Post thread Betting Strategy

NICKx

New member
Joined
Jul 27, 2021
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3
Looking for help to try an make a profitable strategy off the example.
payout= 1:1
Using 4 rounds the strategy wins 1 of 4 rounds 90% of the time, the other 10% it loses all 4 rounds
Rules: if any round is won that game is over, else keep betting till all 4 rounds over
The question is how to I change the betting method to turn positive results
Key= g=game, r=round, b=betting amount, pl=current running Profit/lose
Ex1: 10 games of 4 rounds
g1: r1: b=1, lost, pl=-1 | r2: b=2, lost, pl=-3 | r3: b=4, won, pl=1
g2: r1: b=1, lost, pl=0 | r2: b=2, won, pl=2
g3: r1: b=1, lost, pl=1 | r2: b=2, lost, pl=-1 | r3: b=4, lost, pl=-5 | r4: b=8, won, pl=3
g4: r1: b=1, lost, pl=2 | r2: b=2, lost, pl=0 | r3: b=4, won, pl=4
g5: r1: b=1, won, pl=5
g6: r1: b=1, won, pl=6
g7: r1: b=1, won, pl=7
g8: r1: b=1, won, pl=8
g9: r1: b=1, won, pl=9
g10: r1: b=1, lost, pl=8 | r2: b=2, lost, pl=6 | r3: b=4, lost, pl=2 | r4: b=8, lost, pl=-6
Ex1: results to a -6 profit/lose after winning 9 trades in a row before losing 1
Is it possible?
 

JeffM

Elite Member
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Sep 14, 2012
Messages
6,850
You do not say what the presumed probability of winning is, but I am assuming it is 50%.

The expected value of four rounds, with an initial bet of 1 dollar.

Win in round 1, your profit is 1 dollar with a probability of 50%. 50 cents.

If you lose on the first round, bet 2 dollars on the second round, and win, your profit is
(2 - 1) = 1 with a probability of 25%. 25 cents.

If you lose on the first two rounds, bet 4 dollars on the third round, and win, your profit is
(4 - 2 - 1) = 1 with a probability of 12.5%. 12.5 cents.

If you lose on the first three rounds, bet 8 dollars on the fourth round, and win, your profit is
(8 - 4 - 2 - 1) = 1 with a probability of 6.25%. 6.25 cents.

If you lose in all four rounds, your loss is (1 + 2 + 4 + 8) = 15 dollars or 1500 cents. Probability is 6.25%. Expected value is minus 93.75 cents.

Add them up. 50 + 25 + 12.5 + 6.25 - 93.75 = ZERO.

No gain or loss possible in a fair game with fair payoffs. That is why we say fair. But drive down the Strip in Las Vegas; they flash in neon signs the degree to which the games are unfair. “We are less unfair than our competitors,” they shout. Weird place.

You are correct: the betting strategy has a 93.75% chance of winning, but the size of the loss when you infrequently lose offsets the much more frequent but also much smaller profits.
 

NICKx

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Joined
Jul 27, 2021
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3
Thank you for your reply. Thats great info.
I wasnt looking for a fair game each round.
I was looking for, out of the 4 rounds my probability on winning 1 of them is 90%.
What that boils down to for each round, idk.
I dont think im looking to base it off each fair round but yet a game of the 4 rounds.
 

Cubist

Senior Member
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Oct 29, 2019
Messages
1,142
probability on winning 1 of them is 90%
...therefore the probability of loosing all four is 10%. If we let q = chance of loosing any one round then q^4 = 0.1 => q = 0.1 ^ (1/4) ≈ 0.56234132519. So a little bit more than 50% (therefore most people probably wouldn't notice the difference between this and a coin toss). If we let the betting amounts in each successive round be b1, b2, b3, b4 then the following table can be produced of all possible outcomes...
Code:
Win/loose        prob            return
w                    (1-q)       b1
lw               q * (1-q)       b2-b1
llw              q^2*(1-q)       b3-(b1+b2)
lllw             q^3*(1-q)       b4-(b1+b2+b3)
llll             q^4*(1-q)       -(b1+b2+b3+b4)
The expected return from this is the sum of prob*return for all of the rows above. This simplifies to...
expected return = (1 - 2*q) * (q*(q*(q*b4 + b3) + b2) + b1)

So, as @JeffM correctly points out, if q is exactly 0.5 or 50% then the expected return is zero because (1-2*q)=0. BUT in your game q is actually a bit more than 50% so the expected return becomes...
(1 - 2*q) * [q*(q*(q*b4 + b3) + b2) + b1]
= (negative) * [ positive ]
= negative

The more you bet the more you expect to loose in the long run. In the words of Joshua from the film Wargame, "the only winning move is not to play." Hmmm, actually the only winning move is to be the house (or the street hustler, etc) !
 

Cubist

Senior Member
Joined
Oct 29, 2019
Messages
1,142
Whoops, I made a mistake when I typed in the table above. The highlighted red bit below shouldn't be there (the equation above is correct, I think, this was just a typo in the table)...

Rich (BB code):
Win/loose        prob            return
w                    (1-q)       b1
lw               q * (1-q)       b2-b1
llw              q^2*(1-q)       b3-(b1+b2)
lllw             q^3*(1-q)       b4-(b1+b2+b3)
llll             q^4*(1-q)       -(b1+b2+b3+b4)
 

NICKx

New member
Joined
Jul 27, 2021
Messages
3
Thanks Cubist. You really know your math! So it looks like I’m back to square 1 at trying to make money. Haha
 
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