Progression question

[FreeRoulette]

I want to make a tool that creates a progression.

There will be three inputs
- Max Bet
- Min Profit
- Wins

So for example someone wants to play the Martingal
- Max Bet = 256
- Min Profit = 3
- Wins = 1

This is interpreted as: Play a progression up to a max bet of 256 where the minimum profit is 3 and the progression only has to hit 1 time to win.

The Martingale is super easy, but what is the algorithm to calculate a progression where I want to hit any 2 times and win at least $3 ? And of course I want to do it with the minimum bet amounts possible.

Thanks for you help,
FreeRoulette

[furple]

1) Bet: $ 2 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $2
2) Bet: $ 3 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $5
3) Bet: $ 4 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $9
4) Bet: $ 6 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $15
5) Bet: $ 9 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $24
6) Bet: $ 14 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $38
7) Bet: $ 21 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $59
8) Bet: $ 31 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $90
9) Bet: $ 47 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $137
10) Bet: $ 70 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $207
11) Bet: $ 105 on 1 spots. Profit on a Win: $ 3 Bankroll Needed: $312
12) Bet: $ 158 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $470
13) Bet: $ 237 on 1 spots. Profit on a Win: $ 4 Bankroll Needed: $707

[FreeRoulette]

Haha funny.

That system is a win once progression.

[chrisbis]

@Furple.

He can only go to level 10 on Ur progression statement.

He only has 256 BR.

[furple]

I thought his max bet was 256. Anyway it looks like he's after something else.
A win more than once progression apparently.

[VLS]

Hello dear FR.

More likely you will end up having to make nested loops.

Create each progression step with loop "A", and incrementally rise the unit (+1 at a time) in loop "B", checking it delivers the $3.

Exit loop "A" if $unit > 256.

P.S: Which programming language are you targetting?

Perhaps I can assist with an actual example in your target language is one I can handle.

Cheers!
Vic

[FreeRoulette]

Hey Vic,

Thanks for helping me out. Python will work. If I do two nested loops is that for bet 1 and bet 2?

The difficult part is that suppose the player wants to hit any 5 times. So the progression below shows hits in brackets

2,4,[3],[2],[2]

At this point the player hits 3 times and makes a profit. The problem is that it should only make a profit when it hits 5 times, so betting [3] was too much.

Thanks again

[VLS]

The problem is that it should only make a profit when it hits 5 times, so betting [3] was too much.

Hello dear FR, for a mandatory 5-in-a-row win, You have to generate reversed progression for every unit amount and compare the *NET WIN* after 5th step of the interim progression to the running lost, to check if it does meet the min-profit.

i.e.

Unit = 1

Interim reversed progression = 1 2 4 8 16

Net win = (16 * 2) - 1 = 31

For a running lost of 1 to 28, this progression with 5 consecutive wins covers the condition of Min profit = 3

From 29 on, you must go:

Unit = 2

Interim progression: 2 4 8 16 32

Net win = (32 * 2) - 2 = 62

For running lost of UP TO 59, this one should cover Min. profit = 3

and so on, in an incremental loop.

Kindly try to use this information to come up with that Python script and then shall you get stuck allow me to review it.

In case you don't know to start, just "shout" and allow me to do a basic skeleton code for you to , give it a shot at completing it

Cheers!
Vic

[VLS]

This is INCOMPLETE python code made at 2:40 a.m. does need some debugging, yet you may get the basic idea from it!

Code Select Expand


#!/usr/bin/python

#parameters
maxbet = 5
minprofit = 3
wins = 5

#progression variables
runninglost = 0 #running lost amount
progression = list() #list for the resulting progression
progression.append(1) #add first step to it

#first loop = unit range
for unit in range(1, maxbet + 1): #(iterations = maxbet)
#second loop = caculate a reversed progression of N steps, where N = wins
runninglost += progression[-1] #add last step to runninglost
iprog = list() #declare interim progression

iprog.append(unit) #add "unit" as first value
for I in range(1, wins): #append the rest of the values
tempval = iprog[-1] * 2 #"martingale" it
iprog.append(tempval) #append

#[DEBUG] print the iprog
for item in iprog:
print item

#calculate
netwin = (iprog[-1] * 2) - unit

#[debug] print netwin
print "netwin: " + str(netwin)

#check net wins satisfice the profit >= minprofit condition
if (netwin - runninglost) > minprofit
print "Satisfied"

#the show is over for current unit
# homework: append current unit to progression list
#progression.append()


See ya around! Off to bed!
Vic

[FreeRoulette]

Hello VLS,

Thank you for the python program. I ran it and got the following results.  This does a win once progression. I tried to modify it for a win twice progression, but I'm a little lost. I think the solution is going to end up being some recursion program that always scrambles my brain.

Example of a win any two times progression
1, 1, 2, 3, 6

________________________________________
results from python


1
2
4
8
16
netwin: 31
Satisfied
2
4
8
16
32
netwin: 62
Satisfied
3
6
12
24
48
netwin: 93
Satisfied
4
8
16
32
64
netwin: 124
Satisfied
5
10
20
40
80
netwin: 155
Satisfied