maximum streak of getting lower or higher number.

[J0HNSN0W]

what do you think if we know about how many streak of losses we will get from a specific numerical where we bet for example betting on 40 and we know how many times the rolled dice can go under than i mean max time . and we will be able to calculate a specific ammount needed to bet to recover every loss streek. like if we bet x and accoding to multiplier our loss will gonna get recovered. ya it can be a math type beeting to if you bet low

i do have my notebook filled x pages within x years when i was not betting while betting. you know like no money bets. just i need some good math guys and we will broke this

[utmostactual]

Just flip a coin every bet

[mazzayu]

wen maxwin

[Spicin]

Suppose I start with 0.01 cent and double it on every loss. lets calculate how many tries before I lose $1000 in a 50-50 win chance game.. 

To calculate how many losses it will take before you lose $1000 in a 50-50 chance game where your bet doubles with each loss, let's analyze : 

Initial Bet: $0.0001 (0.01 cent)

Doubling: On each loss, the bet doubles.

Total Losses: The total amount lost is the sum of all bets until the bet exceeds $1000.

The formula for the total losses after n bets is the sum of a geometric series:

Total Loss=0.0001×(2n−1)\{Total Loss} = 0.0001 \times (2^n - 1)Total Loss=0.0001×(2n−1)

You need this total loss to exceed $1000:

0.0001×(2n−1)>10000.0001 \times (2^n - 1) > 10000.0001×(2n−1)>1000 2n−1>10000.00012^n - 1 > \frac{1000}{0.0001}2n−1>0.00011000 2n>100000012^n > 100000012n>10000001 n>log⁡2(10000001)n > \log_2(10000001)n>log2(10000001)

Using the properties of logarithms:

n>log⁡(10000001)log⁡(2)n > \frac{\log(10000001)}{\log(2)}n>log(2)log(10000001)

~ 24 

It will take 24 losses for your total losses to exceed $1000 if you start with a 0.01-cent bet and double it after each loss.

Plus now calculate the probability of this happening!

In a game with a 50-50 win-loss chance, the probability of losing 24 times in a row is calculated as:

P(24 consecutive losses)=(12)24P(\text{24 consecutive losses}) = \left(\frac{1}{2}\right)^{24}P(24 consecutive losses)=(21)24

Let me compute this probability.

The probability of losing 24 consecutive times in a 50-50 game is approximately 0.0000000596 (or about 0.00000596%).

That is like nearly impossible... but in Stake people have seen consecutively 32 losses.. or more than 19 losses in a row.. 
So I doubt stake wouldn't have already accustomed this in their Machine Learning Algorithm or their generic Algorithm..