Topic: Challenge: What's the best way to win 1 BTC with 1 BTC? - page 10. (Read 15989 times)

I say play Blackjack and when the dealer hits Blackjack go all in next hand.

Just change chance to 2x, choose ALL IN, choose HIGH, and then pray for your BTC 

Compare any gambling strategy to simply holding 1 Bitcoin for 6 months.  Chances are best you will double your money that way.

Really? The strategy that won was make 1 bet of 0.0001010101 BTC @ 9900x?   Is it just me or does it seem that using expected value (EV) is not a good way strategize with dice betting?

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.

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.

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).

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

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