Challenge: What's the best way to win 1 BTC with 1 BTC?

[FanEagle]

4 bets of 60% win?
That should fit. I got so nice lucky streaks with 60%.
To be more specific:
Bet size 0.5
60% of win chance.
Over
0.325*4= 1,3btc

[1715546967]

1715546967

[1715546967]

1715546967

[1715546967]

1715546967

[boumalo]

Challenge: The person who provides the best way to win 1 BTC using a bankroll of 1 BTC. I will give 0.1 BTC to (payable to their moneypot account).

Assuming the constraints of a dice site such as bitdice.me or pocketrocketscasino.eu: 1% house edge, 9900x max payout, 1 satoshi minimum bet.  Provide a well-defined betting sequence to maximize EV. If I don't have the skills to analytically analyse your solution, I will be forced to simulate it over a few billion runs. Your betting sequence must be FULLY specified without any ambiguities.

Let me start with the most obvious solution:


Bet 1 BTC @ 2x  ... which gives an EV of -0.01

I don't think you will find better than the most obvious solution

[RHavar]

According to these conditions, maximal EV with 1 BTC bankroll attempting to win 1 BTC is achieved by making 1 bet of 0.0001010101 BTC @ 9900x.

You're right, my language was pretty poorly picked. I should have said: "Maximizing the probability to win 1 BTC" as opposed to "Maximizing the EV".  Your bet does however does use fractions of a satosishis, so I'll give you 0.05 if you tell me your MP Account or bitcoin address.

The core question is still open, and for the people who suggested martingale betting sequences, I'll writes some code to figure out the probabilities a bit later.

[blockage]

Strategy 1:
You're always playing 9900x using a martingale sequence, i.e. your wager for each round is

Code:

ceiling((2 - PLAYER_BR)/9899,1e-8)

you run out of funds after 6861 bets. You lose all of them with a probability of

Code:

0.9999^6861 = 0.5035187605260536

or put differently you have a 49.64812% chance of winning.

Strategy 2:
You're constantly playing 20203 satoshi but increase the payout each round by exactly 1x starting at 4951x which will leave you with 1.00004850 BTC profit. You run out of funds after 4949 games and your last payout will be 9899x. You lose all of them with a probability of

Code:

prod(1-0.99/i, i=4949..9899) = 0.5035283796177312

leading to a 49.64716% chance of winning.

I Hope the math is correct 

[RHavar]

Strategy 1:
You're always playing 9900x using a martingale sequence, i.e. your wager for each round is

Code:

ceiling((2 - PLAYER_BR)/9899,1e-8)

you run out of funds after 6861 bets. You lose all of them with a probability of

Code:

0.9999^6861 = 0.5035187605260536

or put differently you have a 49.64812% chance of winning.

I was hoping someone first provide simpler sequences (e.g. bet 0.5 at 3x, if lose bet the other 0.5 at 4x: which reduces the  house edge to 0.835%), but looks like you might have nailed the ideal (?) betting sequence , with an effective house edge of  0.70376%.

If no one can beat this, you have a well-deserved 0.1 BTC coming your way

[1gallon]

RHavar

following the post from blockage, you can reduce the house edge even more with another website that is 0.5% house edge, thus reducing it to an effective 0.20376% house edge at safedice.com

[RHavar]

RHavar

following the post from blockage, you can reduce the house edge even more with another website that is 0.5% house edge, thus reducing it to an effective 0.20376% house edge at safedice.com

Currently doing one of blockage's bets on safedice now would incur a ~2.5% house edge, due to their floating edge scheme. The point of the thread wasn't "which site is the best", but more "which betting sequence is the best, given a fixed house edge".  =)

[1gallon]

RHavar

following the post from blockage, you can reduce the house edge even more with another website that is 0.5% house edge, thus reducing it to an effective 0.20376% house edge at safedice.com

Currently doing one of blockage's bets on safedice now would incur a ~2.5% house edge, due to their floating edge scheme. The point of the thread wasn't "which site is the best", but more "which betting sequence is the best, given a fixed house edge".  =)

ah i see that now.

This is not a true fibonnaci, but

bet series looks like: 1 2 3 4 7 11 18 after 2 wins back to base.

basebet: 0.00000001
loss 1: increase by 0.00000001
loss 2: increase by 0.00000001
loss 3: increase by 0.00000001
loss 4: increase by previous bet
loss 5: increase by previous bet
loss 6: increase by previous bet
win 7: stay same roll again


I'm not sure how to calculate the EV though.

[Madern]

Go to satoshimines and bet 1 btc on 3 mines and in 8-9 bets you ate good to go

[elephantas1]

Use this technique on primedice :

Multiplier: x 6.262  - Increase on loss : 22.71 % - BaseBet : 15 satoshi.

and this would take a few hours and most probably you will still end up loosing it. i personally would make 1 btc all in bet and pray to win it

[funtotry]

Simplest and most reliable way would just do a 50% bet on primedice. All these betting "strategies" won't make you a profit, the sites design to make you lose. Try doing a 2x bet. You either lose it all or get your goal.

[freedomno1]

Challenge: The person who provides the best way to win 1 BTC using a bankroll of 1 BTC. I will give 0.1 BTC to (payable to their moneypot account).

Assuming the constraints of a dice site such as bitdice.me or pocketrocketscasino.eu: 1% house edge, 9900x max payout, 1 satoshi minimum bet.  Provide a well-defined betting sequence to maximize the chance of winning 1 BTC. If I don't have the skills to analytically analyse your solution, I will be forced to simulate it over a few billion runs. Your betting sequence must be FULLY specified without any ambiguities.

Let me start with the most obvious solution:


Bet 1 BTC @ 2x  ... which gives an EV of -0.01, and a chance of winning at 49.5%

Take a bet on Betmoose and choose the winning side
Rinse and repeat with outcomes that are known but not settled yet or where it's very close to settle

Edit with constraints
Assuming the constraints of a dice site such as bitdice.me or pocketrocketscasino.eu: 1% house edge, 9900x max payout, 1 satoshi minimum bet.  Provide a well-defined betting sequence to maximize the chance of winning 1 BTC. If I don't have the skills to analytically analyse your solution, I will be forced to simulate it over a few billion runs. Your betting sequence must be FULLY specified without any ambiguities.

Buy Clams and invest in just dice XD

But lets see
Multiplier X 12
Bet size 0.01
Repeat till win then multiply X2
If win 0.02 till win
If lose 0.01 till win
Or Bust

[Wendigo]

Bet on over 0.5 goals in soccer games, look at the h2h statistics and pick carefully. The odds are usually 1.1 and you need a couple of bets.

[MadZ]

According to these conditions, maximal EV with 1 BTC bankroll attempting to win 1 BTC is achieved by making 1 bet of 0.0001010101 BTC @ 9900x.

You're right, my language was pretty poorly picked. I should have said: "Maximizing the probability to win 1 BTC" as opposed to "Maximizing the EV".  Your bet does however does use fractions of a satosishis, so I'll give you 0.05 if you tell me your MP Account or bitcoin address.

The core question is still open, and for the people who suggested martingale betting sequences, I'll writes some code to figure out the probabilities a bit later.

I made a Moneypot account with the username MadZ, deposit address 1LnQYjwRhsZkp21jaMx6PNJynBw6w2BrKR, thanks!

Public note: If my btctalk account is compromised in the future, I believe I will be unable to sign this address to prove ownership.

I believe blockage's two solutions will probably yield the highest probability of achieving 1 BTC, although I will make an attempt at it later. The key is to minimize the amount you wager, which is done at the lowest odds, since the more you wager the more you lose out to the house edge.

[1gallon]

According to these conditions, maximal EV with 1 BTC bankroll attempting to win 1 BTC is achieved by making 1 bet of 0.0001010101 BTC @ 9900x.

You're right, my language was pretty poorly picked. I should have said: "Maximizing the probability to win 1 BTC" as opposed to "Maximizing the EV".  Your bet does however does use fractions of a satosishis, so I'll give you 0.05 if you tell me your MP Account or bitcoin address.

The core question is still open, and for the people who suggested martingale betting sequences, I'll writes some code to figure out the probabilities a bit later.

I made a Moneypot account with the username MadZ, deposit address 1LnQYjwRhsZkp21jaMx6PNJynBw6w2BrKR, thanks!

Public note: If my btctalk account is compromised in the future, I believe I will be unable to sign this address to prove ownership.

I believe blockage's two solutions will probably yield the highest probability of achieving 1 BTC, although I will make an attempt at it later. The key is to minimize the amount you wager, which is done at the lowest odds, since the more you wager the more you lose out to the house edge.

Here's something that no one else really do, but i abuse it every so often. If you bet with 2 satoshi at 1.33% payout on primedice, you'll 'win' 1 satoshi how's that for beating house edge!! if you bet 5 satoshi, you'll win 2 saotoshi back ;P and if you bet 8, you'll win 3 satoshi

[Aemon]

I am quite certain there is no better odds than Martingale. 

You will probably want to have enough money for say 20 losses. 

The second safest best is to YOLO it all.  These are the 2 highest chances you will have in order to win that much BTC.

[MadZ]

According to these conditions, maximal EV with 1 BTC bankroll attempting to win 1 BTC is achieved by making 1 bet of 0.0001010101 BTC @ 9900x.

You're right, my language was pretty poorly picked. I should have said: "Maximizing the probability to win 1 BTC" as opposed to "Maximizing the EV".  Your bet does however does use fractions of a satosishis, so I'll give you 0.05 if you tell me your MP Account or bitcoin address.

The core question is still open, and for the people who suggested martingale betting sequences, I'll writes some code to figure out the probabilities a bit later.

I made a Moneypot account with the username MadZ, deposit address 1LnQYjwRhsZkp21jaMx6PNJynBw6w2BrKR, thanks!

Public note: If my btctalk account is compromised in the future, I believe I will be unable to sign this address to prove ownership.

I believe blockage's two solutions will probably yield the highest probability of achieving 1 BTC, although I will make an attempt at it later. The key is to minimize the amount you wager, which is done at the lowest odds, since the more you wager the more you lose out to the house edge.

Here's something that no one else really do, but i abuse it every so often. If you bet with 2 satoshi at 1.33% payout on primedice, you'll 'win' 1 satoshi how's that for beating house edge!! if you bet 5 satoshi, you'll win 2 saotoshi back ;P and if you bet 8, you'll win 3 satoshi

Your actual balance is rounded beyond what you see on your screen, this does not work.

[Aemon]

According to these conditions, maximal EV with 1 BTC bankroll attempting to win 1 BTC is achieved by making 1 bet of 0.0001010101 BTC @ 9900x.

You're right, my language was pretty poorly picked. I should have said: "Maximizing the probability to win 1 BTC" as opposed to "Maximizing the EV".  Your bet does however does use fractions of a satosishis, so I'll give you 0.05 if you tell me your MP Account or bitcoin address.

The core question is still open, and for the people who suggested martingale betting sequences, I'll writes some code to figure out the probabilities a bit later.

I made a Moneypot account with the username MadZ, deposit address 1LnQYjwRhsZkp21jaMx6PNJynBw6w2BrKR, thanks!

Public note: If my btctalk account is compromised in the future, I believe I will be unable to sign this address to prove ownership.

I believe blockage's two solutions will probably yield the highest probability of achieving 1 BTC, although I will make an attempt at it later. The key is to minimize the amount you wager, which is done at the lowest odds, since the more you wager the more you lose out to the house edge.

Here's something that no one else really do, but i abuse it every so often. If you bet with 2 satoshi at 1.33% payout on primedice, you'll 'win' 1 satoshi how's that for beating house edge!! if you bet 5 satoshi, you'll win 2 saotoshi back ;P and if you bet 8, you'll win 3 satoshi

Your actual balance is rounded beyond what you see on your screen, this does not work.

Yes, there are fractional bits at the end of those, so you won't profit that much every time, it will just be some of the time so you won't be the house edge that way

[RHavar]

I made a Moneypot account with the username MadZ, deposit address 1LnQYjwRhsZkp21jaMx6PNJynBw6w2BrKR, thanks!

Sent you 0.05: 8983f965ddd3d6e5b42a4b035f6b3072ff4984ce726f13a6b687e79260fa3ada

and added 0.1 to blockage's account. If someone can beat his solution, I'll give them 0.1 as well.

[srgkrgkj]

Here's the strategy:

1. Choose a game with a particular price multiplier to play. Let r = the price multiplier
2. Let total number of times you want to make a bet on this game = "bets".
3. Let maximum amount of coin willing to lose = "max"
4. p is probability of winning the game you're playing, q is the probability of losing the same game, q = 1 - p
5. Let the amount by which you multiply each bet after a loss be "m", m = 1/(1 - 1/r)
6. Let expected losses in a row be "n", n  = -log(bets*p + 1)/log(q)
7. Let "init" be the ideal starting amount to gamble (in the absence of transaction fees and 0.5% return on loss = max/sum(m^(1:n))

Generally, transaction fees and return on loss have little effect on the strategy. However, if you would like to take them into account:

7. Let "init" be the starting amount to gamble when taking account fees and 0.5% return on loss,
init = (max - n * 0.0005)/(sum(m^(1:n))*(1 - sum(m^(1:(n-1)))/(sum(m^(1:n)))*0.005))
8. Calculate:
 (1-(init*m^n*r - 0.0005)/(init*m^n*r))*100

This is the percentage of winnings lost to fees after an expected win (after n losses in a row). If this is too high, choose either higher max or lower number of bets and try again.


Warning: This strategy will only prevent you losing more than a maximum amount in the same order of magnitude as the one you selected. There's a lot of variance in the game, and sometimes you may lose much more - especially if p > 0.5 or your maximum is large. It works quite well with small maximum btc amounts.

For example, playing a the price multiplier = 8x  game, wanting to bet 100 times and wanting to lose a maximum of 1 btc, not taking into account fees or returns on losses:


1. r = 8
2. bets = 100
3. max = 1 btc
4. p = 8000/65536, q = 1 - p
5. m = 1 / (1 - 1 / r)
6. n = round(-log (bets * p + 1) / log( q ))
7. init = max / sum( m ^ ( 1 : n ) )
8. % fee loss = (1-(init * m ^ n * r - 0.0005)/(init * m ^ n * r)) * 100


then
m = 1.142857
init = 0.009294345
% fee loss = 0.04653956%

As above, with fees and return on loss:

7. init = max / sum( m ^ ( 1 : n ) )
8. % fee loss = (1-(init * m ^ n * r - 0.0005)/(init * m ^ n * r)) * 100

then
m = 1.142857
init = 0.009241403
% fee loss = 0.04680618%

Below are some simulations of martingale betting on Satoshi Dice using the strategy  outlined in this post. They show the usefulness of the strategy when you use a low max, as compared with the standard nightingale (start with 1btc, double each time you lose).

Courtesy of organofcorti but i wouldent mind some BTC for pointing you in the right direction -->https://bitcointalk.org/index.php?topic=94481.0