Is there any difference?
Player A)Betting black, doubling bet size once if a loss for a max of 2 bets total (75% chance of winning 1 unit, 25% chance of losing 3)
and
Player B)Betting 27/36 numbers each time player A is making their first bet of series (75% chance of winning 1 unit, 25% chance of losing 3)
Is one of these bets more likely to be closer to the expected probability after a large number of trials?
If so, why?
Player "B"'s results will always be utterly and absolutely independent of the results of Player A's activities.Quote: Stats
Player B)Betting 27/36 numbers each time player A is making their first bet of series ?
Quote: Stats
Player B)Betting 27/36 numbers each time player A is making their first bet of series
The reason I say each time player A is making their first bet is so that player B does not make more bets (it takes player a upto 2 bets to have a 75% probability, player b only takes 1 bet to have same probability)
Quote: StatsTo make what Im trying to find out simpler to explain, lets say that there is a fair roulette game with the same pay and no zero's on wheel.
Player A)Betting black, doubling bet size once if a loss for a max of 2 bets total (75% chance of winning 1 unit, 25% chance of losing 3)
and
Player B)Betting 27/36 numbers each time player A is making their first bet of series (75% chance of winning 1 unit, 25% chance of losing 3)
Is one of these bets more likely to be closer to the expected probability after a large number of trials?
If so, why?
Is B betting 1/3 of a unit per number?
If so, then, as you point out, their probability distributions are exactly the same and completely independent, assuming that B picks his 27 numbers randomly.
I am working on a spreadsheet to test your theory, both directly (no zeros) and real world (double zero).
ETA - with no zeros, B is way ahead. Toss in a double zero and B ends up way behind.
Quote: AxiomOfChoice
If so, then, as you point out, their probability distributions are exactly the same and completely independent, assuming that B picks his 27 numbers randomly.
Yes, both players are risking the same amount of units for the same payoff.
Quote: TerribleTomI think Player B ends up way ahead.
I am working on a spreadsheet to test your theory, both directly (no zeros) and real world (double zero).
ETA - with no zeros, B is way ahead. Toss in a double zero and B ends up way behind.
Why do you think that will be the case?
Quote: StatsWhy do you think that will be the case?
I built a spreadsheet to test the theory in the OP.
With a 36 number wheel, even/odd always ended up down, 1-27 always ended up winning big.
Add two zeros (37 & 38, always losers) and both plans ended up losing and 1-27 lost A LOT.
ETA - I was not betting 27 units on the outside. If I bet it that way, Plan A always (er, usually) loses more than Plan B.
Quote: TerribleTom
Add two zeros (37 & 38, always losers) and both plans ended up losing and 1-27 lost A LOT.
ETA - I was not betting 27 units on the outside. If I bet it that way, Plan A always (er, usually) loses more than Plan B.
So when your risking the same amount both sides eg bet 9 units and then if required 18 on outside- it was worse than betting 27 units inside each time player a was making the 9 unit bet?
Thats interesting, I wonder what the reason is that one method works better than the other?
Quote: StatsSo when your risking the same amount both sides eg bet 9 units and then if required 18 on outside- it was worse than betting 27 units inside each time player a was making the 9 unit bet?
Thats interesting, I wonder what the reason is that one method works better than the other?
He is clearly mistaken. The bets are all 0 EV. That means that, regardless of betting system, your results will approach 0% of the amount bet as your number of bets increases. It is not possible for one to be "way up all the time" and the other to be "way down".
The question you asked was more about variance than EV. Both systems have the same variance. They are essentially identical.
Quote: AxiomOfChoiceHe is clearly mistaken. The bets are all 0 EV. That means that, regardless of betting system, your results will approach 0% of the amount bet as your number of bets increases. It is not possible for one to be "way up all the time" and the other to be "way down".
The question you asked was more about variance than EV. Both systems have the same variance. They are essentially identical.
Would this also be true on a real wheel with zeros? As one player is placing more bets but at less value.
Quote: StatsWould this also be true on a real wheel with zeros? As one player is placing more bets but at less value.
It sounds like both players are putting down the exact same amount of action at the same house edge so the EV is the same, so the results will converge to the same value (same % of same amount bet)
It also sounds like, per set of spins (which can either be one or two) both players bet with the same probability of winning and losing the same amounts, so the probability distributions are identical -- both should be equally likely to be within any particular distance of mean after any particular number of trials (where a trial is 1 set of spins).
Quote: AxiomOfChoiceHe is clearly mistaken. The bets are all 0 EV. That means that, regardless of betting system, your results will approach 0% of the amount bet as your number of bets increases. It is not possible for one to be "way up all the time" and the other to be "way down".
The question you asked was more about variance than EV. Both systems have the same variance. They are essentially identical.
There was definitely a problem with the spreadsheet. Once corrected, results would vary from refresh to refresh but no one system showed any advantage over the other.
I would have been suprised if there had been a consistant difference between the 2.
https://wizardofvegas.com/forum/questions-and-answers/math/16579-martingale-changes-the-house-edge/
If the game is fair, there is no difference. If it is house edge, there is.
Quote: kubikulannI have mentioned this in a similar thread.
https://wizardofvegas.com/forum/questions-and-answers/math/16579-martingale-changes-the-house-edge/
If the game is fair, there is no difference. If it is house edge, there is.
Thanks
