times, people making bets will call them off after only a few or maybe even just one attempt.

Thanks for the awesome life hack!!

I know this one! ðŸ™‚ In any scenario with limited amount of money, sooner or later you are going to hit a losing streak long enough that will make you lose your whole stack. That will leave you dry and unable to recover your previous losses. Any loop that you eventually win will only make you your "base bet" (2 coins in the example), but to be able to afford longer losing streaks, you need exponentially more money. If you start off with 6 coins, you can afford to lose once, but then you have to win. In order to be able to lose twice in a row, you need a winning loop four times in a row to accumulate additional 8 coins. To survive 3 losses in a row, you need another 16 coins.

A few years back I even made a spreadsheet with 10000 random coin tosses (0 and 1 values) to see how likely you are to get k-long sequences in n tries (which represent losing series if say you keep betting black but the wheel keeps coming up red). The RNG produced longest streaks of 17 and 12 consecutive values for 0 and 1 respectively. Quite in agreement with prediction, as you can expect a k-long sequence of zeroes or ones in 2^k tries long chain (conveniently, 10 000 is 2^13.29… I really have no idea why I chose that number at that time). In other words in a 2^k tries, you are as likely as not to encounter a k-long streak (the longest losing streak you can afford). As the number of tries grows, the probability creeps up to certainty.

But if you have unlimited amount of money, and find a casino (with finite amount of money, so that we can avoid the infinity-on-infinity scenario, and the casino will eventually go bankrupt) that will let you play this system, then go ahead. But it would seem kinda waste of time… if you already have unlimited money, what is the point of playing?

the 'house' loves the 'martingale' player. as you've suggested their are barriers – the house puts limits on you – your bank roll has to be significant and as you pointed out that if you only wanted to win $1 and hit a losing run so by play 20, you are putting over $1,000,000 on the table — by bet 25 you are up to $33,000,000 … for $1 — maybe good advice not to use this strategy ðŸ™‚ .

1. the roulette is not 50/50 there is 0 and on some 00 too.

2. this strategy is winning only if you have an infinite amount of money to gamble (or at least, more than the Casino) or else the game ends when your wallet is empty, independent of the amout you have to gamble next

3. the majority of the roulettes put a limit on the maximum amout that you can gamble, when you reach this limit you can only lose

and think as the Casino has an infinite amount of money

I bet you are wrong, or right… Doubling my previous bet and adding 2 every time I lose. Wanna bet?

I started by limiting the process to only a particular number of recursions. If you look at what would happen if you applied the strategy 3 times in a row before restarting, this is what you might find. If each stage has a 50/50 chance of paying out a net $2 or continuing to the next phase, that means you have a 50% chance of winning on the first phase, 25% chance of winning on the second phase, and 12.5% chance of winning on the last phase, leaving 12.5% chance of losing. That means you have an expected return of 7/8 x $2 – 1/8 x $8, because $8 is what you bet on the last round before giving up. This adds to an average return of $0.75, meaning you only need to bet a maximum of $8 to make a profit on average.

Now the explanation of why this doesn't work. It's true that you have a 12.5% chance of losing the $8 bet, but the net loss when you make that loss is actually the sum of the previous 3 bets, $2+$4+$8, so in actual fact you have an expected average return of 7/8 x $2 – 1/8 x $14, which is equal to $0, the same as the simple 50/50 game. You can generalize this to use any number of rounds, the average win is equal to (2^x-1)/(2^x)

2, and the average loss is equal to 1/(2^x)(2^(x+1)-2), and these expressions are equal for all values of x, meaning this strategy will never be profitable, even infinite money available to bet with.I found the answer! It's funny because someone told me this paradox 4 years ago and it baffled me. I kept it in the back of my mind and never thought of it again. But it's even not that hard, sometimes you need a deadline to do something ðŸ˜‰

Do you ever think about making your Reddit active?

I think this was based on a fallacy produced by our cognizance, that past results influence future results. For example if the probability that you get red is 50/50 and you got black three times before we perceive that the effective probability of getting red next is higher than when we started, because we perceive our own string of spins as a single unit or continuous action, and we analyse it as such.

"I love deadlines. I love the whooshing noise they make as they go by." Douglas Adams (my idol)

It seems like you're just playing a reverse lottery. Get paid $2 for a really small chance of losing $1,000,000, rather than the other way around. Since almost everyone loses money on lottery tickets, almost everyone wins money with that strategy. You're just risking being burnt really badly.

* Spoiler Possibly **

I wrote a (BASIC) program that employed the double down strategy in a computer science class back in the early 90s. I had heard about this strategy from shipmates in the Navy. If I recall correctly, you'll need about $45,000 (or was it $450,000) to confidently bet with this strategy.

A bit late I know, but this betting system is called a Martingale, it's three hundred years old, and it doesn't work. The log of your debt less your actual winnings follow a brownian path which crosses zero sqrt(n) times on average but also gets to sqrt(n) in magnitude on average over n trials. trouble for the gambler is that his losses vastly exceed his winnings on average, so the house can cut and run when they feel fat.

The nice thing is, for a while, online casinos had a limit on the number of times a color could be rolled in a row on roulette. So you could pull this out. The only problem is though nearly all online casino sites were scams to get your credit card info ðŸ˜€

Since every bet is independent of the other the chances for you to lose every time is 50%, so there is no let go by gods of chance to let you off the hook, without you getting "lucky". You have as much chance of winning after 8 rounds as the person playing his first round. Experience of losses or even victory shouldn't determine your next result. Hence you need infinite capital to break even or +2. So if you run out of your chances at the Casino you come back the next day and start playing again. The markets will ensure you unlimited supply of money by printing or giving you back your losses (as losses goto banks via casino and they give it back). If you can make infinite binary bets in an instant then you break even and banks earn their return :).

A roulette table with two green spots (0 & 00) offers the worst odds for the player of any table game in the casino with an insane 5.26% house edge. You'd be much better off studying basic blackjack strategy guides, which when mastered, can reduce the house edge to as little as 0.44%.

I have worked several years in a casino. I actually know of one bet (and there is only one) that offers the player a true-odds payoff with no house advantage. If you can name the game where this bet can be made and the two different ways on this game where this bet can be made, you will win a prize.

wrong, it pays to gamble. should. and take risk.

everything can be viewed as gamble actually

I figured this out independently while watching the movie the Gambler

The most you can win in any series is 2. However the most you can loose is everything. All that you've done is shifted the odds to win more often or loose everything.

What if you had an infinite bank roll? You would need to be immortal as well.

What if you were immortal and had an infinite bankroll, well, not that infinity is valid in the real world but 2/âˆž = 0

It would be interesting if someone graphically, showed a Monti-Carlo simulation with a graph.

You are sooooo Dumb,….

Excellent!

only 1 out of 1,000 Youtubes that tells you the truth about Roulette

Congratulations!

What gamblers don't understand is the difference between odds and probability.

On a French wheel, there are 37 numbers

the odds of getting any number are all the same 1/37

that's correct

So when we spin the wheel 37 times

all 37 numbers show up? 0 – 1 – 2 – 36?

no – you know that

in fact, 13 numbers will not spin out –

Math says that in the next 37 spins 13 numbers will not spin out

(36/37)^37*37 =13.4

why?

Because there will be duplicates, triplicates, quads, spun out instead of them

In the next 100 spins, 6 numbers still won't show up

(36/37)^100*100=6.5

in the next 1,000 spins, 1 number still won't show up

(36/37)^1,000*1,000=1.3

Now you see that the odds are based on math – verified by billions of spins of millions of Roulette wheels

This is a perfect example of math telling you that THE WORST BET IN THE CASINO IS SINGLE NUMBER ROULETTE BETS

So here's a little test for you:

You walk up to a Roulette table and see 10 Reds just spun out

You buy your chips and which would you do:

1) Bet all chips on Red

2) Bet all chips on Black

3) Order a drink and take time to decide what kinds of bets you want to use in this session

#3 is correct because the odds haven't changed for Red or Black and there is no rush to put down a bet for either

Great article – congratulations

Einstein came up with this. It's called the Martingale system. John Scarne, gambling expert, points out that it is very common to have red or black come up eight times in a row several times per day, and with the minimums and maximums, as others have noticed, this will take you to the cleaners– like EVERY gambling system.

ItÂ´s all said. In addition:

Ignoring casino limits, in fact, you donÂ´t need an infinit amount of money for this strategy, a very, very large amount of money is enough. Why? Because a very, very unlikely event is actually impossible in human life time.

Nevertheless, the necessary amount of money is way too large for a normal person.

But Central banks are able to play this game successfully via commercial banks, because they can print every amount of money they need. And, blieve me or not, they are playing this game successfully in real life. The lender of last resort …

I once did the math for this, and it turns out that if you take into account how much money you have to bet with, and if you assume that the casino never runs out of money (a realistic assumption i'd say), eventually you'll lose it all. The odds are still stacked against you. Having more money to begin with only delays the inevitable loss.

Martingale? Plus there is a '0' in the wheel where you lose on both options.

Because you'll win eventually. After you won, you've only won two. Then you bet those two and lose so now you're down to nothing.

The people who understand why this doesn't work demonstrate why the so-called "gambler's paradox" is better known as the "gambler's fallacy."

So waht's the answer ? I read comments under here and none covered the question. The upper limit in bets is not a problem here since we don't bet more than 7 euros every 3 rows.

Even with 300 losing tosses in a row (which never happened in casion's history), you would just need 700 euros. So what's the explanation ?

I know how to win at gambling: don't play the games, bet against the people playing the games! These people just love to gamble, and you can use the house advantage in your favor.

This is not a paradox because probability says you are not gonna win any money is based on the assumption that you have to keep playing the game again and again until infinity.

But the core concept of this strategy is that you must stop at a certain point, and do not play any more.

If on the other hand you apply this strategy but keep playing again after winning, it will be true that you are not gonna win any money in the end, which is of course inifinity.

There's nothing stopping you from losing forever

Just found your channel and I think your content is really interesting ðŸ™‚

Casinos have maximum bets for this reason.

I also did this with a friend and spend some time on it some years ago. The paradox comes from the assumption that you have an unlimited amount of money to chase your bet with. And you don't.

Let's say you have a 1000$ and bet 1$ per round. As you said you will always get 1$ with this strategy per round after doubling the bet it enough times. It very unlikely that you will lose so many time in a row that you can't afford yet another bet; but the chance is not zero (in my case it's about 10 times 1$ * 2^10=1024$). But earning 1$ from a pool of 1000$ is probably not that interesting, so you want to play several rounds each earning you 1$. And the more times you want to make one more dollar the more likely you are to run into that small probability to lose enough times in a row that you can't afford to double your bet, thus losing at least half you pool.

I can't remember if we did the math, but did computer simulate it. As I remember it the odds felt sort of like playing normally.

So, if anyone's interested, I threw together a simple python script to simulate this betting game. https://gist.github.com/jonnyrobbie/a5bfc6de378bc1a8edc10c8d5f08728f

The casinos set their payouts for games of chance based on the odds of winning. You use roulette as an example, but even with the best odds of any game it is not 50/50 because of the 2 green slots. You need to find a casino that lets you bet on coin flip and pays out 2:1…

…like the one in National Lampoon's Vegas Vacation. Otherwise your theory won't work in real life.

Black or red has equal probability, but equally losing probability. Hence, no way to win in the long run. It's not 50/50. It's slightly more than 50 in favor of the casino.

Best strategy if you ever gamble:

Stop if you have won, and never do it again unless you are willing to lose your earnings ðŸ™‚

(Added bonus: you look bad ass if you leave the poker table after you've won by bluffing)

Your voice is very sexy.

Gamblers fallacy

obviusly you have never been to a casino

Just a tip: past tense of bet is bet. Otherwise, great video.

That strategy is called a Martingale and it ultimately doesn't work. In Forex trading, price can only do 2 things; it can go up or it can go down. You have a 50/50 percent chance of guessing it every time. People have applied the Martingale strategy and people have failed. Price never performs the way you want it to or need it to when your money is on the line.

You could lose all your money

This only works if your goal is to make 2 dollars, it's not worth the risk, I've seen it go one color for 12 spins in a row, which if you're betting only 1 dollar would be $4,096 buy in to profit 2 dollars, you would eventually win, but it's a stupid risk to make only 2 dollars, also the table limits would have stopped you long before you even got to that point. If you win then stop it could work theoretically. If everyone did this and stopped it could screw the casinos, but people like to play longer, make more money and change up their bets, so the house will win.

you are an idiot. this betting system is called the Martingale and it does not work

What a load of shit.

There is no paradox. The system works. What fails is people who don't do the math and take the necessary precautions. For instance, the amount of money you have determines your stop loss. The stop loss is the limit you impose to determine when you will accept a loss in a losing streak. When you hit that limit, you simply accept the loss and start over again as you would if you had won. Eventually, you will make back your loss if you don't keep hitting your stop loss which you won't do if you've done the math and calculated things correctly. In addition, you don't just bet on a single color all of the time. Instead of betting black 8 or more times in a row, bet black 4 times in a row, then a red, then black 4 more times in a row. Then bet red 4 times in a row, then a black, then red 4 more times in a row. Repeat. This makes an eventual loss far more unlikely to occur twice in a short amount of time before you've recovered your loss from hitting your stop loss. Again, the amount of money you have will determine the string of losses you can allow before imposing your stop loss. The lower the value of your initial bet, the better your chances. You can survive a greater string of losses starting with a dime rather than with a dollar or more. There is also a limit as to how long a losing streak can be by mere chance. Find that limit and have the amount of money required to survive it and you will never hit your stop loss.

It is the Gambler's Fallacy, in an honest game of chance, no matter how the probability comes out, it is the same every time: Your wins and losses do not influence the probability at all. In state lotteries, the state runs it like a bookie, they get a percentage (expect up to 50%) to run the lottery, thus it can be honest, and we should expect the honest bookie who is getting paid, because he is not gambling; therefore, what I do is pick one set of lottery numbers and stay with them, because no matter the draw, it is the same probability of winning, no matter how many times you lost or won.

If there is no Zero/green number in roulette the expected long term value in using a martingale system is 0%. You're not expected to earn or lose anything on average. So there is no benefit. You'll probably earn small amounts for a while then lose everything you've gained. Normally there is a zero or also a double zero which means the average roi/expected value is less than 0%.

Roulette wheels have a 0 and sometimes a 00 as well which are colored green. This gives the house an advantage. Roulette is one of the worst if not the worst casino games. The house has over a 5% advantage.

Its illusion that big lose series are not often happens, each try is independent so there is always this same chance to get particular color and it is 50%

It would work if you stop at your first win or you have a can of red paint or black.

Return a the school. Look in your teacher paper. This system doesn't work.

If you think this takes too much cash to keep up, check out the Numberphile video "Infinity Paradoxes."

Actually, the numbers are 0-48 and 0 isn't considered as black or red or even or odd(in roulette). So the chances are a little less than 50%.

i dont understand why this is a paradox..its a completely different story if your odds are 1/10 vs 50/50. I personally have found a lot of success using this strategy at 50/50 odds(albeit not with real money). This is like saying why is it that a 8 gauge steel plate cant stop a bullet but a 3 inch thick steel plate can. just completely different scenarios why is it a paradox

On a 5/500 table, you hit table maximum after 7 losses in a row. The chances of that happening are 1 in 89.38 "rounds," at a total loss of $635….the 8th round maximum bet is $500, so you will still be $135 negative even if you win that one. Consistently doing this means that you have 88.38 rounds of winning $5 ($441.90), but 1 round of a $635 loss….net profit is -$193.10 in the average 89.38 rounds of play. On a table that gets 40 spins an hour, with an average ratio of 3 to an average round, you're looking at net -$28.80 per hour on average…..with a comp rating of $15 average, gives you a breakfast buffet in about 50 hours (after $1440 loss), that's an expensive breakfast.

I prefer the personal method of the stock market. You graph out black/red (or any pair of opposites, even/odd, high/low), and when the "stock" falls too much, you "buy" it (play X on it every spin). If it goes up, you "sell" (stop betting it)….if it goes down further, you "buy more" (play 2X, then 3X, etc. at each buying level) and you average down your higher sell points. If you set your levels just right, you could play an entire weekend with little bankroll and get a lot of comps.

For example, 5-tick levels, 5 red show up in a row. You bet $5 on black, if you win $25 (5 black more than red), you stop. If you lose $25 (5 red more than black), you bet $10 to try and win $50 before going back to $5 betting to win $15 more (averaged down). If you lose $50 more, you go to $15 for $75 before $10 for another $20 before $5 for another $10. This continues, as you follow the "market"…..$525 gives you a range of 30, giving you I think 30 hours of play? (give or take)

There is no paradox. At some point you are bound to lose all your money. Period. Every gambler will go broke.

The seeming "paradox" is from confusing the mathematical world (which is scale invariant), with the physical world (which is not). There is a limit to the amount of physical dollars you can hand over. In the mathematical world, one would never lose. In the physical world, one would always lose.

The St.Petersburg paradox is a better, more historic example of this. First stated in the early 1700's….it still is debated today.

gamble but just for fun .without betting money

Iâ€™ve worked in a casino & witnessed people lose their homes & pretty much their life playing the Martingale strategy. The numbers get massive so quickly. Itâ€™s almost a guarantee that every roulette table will have a run of 7-10 straight black or red within a 24 hour period. The most I remember was 17 straight black at a roulette table. The Martingale seems like itâ€™s the answer & most of the time it will work, but occasionally it will destroy you

ingenius facts

the way the casino wins is due to the fact that they impose a maximum bet. So over the course of a long time, the odds are in their favor that at least once you will have reached the table maximum and lost.

That is why casinos have a betting limit. You canâ€™t keep increasing your bet till you win even if you have sufficient funds to do so.

I'm gonna go bet my wife and kids…

â˜¹

The odds are not 50/50 in roulete for red black. They are closer to 47.5/52.5 that is a huge difference from 50/50 especially after thousands of spins.

I know a bigger one : 'What would be the point in the effort of producing the first baby who earns â‚¬ 10^500 per second at age 16?!' There are several paradoxes : first, obvious one is : inflation. But what is the point if one person owns ALL the money?! Neither other person would be economically involved anymore, and there would be no more point anymore either! Hypothetically, the economic system races to a situation in which 1 company is left, owning all of the assets, leaving everyone else in debt, to that one company. But that would have become irrelevant, as underway, everyone involved (the whole world), would have stopped caring about money all together, because there would have been no point in one egotripper worrying about everyone owing him.

This strategy doesn't get worse when you have lower chance of winning than 1/2, since you can get more as you win(at least this is how gamble usually works). So you need less to cover your total losses.

This is not a paradox. It's a Martingale strategy. It simply shifts the risk profile from 50:50 for each roll played individually to a risk of an occasional huge blowup when you reach the casino limit or run out of money.

Martingale strategy- Common Gamblers fallacy. You hit the table limit very quickly and runs of 9-10 can happen.

It is still 0 actually. It seems you are gonna win all the ways but it is not like it seems. Read all very carefully. Lets say you won at 'n.' try. Probability of winnings at p(n=1)=0.5 right. Won at p(n=2)=0.5*0.5 (lose-win). p(n=3)=0.5*0.5*0.5=0.125 (lose-lose-win).Seems you win anyways but you are gonna quit somewhere n=?. It doesnt matter when. Maybe n=10. Maybe n=100. Maybe n=infinity. But ofc you are gonna quit somewhere. Lets look at all the probabilities. Lets look at losing probability if you quit at n=10. You must loose 10 time in a row right. it means p=(n=10)= 0.5^10. So how much money did u lose at this point. 2+4+8+…+2^10 = 2046. You are -2046$ with 0.5^10 probability. You are +2$ with 1-(0.5^10)probability. So lets gather this two together with multiplying their probabilities. Remember, you started with 2$. -2046*(0.5^10) + 2*(1-(0.5^10)). At the end this is going to be = 0. It means you are getting nothing with this :). n=10 was just a simple example. It can be n=2,3,4,5,6,7,….,infinity. The result will be the same 0 :). You can try with n=3,4 with your calculators. When n starts to getting high calculators may not work so correctly. It s about the limited bits of your computer.

FIND FIVE OR MORE CONSECUTIVE LOSSES VIRTUALLY AND ENTER REAL POSITION SIXTH BET ..IF SIXTH BET ALSO LOSS. THEN MARTINGALE ON SENENTH.. IF WIN ON SEVENTH.. AGAIN FIND FIVE OR MORE CONTINUOUS LOSS..

I'm not sure which paradox you are referring to. I know someone mentioned Martingale, so there's that, but even with unlimited cash, casinos set max limits on bets for this exact reason.

You could lose 20 times in a row, and on top of this you only gain a small margin, which means you end up wasting time for a small marginal win, if you invested that time in something far more productive, you would probably end up with far more consistent gain, with very little lose. On top of this you will never have more money then the house, and so long term you will always lose, because they have more capital then you. Unless you get lucky, but that again would only be short term gain. But most people don't have this kind of capital to keep betting if they have a really bad streak, so because they don't have more money then the casino, they will end up losing long term, and because the odds are stacked in the casino's favour they will stop you being able to bet at a certain amount meaning you couldn't get as large as you wanted to recover loses. Meaning you would always be at a disadvantage. The only way to win is to cheat the casino or rig the game, because it's all set up in their favour.

This is how much money you would need to recover bets, if you lost 20 times in a row: $1,048,574. Each gamble is also 50/50. It resets each time you bet. So you only have 50/50 chance of winning. At that point in time the odds are 50/50, but it doesn't seem likely that 21 times in a row it would be one colour. But it can happen, so likely or not at that betting point it's still 50/50, and the odds are also stacked in the favour of the casino.

Roulette can be beaten. It comes down to the player cannot play every spin and and never double up. It is mathematically a losing slope when you chase down a win by increasing your wager. Stick to a consistent point, such as wait for a couple blacks, then wager on red. Stay consistent , stay patient, and yes you will overtime win more than your loses. That easy.

It's pretty simple. In your roulete game you have an expected value of zero which makes it a waste of time to play at best and at worst you still have a risk of ruin greater than 0.

In real roulette you have negative EV which makes it unprofitable to begin with in the long run, no matter the betting strategy.

There are very few gambling games that can be played profitably and when the casino finds out you do they will just kick you out.

An example would be blackjack card counting, but even when you do everything perfectly you need a huge bankroll to have a decent risk of ruin and a decent expectation.

For example to have 2% risk of ruin and 10$ per hour EV you need a $10000 bankroll.

SO to conclude: Don't gamble unless you know exactly what you are doing and by that I mean you did the math and preferably also ran some simulations just to make sure.

yeah, this type of betting strategy is called martingale, just double your money every time you lose. This is not a working strategy that you'll eventually lose all your money or else hit the maximum bet and that is you loos all your money you have bet

You forgot 0 and 00. The green numbers. Casinos never give you a 50/50 wager.

Many times I've seen red hit 15 times in a row. Martinggale is sucker bet

This would be fantastic if it weren't for those blasted pesky green slots on the roulette table. 18Red, 18Black, 2Green. The probability of landing green is 5.2%. If you land green, you lose everything. 1,2,4,8,16,32,Green. F*********K!!!! Eventually the green will get ya! 1 in 20 each spin. I'd like to see the same experiment but with the 5.2% odds of losing every time. hmm.. maybe you could find an optimal strategy for the game!

After watching this I tried it in Vegas and lost everything. Now I'm homeless. Thanks a lot!