After
this discussion with dooglus about the probability of 'n' coin tosses in a row, and then
watching a forum member getting excited abut winning on Satoshi Dice using the Martingale strategy and subsequently losing it all, I thought it a good idea to present a safe way to use the Martingale strategy to minimise your losses, and play the game for longer.
This is not an endorsement of gambling, but a way to gamble more safely and longer. The method does not guarantee a win, but it does make sure you're less likely to lose more than you'd want to.
The basic idea is to determine the number of times you want to gamble and the maximum amount of coin you can stand to lose. Then, using the probability of a win for the game you're playing, calculate the starting gamble and multiplier which will allow you the best chance of lasting the number of bets you'd like to play.
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).
