24Bingo
Joined: Jul 4, 2012
• Posts: 1348
August 30th, 2014 at 6:07:03 PM permalink
Quote: JyBrd0403

Here's the problem, the math these guys are giving are saying that the losses somehow magically keep up with the wins the D'Alembert produces. I haven't a clue how they think that's possible. The only thing I can see is they are basing the math on streaks ( losses in a row). The losses in a row, can grow the more trials you have, but there's no reason why the total losses would grow any higher than the losses in a row. I mean if the 50/50 game is on the positive side of wins and losses you can play a million trials and only go down -18 units. Again, my only point is that the D'Alembert wins should easily outpace the losses eventually, because the losses in a row are nothing to the D'Alembert.

Nope. Nothing to do with streaks. Remember a few simple facts (all these on a 50/50 game):

1. You will not necessarily be up, except at the end of each progression.
2. Although the chance of an individual progression lasting forever diverges to zero, the length has a nonzero chance of being arbitrarily high. A million, a billion, A(g64,g64), doesn't matter, it has a finite chance of happening.
3. In fact, the chances of large progressions decrease slowly enough that the expected length diverges - it's infinite.
4. Because it's infinite, a stop condition of "I'll play until my progression ends!" plugged into a calculation gets you a dope slap from Athena.
5. If instead you're stopped after an arbitrary number of hands t, the expected value of the bet you were about to make will be 1/2 + sqrt(t + 1/4).

So let's say, for the sake of argument, you can only play 1,332 hands. You happen to be having perfectly average luck as concerns both the game itself and the formula in 5, so you've won 666 hands and lost 666 hands, but of those 666 hands you won, 36 were on the first bet of their respective progressions, so not having counted them in your D'Alembert, you're about to bet 37 units. Tell me... when you get the tap on the shoulder, how much have you won?
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
JyBrd0403
Joined: Jan 25, 2010
• Posts: 548
August 31st, 2014 at 2:40:03 AM permalink
Quote: 24Bingo

Nope. Nothing to do with streaks. Remember a few simple facts (all these on a 50/50 game):

1. You will not necessarily be up, except at the end of each progression.
2. Although the chance of an individual progression lasting forever diverges to zero, the length has a nonzero chance of being arbitrarily high. A million, a billion, A(g64,g64), doesn't matter, it has a finite chance of happening.
3. In fact, the chances of large progressions decrease slowly enough that the expected length diverges - it's infinite.
4. Because it's infinite, a stop condition of "I'll play until my progression ends!" plugged into a calculation gets you a dope slap from Athena.
5. If instead you're stopped after an arbitrary number of hands t, the expected value of the bet you were about to make will be 1/2 + sqrt(t + 1/4).

So let's say, for the sake of argument, you can only play 1,332 hands. You happen to be having perfectly average luck as concerns both the game itself and the formula in 5, so you've won 666 hands and lost 666 hands, but of those 666 hands you won, 36 were on the first bet of their respective progressions, so not having counted them in your D'Alembert, you're about to bet 37 units. Tell me... when you get the tap on the shoulder, how much have you won?

Were you able to understand why the Marty is not considered a winning game? Was I able to explain that so you understand?

One question for you, if streaks don't matter, what exactly pushes the losses lower? If you're only losing 18 in a row in a million trials, how far do you think you can actually go down in a million trials? In other words, what's the loss limit for a million trials? If you're suppose to only lose 18 in a row in a million trials, how much do you think the total losses are suppose to be? What's mathematically pushing the game down? For losses in a row it's simply double the trials add 1 loss in a row. How do you think the total losses are mathematically determined?
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
August 31st, 2014 at 8:25:25 AM permalink
Quote: JyBrd0403

Were you able to understand why the Marty is not considered a winning game? Was I able to explain that so you understand?

I understand that you understand how quickly Martingale progressions increase, and correctly recognize that this means you could be playing quite a long time, and still be wiped out very quickly by a streak of losses. But a single win will still put you back on your merry way, rather than the slog (of mean length ∞) you'll need to get back to the point where you're genuinely ahead if you do fall behind on the D'Alembert. In fact, unlike the D'Alembert, "I'll play until I win" is a strategy with a finite expected time at table (but an infinite expected nadir to your progression).

Quote: JyBrd0403

One question for you, if streaks don't matter, what exactly pushes the losses lower?

As I said, wins on the first bet of your progression.

Quote: JyBird0403

If you're only losing 18 in a row in a million trials, how far do you think you can actually go down in a million trials?

If you're stopped after a million trials, sight unseen, you will on average be stuck in your progression for 500,000 units, which is how much you'd be stuck for after about 1,000 losses in excess of wins not sustained on the first bet of a progression. As for how much you "can" go down - well, ignoring wins on the first bet of your progression, you'd have a standard deviation of 1,000 out from the origin, so about a 1 in 6 chance of having taken more than 1,000 more losses than wins - about a 1 in 40 chance of having sustained 2,000. You might notice those numbers are a bit larger than 18. Unlike the Martingale, it doesn't matter if your losses are lined up in a neat little row, only that they're not offset by enough wins.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
JyBrd0403
Joined: Jan 25, 2010
• Posts: 548
August 31st, 2014 at 2:43:18 PM permalink
Let me ask you this. If the downside is Infinite, wouldn't the Upside be Infinite as well? We used to always say for +/- 300 or +/- 500 for 50% games. Meaning you might be up +300 or -300. All, you stated above is that you could win for Infinity or lose for Infinity +/- Infinity. You seem to only look at the loss side and ignore that you will also be up on a 50/50 game. IF you would take into account the upside you would see that a 50/50 game would have a range of how high or low it will go.

You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.
RS
Joined: Feb 11, 2014
• Posts: 8624
August 31st, 2014 at 2:59:59 PM permalink
Quote: JyBrd0403

Let me ask you this. If the downside is Infinite, wouldn't the Upside be Infinite as well? We used to always say for +/- 300 or +/- 500 for 50% games. Meaning you might be up +300 or -300. All, you stated above is that you could win for Infinity or lose for Infinity +/- Infinity. You seem to only look at the loss side and ignore that you will also be up on a 50/50 game. IF you would take into account the upside you would see that a 50/50 game would have a range of how high or low it will go.

You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

You're deluding yourself.
24Bingo
Joined: Jul 4, 2012
• Posts: 1348
August 31st, 2014 at 3:31:07 PM permalink
Quote: JyBrd0403

Let me ask you this. If the downside is Infinite, wouldn't the Upside be Infinite as well? We used to always say for +/- 300 or +/- 500 for 50% games. Meaning you might be up +300 or -300. All, you stated above is that you could win for Infinity or lose for Infinity +/- Infinity. You seem to only look at the loss side and ignore that you will also be up on a 50/50 game. IF you would take into account the upside you would see that a 50/50 game would have a range of how high or low it will go.

I literally don't know what you're trying to say here. All I'm saying is that the expected length of each progression is infinite. I don't know what you mean by "win for Infinity or lose for Infinity +/- Infinity." I don't know what the "upside" is, unless you mean that that means that the expected return at the end of that progression is also infinite. That's true, but that's only a further illustration of how impossible your strategy is.

Quote: JyBrd0403

You stated that you could be down -1000 after a million trials. That's fine.

No, I said you were reasonably likely to have 1000 or more losses than wins, or 1000 more wins than losses. I said you were likely to be stuck for an expected value of 500,000 units, i.e., as though you'd just lost a 999-unit bet and were about to make a 1,000-unit one (or won a 1,000-unit bet and about to make a 999-unit one).

Quote: JyBrd0403

But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001.

That's how it works when you're talking about wins vs. losses, but I wasn't talking about wins vs. losses, I was talking about the amount you were likely to be stuck by if you played exactly that number of hands, then got the tap on the shoulder.

The bet you were about to make after two million, that's 1400.

The bet you were about to make after four million, that's about 2000.

Of course, they could be much higher or much lower, but that's the average.

It keeps going up because every time you get back to the beginning of your progression, you have a 50/50 chance of winning, and those wins don't count. Over time, they add up.

Quote: JyBrd0403

In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

What the hell is a "loss limit," anyway?
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
DeMango
Joined: Feb 2, 2010
• Posts: 2958
August 31st, 2014 at 3:34:13 PM permalink
Can we run a sim and get over this thread?
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5745
August 31st, 2014 at 3:39:55 PM permalink
Quote: JyBrd0403

You stated that you could be down -1000 after a million trials. That's fine. But the next million trials you could be +1000. So, that's 2 million trials and the D'Alembert won 1 million. Then, you can have a million trials down -999, and the next million +999. That's 4 million trials the D'Alembert won 2 million. Now a million down -1001, you know what the difference is this time? Your in the Black, your playing on the houses money at -1001. In other words, on those favorable runs that WILL occur over time, is where the D'Alembert WILL outpace the loss limits.

I just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).
At one point, I got a run that consisted of the following before it eventually turned a profit:
621,937,312 bets
Max bet was 70,120x the initial bet
Lowest loss point was 2,147,438,594 initial bets
So if the minimum bet is \$1, you had to have had a bankroll of at least \$2.147 billion - otherwise you had to stop with an overall loss at some point.
Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet \$5; maximum bet \$350,000). I haven't seen many of those around.
Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.
JyBrd0403
Joined: Jan 25, 2010
• Posts: 548
August 31st, 2014 at 3:52:28 PM permalink
Quote: ThatDonGuy

I just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).
At one point, I got a run that consisted of the following before it eventually turned a profit:
621,937,312 bets
Max bet was 70,120x the initial bet
Lowest loss point was 2,147,438,594 initial bets
So if the minimum bet is \$1, you had to have had a bankroll of at least \$2.147 billion - otherwise you had to stop with an overall loss at some point.
Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet \$5; maximum bet \$350,000). I haven't seen many of those around.
Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Legitimate claim that the D'Alembert always wins. Remember you said that.

The D'Alembert doesn't have to be used for gambling alone. If I run simulations and it always wins, then the D'Alembert always eventually wins.

Congratulations, Don. You just proved every mathematician on the planet WRONG. How does it feel?

Also, I finally broke down and got one of these simulator things that I hate for myself. Already got the D'Alembert programmed in. I'll run some simulations tonight or tomorrow.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5745
August 31st, 2014 at 4:12:56 PM permalink
Quote: JyBrd0403

Quote: ThatDonGuy

I just ran another simulation, but this time, the stop point was not when losses = wins, but when you started to make a profit (so losses could still exceed wins).
At one point, I got a run that consisted of the following before it eventually turned a profit:
621,937,312 bets
Max bet was 70,120x the initial bet
Lowest loss point was 2,147,438,594 initial bets
So if the minimum bet is \$1, you had to have had a bankroll of at least \$2.147 billion - otherwise you had to stop with an overall loss at some point.
Also, you had to be at a table that would allow a bet of 70,000x your initial bet (e.g. minimum bet \$5; maximum bet \$350,000). I haven't seen many of those around.
Finally, even if you are playing a game that has one resolution every 5 seconds (not many of those around, either, except maybe electronic ones), that would require playing 24 hours a day for 35,991 days, so even if you had a team playing that could cover 24 hours a day, you have to play nonstop for 98 consecutive years.

If you can meet all three of those conditions, then you have a legitimate claim that D'Alembert always wins. Otherwise, such claims are meaningless.

Legitimate claim that the D'Alembert always wins. Remember you said that.

I don't remember ever saying that it didn't.

Quote: JyBrd0403

Congratulations, Don. You just proved every mathematician on the planet WRONG. How does it feel?

Name one mathematician, much less "all of them," that ever said that D'Alembert doesn't always "eventually" win given infinite time and bankroll.

What I want to know is, how is the claim "D'Alembert always wins eventually" useful in any way, shape, or form in real life? Certainly not in gambling.