[Guest]

Imagine you are playing a game at a casino. You give up some amount of money.

The dealer flips a (fair) coin. If it is heads, you win back your money double.

If it is tails, you get no money.

Since the game is 50% - 50%, you can't expect to just win money automatically. On average, if you bet $1 every time and play the game over and over, you can expect to break even.

However, consider this strategy:

Start here:

Bet $1. If you win, start over. If not, continue:

Bet $2. If you win, start over. If not, continue:

Bet $4. If you win, start over. If not, continue:

Bet $8. If you win, start over. If not, continue:

Bet $16. If you win, start over. If not, continue:

etc...

So - what happens? Eventually, you have to win. It is completely improbable that you would lose forever.

What happens if you win at step 5? You win back 16. What did you lose (steps 1-4)?

8 + 4 + 2 + 1 = 15

So you actually won an entire dollar.

If you win at step 6, you get $32, and in steps 1-5 how much did you lose? Only $31.

So it seems like this strategy is 100% fool-proof. You always win $1 more than whatever you lost in all the previous steps. And when you repeat this strategy over and over, you can win an endless amount of money.

What is the flaw in this strategy? There are two flaws that I know of, although they are more-or-less the same, so you would probably come up with both of them I think.

[HoustonHokie]

Actually, I don't think the strategy is flawed at all. I set up a simulation to run 10000 games with each strategy. With consistent $1 bets, you break even. With the doubled bets, you expect to win $0.50 every game. So after 10000 games, my simulation was up $5000 most of the time. All you need is enough seed money to cover a string of bad flips.

A better strategy would be to triple your bet if you lose. Or go even higher... The results vary pretty widely, but after 10000 games using tripled bets, my simulations were up anywhere from $100,000 to $10,000,000!

I don't look for this game in Vegas anytime soon. But let me know if you're going to start a casino with it - I'll come over and play a couple hundred times!

[Guest]

1. Since there is a 50-50 probability the coin could land 999,999 times on the wrong side making you lose, before it actually lands on the winning side. You will need to have a huge amount of money to gamble in order to be 100% that you will eventually win.

2. Casinos have a maximum bet on most of the games. That way even if you have a huge amount of money to gamble, you can't actually practise your method once you reach the maximum bet.

And a bonus: If you actually have such a huge amount of money, you shouldn't be gambling it!

Edited November 17, 2010 by Lemeshianos

[Guest]

The flaw is that whaterver you do, you will win only one dollar...there must be a better strategy to win more dolars.You can't just go to the casino to win one dollar!

[witzar]

If you have infinite amount of money, then you cannot win any more, since infinity plus your eventual winnings is still infinity.

In other words: what's the point of gambling when you have all the money in the world, can't loose it and can't get richer?

If above argument is not convincing for you, than accept that it is not possible to have infinite amount of money

I you have a finite amount of money, you can afford only a finite number of consecutive bets. Say you have $2^n-1 so you can afford n consecutive bets.

In this case there is a 2^-n chance to loose all the money and 1-2^-n chance to win $1 in one "session". Therefore the expected value of one session is:

EV = +$1*(1-2^-n) -$(2^n-1)*2^-n = $1 - $1 = $0

This explains why you should not expect to get rich playing this game.

And you should to expect to loose in real life casino with this strategy since you are not going to get even-money bet (EV=0) there.

Edited November 17, 2010 by witzar

[Guest]

The flaw is that whaterver you do, you will win only one dollar...there must be a better strategy to win more dolars.You can't just go to the casino to win one dollar!

I double my bet plus the initial bet...ie $1, $3, $7, $15...

when I win I win not the initial bet, but the initial bet times the number of hands I played, win or lose.

But yeah you run out of money (or courage) or you reach the bet limit.

The other flaw is that if Vegas developed this game, trust me it is designed to not have a negative outcome for those guys. I'd still play it mind you, for the excitement. But Vegas usually looks for any kind of loopholes, hence bet limits, etc.

Edited November 17, 2010 by maurice

[Guest]

Well-known technique

This is known as a "martingale" and is well-known and documented in gambling literature. It can be used in any game including today's casino games. Howeever the flaws are real. Doubling your bet can run the bet up a lot more quickly than you realize and you will quickly be broke before you win your dollar. Also the house limit is real, and pit bosses know about the martingale and will enforce the limit strictly if you try to use it in blackjack, crape, or roulette.

[Guest]

Imagine you are playing a game at a casino. You give up some amount of money.

The dealer flips a (fair) coin. If it is heads, you win back your money double.

If it is tails, you get no money.

Since the game is 50% - 50%, you can't expect to just win money automatically. On average, if you bet $1 every time and play the game over and over, you can expect to break even.

However, consider this strategy:

Start here:

Bet $1. If you win, start over. If not, continue:

Bet $2. If you win, start over. If not, continue:

Bet $4. If you win, start over. If not, continue:

Bet $8. If you win, start over. If not, continue:

Bet $16. If you win, start over. If not, continue:

etc...

So - what happens? Eventually, you have to win. It is completely improbable that you would lose forever.

What happens if you win at step 5? You win back 16. What did you lose (steps 1-4)?

8 + 4 + 2 + 1 = 15

So you actually won an entire dollar.

If you win at step 6, you get $32, and in steps 1-5 how much did you lose? Only $31.

So it seems like this strategy is 100% fool-proof. You always win $1 more than whatever you lost in all the previous steps. And when you repeat this strategy over and over, you can win an endless amount of money.

What is the flaw in this strategy? There are two flaws that I know of, although they are more-or-less the same, so you would probably come up with both of them I think.

You may be repeating each step several times adding up your losses. (consider repeating loosing at step 4 for 5 times before moving on to step 5). Not very profitable.

[Guest]

You may be repeating each step several times adding up your losses. (consider repeating loosing at step 4 for 5 times before moving on to step 5). Not very profitable.

Are you saying the process should consider repeating step 4 5 times? If so, then that would take away the "guarantee" of being ahead when you finally win.

Or are you saying that repeating step 4 5 times would take away from the profitability? If so, then that is not what the process calls for. You start with step one. If you lose procedd to next step. If you win return to step 1.

[Guest]

1. There are more games out there designed to win big bucks to discourage you doing this. (WELL I'M FOCUSED TO DO THIS OVER AND AGAIN).

2. They start with minimum and maximum bet limits. (I WILL CARRY LOT OF MONEY SO I CAN HAVE MORE TURNS)

3. They will tell that the exact denomination chips you wish to buy, are currently not available and insists that you buy bigger denomination chips.

4. They will have a service charge set aside (say $2 per table/turn ) so THEY invariably gain 1$.

As per your logic, you are continuing to stay on the game and keep increasing the bets if you lost the previous round. In order to play with $16, you will lose 1,2,4,8 (=15 already). Still, you have equal chance of losing the round 5 also, which will make it -$31 for you or gain $1. That has extreme risk of losing more than winning back and the risk worth AIN'T the ROI.

Consider starting here:

Bet $16. If you win, GET OUT. If not, GET OUT:

Lose: $16 or Gain: 32. Risk is huge but Rewards are double, so worth taking the risk.

Isnt -16 or +32 better than -31 or +1?

Edited November 18, 2010 by aaronbcj

[Guest]

Pls ignore previous reply ...

1. There are more games out there designed to win big bucks to discourage you doing this. (WELL I'M FOCUSED TO DO THIS OVER AND AGAIN).

2. They start with minimum and maximum bet limits. (I WILL CARRY INFINITE/LOT OF MONEY SO I CAN HAVE MORE TURNS)

3. They will tell that the exact denomination chips you wish to buy, are currently not available and insists that you buy bigger denomination chips.

4. They will have a service charge set aside (say $2 per table/turn ) so THEY invariably gain 1$.

As per your logic, you are continuing to stay on the game and keep increasing the bets if you lost the previous round. This works only if you have infinite money. Let's say you are limited to play only till round-5 (betting @ $16). Means, you have to start with $31 in your pocket (compensating loses in previous rounds (1,2,4,8 =$15). Still, you have equal chance of losing the round 5 also, which will make it -$31 for you or gain $1. That has extreme risk of losing more than winning back and the risk worth AIN'T the ROI.

Consider starting here:

Bet $16. If you win, stay. If not, GET OUT:

Bet $8. If you win, stay. If not, GET OUT:

Bet $4. If you win, stay. If not, GET OUT:

Bet $2. If you win, stay. If not, GET OUT:

Bet $1. If you win, GET OUT. If not, GET OUT:

This way, you are not taken by emotion to keep pumping more money, instead, you already set yourself to possibly lose only 16 or win possibly 32+16+8+4+2 ?. You can start with only $16 in your pocket and use winning money on subsequent bets.

Isnt -16 or +62 better than -31 or +1?

[Guest]

Pls ignore previous reply ...

1. Your logic works when you have infinite money. Considering that much money, it is very inefficient way of earning only +1 per game.

2. They may have a service charge set aside (say $2 per table/game ) so THEY invariably gain 1$.

3. Lets say they impose a limit of only certain number of games per player a day (Means, they make it a finite money to be involved from a player) or put in other words, we only start with finite amount (practical).

As per your logic, you are continuing to stay on the game and keep increasing the bets if you lost the previous round. Let's say you are starting with only $31 in your pocket, so you are limited to play only till round-5 (betting @ $16 + compensating loses in previous rounds 1,2,4,8 =$15). Still, you have equal chance of losing the round 5 also, which will make it -$31 for you or gain $1.

Consider starting here:

Bet $16. If you win, stay. If not, GET OUT:

Bet $8. If you win, stay. If not, GET OUT:

Bet $4. If you win, stay. If not, GET OUT:

Bet $2. If you win, stay. If not, GET OUT:

Bet $1. If you win, GET OUT. If not, GET OUT:

This way, you are not taken by emotion to keep pumping more money (if they allow more rounds), instead, you already set yourself to possibly lose only 16 or win possibly (32-16)+(16-8)+(8-4)+(4-2)+(2-1)=31 at end of round-5. You can start with only $16 in your pocket and use winning money on subsequent bets or start with 31 in pocket. Either case, your profit can be 31.

Big Q is will you prefer -1 or +1 (in your way) in R1 or -16 or +16 (in my way).

If we play 10 games (each starting with $31) of each 5 rounds, for each game

round-1: W/L is 50% (1/2)

round-2: W/L is 25% (1/2 * 1/2)

round-3: W/L is 1/8

round-4: W/L is 1/16

round-5: W is 1/32

winning at R5 (after losing R1,R2,R3,R4) is same as winning at R5 (after winning R1,R2,R3,R4). In that case, shdn't we put the highest stake on R1? Isn't ending up with -16 or +31 better than -31 or +1? It all depends on risk perspective one takes as per each round i guess. And also when a player can "stop" and not dragged further by emotions.