Moneypot just took a huge loss? - page 6.

Dannie

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Quote from: Quickseller on January 30, 2016, 11:07:23 PM

if it is your opinion that this bet would allow excessive risks, then it is your responsibility as the owner/operator of MoneyPot to disallow such bets from being placed. As the owner/operator of MoneyPot, it is your responsibility to ensure that your bankroll investors are not exposed to excessive risk.

The problem is he (and probably all the investors) didn't know. In fact, even dooglus's first post in this thread said "I don't understand why this bet was allowed to go ahead...I never looked in detail at how the kelly criterion works when there are multiple mutually exclusive outcomes like in the case of plinko bets" and RHaver said "I'm quite surprised it would allow 0.4 BTC bet -- that seems a bit much to me". They are only aware of what Kelly formula suggested in this case now, after seeing that bet happened and re-visiting the math behind it.

Quickseller

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Quote from: ranlo on January 30, 2016, 04:39:48 PM

To me, risking up to almost 100% of the bankroll is ludicrous. Regardless of the CHANCE of that happening, it's essentially taking the risk that at some point (which, statistically speaking, would happen given enough time) someone completely wipes it.

Regardless of if this bet was kelly compliant, regardless of the question of if it was appropriate to allow this bet, and regardless of what the setters of MoneyPot were under RHaver, if it is your opinion that this bet would allow excessive risks, then it is your responsibility as the owner/operator of MoneyPot to disallow such bets from being placed. As the owner/operator of MoneyPot, it is your responsibility to ensure that your bankroll investors are not exposed to excessive risk.

I would tend to agree that allowing this much of the bankroll to be potentially lost in a single bet is probably not a good idea. According to dicesites.com, many of the top gambling sites have more then ~32k bets placed every day, which puts an excessive risk on the bankroll, as if bets are large enough, then the bankroll could potentially be wiped out multiple times per day. I am not exactly sure why the kelly suggests that this would be an appropriate risk, however this would probably be an example when what the kelly say should be 'overridden' and to allow less of a risk.

ranlo

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Quote from: dooglus on January 30, 2016, 01:15:13 PM

Quote from: EngiNerd on January 30, 2016, 11:23:11 AM

Should they be risking that much of their bankroll in a single bet? It seems like as we've just seen, that leaves MoneyPot and its investors exposed to a whale getting lucky and wiping them out.

According to the Kelly criterion, yes they should be. You need to me risking that much to maximum the expected log of your bankroll. If you only risk 1% of your bankroll on the highest payout, you're risking much less than 1% of your bankroll on the other payouts, and those are the most common. The big payouts are rare enough that apparently it's worth risking more than 1% and relying on the fact that they don't come up very often.

Quote from: elm on January 30, 2016, 11:57:35 AM

if you are saying that "the chance is 1 in 32768 to end up in either or the outside positions" do you mean that for all color bet options?

in case of green it is 3X
in case of orange it is 23X
in case of red it is 121X

Those are the payouts. I'm talking about the chance of them coming up, not the payouts when they do. The payouts are much lower than the odds against them happening because you also get paid when they don't come up. If you could set the payouts to 32768 in the two outsides and 0 everywhere else, that would be a zero house edge game.

Quote from: Blazed on January 30, 2016, 12:50:15 PM

So if I understand this correctly... tl;dr That is how plinkopot works and this guy just got really lucky? or is their an error with kelly?

The guy hit a 1-in-32k shot on his 27th attempt betting 0.4 BTC on the red line. That's about a 1 in 1200 chance of happening. So he got kind of lucky.

It appears that the bet was less than a full Kelly, and that the proportion of the bankroll that Kelly tells us to risk is much higher than our intuition would suggest.

Dooglus the math wizard comes through again, :p.

To me, risking up to almost 100% of the bankroll is ludicrous. Regardless of the CHANCE of that happening, it's essentially taking the risk that at some point (which, statistically speaking, would happen given enough time) someone completely wipes it.

While the house could win 10,000 BTC before that happens, the risk of an all or nothing is still fairly great. From the player's perspective, it's a great deal -- it essentially says that if you lose enough, you can keep your bets in line with the entire bankroll and at some point win back not only ALL your losses so far, but everyone else's losses + all investments that were made.

It's definitely a system that needs to be re-evaluated risk-wise. Plinko is an interesting game due to how it works, though. Ideally, what seems the best from a non-mathematically-inclined perspective is to take the maximum win allowed (say 1%, 2%, whatever) and divide it by the highest multipliers to determine the maximum bet allowed. But I'm sure I'm missing something here...

elm

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Quote from: dooglus on January 30, 2016, 01:15:13 PM

According to the Kelly criterion, yes they should be. You need to me risking that much to maximum the expected log of your bankroll. If you only risk 1% of your bankroll on the highest payout, you're risking much less than 1% of your bankroll on the other payouts, and those are the most common. The big payouts are rare enough that apparently it's worth risking more than 1% and relying on the fact that they don't come up very often.

1 in 32768 is not rare imo and it would depend on how much you will earn with those other payouts to justify such an high payout (in relation to the Bank Roll) for the outside chance.

elm

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Quote from: EngiNerd on January 30, 2016, 12:57:39 PM

Quote from: elm on January 30, 2016, 11:57:35 AM

Quote from: dooglus on January 30, 2016, 03:54:22 AM

Quote from: Game_Seller on January 30, 2016, 03:06:48 AM

I see so plinkopot max at 999x and betterbets has 9999x which means a much smaller bet could win large part of bankroll now the odds must be astronomical to win this.

It's 1 in 65536 to end up in the left hand position, and 1 in 32768 to end up in either of the outside positions. So not really "astronomical".

If you bet once per second you would hit one of the outside pockets every 9 hours or so.

Edit: on average, of course! I don't mean every 9 hours. I mean the average time between hits would be 9 hours in the (very) long run.

if you are saying that "the chance is 1 in 32768 to end up in either or the outside positions" do you mean that for all color bet options?

in case of green it is 3X
in case of orange it is 23X
in case of red it is 121X

Yes, the odds are "1 in 32768" to end up in either of the outside positions for all color bet options. The color of the row has no impact on the odds at all, just the payout multiplier.

are you sure? because this would be a terrible bet for a player if I see it as a standalone bet. for example I would bet 1 satoshi 32768 times (=32768 satoshi) to win a bet to get 3 satoshi - 23 satoshi or 121 satoshi

edit: I see now that dooglus answered it

dooglus

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Quote from: EngiNerd on January 30, 2016, 11:23:11 AM

Should they be risking that much of their bankroll in a single bet? It seems like as we've just seen, that leaves MoneyPot and its investors exposed to a whale getting lucky and wiping them out.

According to the Kelly criterion, yes they should be. You need to me risking that much to maximum the expected log of your bankroll. If you only risk 1% of your bankroll on the highest payout, you're risking much less than 1% of your bankroll on the other payouts, and those are the most common. The big payouts are rare enough that apparently it's worth risking more than 1% and relying on the fact that they don't come up very often.

Quote from: elm on January 30, 2016, 11:57:35 AM

if you are saying that "the chance is 1 in 32768 to end up in either or the outside positions" do you mean that for all color bet options?

in case of green it is 3X
in case of orange it is 23X
in case of red it is 121X

Those are the payouts. I'm talking about the chance of them coming up, not the payouts when they do. The payouts are much lower than the odds against them happening because you also get paid when they don't come up. If you could set the payouts to 32768 in the two outsides and 0 everywhere else, that would be a zero house edge game.

Quote from: Blazed on January 30, 2016, 12:50:15 PM

So if I understand this correctly... tl;dr That is how plinkopot works and this guy just got really lucky? or is their an error with kelly?

The guy hit a 1-in-32k shot on his 27th attempt betting 0.4 BTC on the red line. That's about a 1 in 1200 chance of happening. So he got kind of lucky.

It appears that the bet was less than a full Kelly, and that the proportion of the bankroll that Kelly tells us to risk is much higher than our intuition would suggest.

EngiNerd

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Quote from: elm on January 30, 2016, 11:57:35 AM

Quote from: dooglus on January 30, 2016, 03:54:22 AM

Quote from: Game_Seller on January 30, 2016, 03:06:48 AM

I see so plinkopot max at 999x and betterbets has 9999x which means a much smaller bet could win large part of bankroll now the odds must be astronomical to win this.

It's 1 in 65536 to end up in the left hand position, and 1 in 32768 to end up in either of the outside positions. So not really "astronomical".

If you bet once per second you would hit one of the outside pockets every 9 hours or so.

Edit: on average, of course! I don't mean every 9 hours. I mean the average time between hits would be 9 hours in the (very) long run.

if you are saying that "the chance is 1 in 32768 to end up in either or the outside positions" do you mean that for all color bet options?

in case of green it is 3X
in case of orange it is 23X
in case of red it is 121X

Yes, the odds are "1 in 32768" to end up in either of the outside positions for all color bet options. The color of the row has no impact on the odds at all, just the payout multiplier.

Blazed

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So if I understand this correctly... tl;dr That is how plinkopot works and this guy just got really lucky? or is their an error with kelly?

elm

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Quote from: dooglus on January 30, 2016, 03:54:22 AM

Quote from: Game_Seller on January 30, 2016, 03:06:48 AM

I see so plinkopot max at 999x and betterbets has 9999x which means a much smaller bet could win large part of bankroll now the odds must be astronomical to win this.

It's 1 in 65536 to end up in the left hand position, and 1 in 32768 to end up in either of the outside positions. So not really "astronomical".

If you bet once per second you would hit one of the outside pockets every 9 hours or so.

Edit: on average, of course! I don't mean every 9 hours. I mean the average time between hits would be 9 hours in the (very) long run.

if you are saying that "the chance is 1 in 32768 to end up in either or the outside positions" do you mean that for all color bet options?

in case of green it is 3X
in case of orange it is 23X
in case of red it is 121X

EngiNerd

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Quote from: dooglus on January 30, 2016, 03:54:22 AM

Somewhat more practically, the default 'orange' row at plinkpot has these payouts:

[23, 9, 3, 2, 1.5, 1.2, 1.1, 1, 0.4, 1, 1.1, 1.2, 1.5, 2, 3, 9, 23]

and moneypot will risk up to 83.90% of its bankroll per bet on that line.

To win 150 BTC at 23x you need to bet just 150.0 / (23 - 1) = 6.818 BTC per ball. (And get lucky).

Should they be risking that much of their bankroll in a single bet? It seems like as we've just seen, that leaves MoneyPot and its investors exposed to a whale getting lucky and wiping them out.

dooglus

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Quote from: Game_Seller on January 30, 2016, 03:06:48 AM

I see so plinkopot max at 999x and betterbets has 9999x which means a much smaller bet could win large part of bankroll now the odds must be astronomical to win this.

It's 1 in 65536 to end up in the left hand position, and 1 in 32768 to end up in either of the outside positions. So not really "astronomical".

If you bet once per second you would hit one of the outside pockets every 9 hours or so.

Edit: on average, of course! I don't mean every 9 hours. I mean the average time between hits would be 9 hours in the (very) long run.

Game_Seller

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Quote from: dooglus on January 30, 2016, 02:58:08 AM

Quote from: Game_Seller on January 30, 2016, 02:24:50 AM

So this means if I bet on any of the moneypot plinko games then there is a very tiny chance to win big money? Not a bad thing for us players but looks like odds are still very tough to beat. These bets are crazy look at dem odds.

At plinkopot you can set your own payout table, so long as the house edge isn't under 0.9%. So whether you have a tiny chance of winning big money really depends on the payout table you select. MoneyPot allows you to set the payout multiplier to over 1000x, but plinkpot apparently doesn't.

If you set the payout table to:

  [1.0656, 1.0585, 1.0368, 1.0155, 0.9616, 0.9841, 0.9895, 0.9918, 0.9962,
   0.9918, 0.9895, 0.9841, 0.9616, 1.0155, 1.0368, 1.0585, 1.0656]

then the house edge is 0.9997% and I believe moneypot will risk up to 99.974% of its bankroll per bet.

So in theory it is possible for someone to win almost all of the bankroll in a single bet.

But in practice to win 150 BTC when the highest payout is only 1.0656x you would have to be betting
  150 / (1.0656 - 1) = 2286.58
over 2.2k BTC per bet...

Somewhat more practically, the default 'orange' row at plinkpot has these payouts:

  [23, 9, 3, 2, 1.5, 1.2, 1.1, 1, 0.4, 1, 1.1, 1.2, 1.5, 2, 3, 9, 23]

and moneypot will risk up to 83.90% of its bankroll per bet on that line.

To win 150 BTC at 23x you need to bet just 150.0 / (23 - 1) = 6.818 BTC per ball. (And get lucky).

I see so plinkopot max at 999x and betterbets has 9999x which means a much smaller bet could win large part of bankroll now the odds must be astronomical to win this.

dooglus

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Quote from: Game_Seller on January 30, 2016, 02:24:50 AM

So this means if I bet on any of the moneypot plinko games then there is a very tiny chance to win big money? Not a bad thing for us players but looks like odds are still very tough to beat. These bets are crazy look at dem odds.

At plinkopot you can set your own payout table, so long as the house edge isn't under 0.9%. So whether you have a tiny chance of winning big money really depends on the payout table you select. MoneyPot allows you to set the payout multiplier to over 1000x, but plinkpot apparently doesn't.

If you set the payout table to:

  [1.0656, 1.0585, 1.0368, 1.0155, 0.9616, 0.9841, 0.9895, 0.9918, 0.9962,
   0.9918, 0.9895, 0.9841, 0.9616, 1.0155, 1.0368, 1.0585, 1.0656]

then the house edge is 0.9997% and I believe moneypot will risk up to 99.974% of its bankroll per bet.

So in theory it is possible for someone to win almost all of the bankroll in a single bet.

But in practice to win 150 BTC when the highest payout is only 1.0656x you would have to be betting
  150 / (1.0656 - 1) = 2286.58
over 2.2k BTC per bet...

Somewhat more practically, the default 'orange' row at plinkpot has these payouts:

  [23, 9, 3, 2, 1.5, 1.2, 1.1, 1, 0.4, 1, 1.1, 1.2, 1.5, 2, 3, 9, 23]

and moneypot will risk up to 83.90% of its bankroll per bet on that line.

To win 150 BTC at 23x you need to bet just 150.0 / (23 - 1) = 6.818 BTC per ball. (And get lucky).

Game_Seller

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So this means if I bet on any of the moneypot plinko games then there is a very tiny chance to win big money? Not a bad thing for us players but looks like odds are still very tough to beat. These bets are crazy look at dem odds.

blockage

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Quote from: blockage on January 30, 2016, 01:01:52 AM

Quote from: dooglus on January 30, 2016, 12:09:50 AM

Edit2: I tried reading the Haskell code. (I don't know Haskell. I don't even know how many L's it has):

Code:

table =
  [ ( 65536     / (2^32) , 1 - 121.0 ), ( 749731840 / (2^32) , 1 - 0.5   )
[...]
fun x = sum [ p * log (1 + b * x) | (p,b) <- table ]

It looks like you're using newton's method to maximize the sum of the products - but you're using the payout multipliers from the player's point of view. You need to look at it from the other side. The house never wins 121x. The player's big wins are the house's big losses. We need to calculate each house profit or loss as a multiplier of the amount the house is risking.

No, it's from the house's point of view. Notice the 1 - 121.0 in the second component of the first entry? It takes the stake of 1 into account and subtracts the payout to get the profit (-120) for the house in that case. Sorry I should've normalized the table a bit more.

Edit: Oh I see what you're doing. You're normalizing the stake for the house. Let me ponder about that some more.

Ok, I now believe that you are correct and my initial calculation is invalid. The payouts need indeed be normalized by the house's risk, i.e. the biggest amount the house can lose, which is 120 in this case. So the code isn't faulty per se, but the payout table was.

TwitchySeal

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Quote from: RHavar on January 30, 2016, 12:58:43 AM

Quote from: Emerge on January 30, 2016, 12:34:20 AM

So in layman's terms, did the bet go through due to an error or was it simply just luck? A huge heap ton of luck?

The bet going through or not is something completely deterministic. As in, if it's "kelly compliant" it will go through. If it's not, it won't. So there's no luck involved. The greater question is, was the bet "kelly complaint" or not? I'm not really sure. Both Dooglus and I have independently calculated it, and are getting the same result (which would indicate it is). Blockage has calculated it, and believes the house is risking ~120x too much.

Honestly, I don't know which is right. I have very little confidence in the hacked up code I wrote and the results are rather unintuitive (e.g. in some cases, Dooglus found it's recommending the casino to risk 90% of it's bankroll for a plinko bet with < 1% edge). Perhaps Dooglus and I are making the same mistake and getting the same result. But I also feel strongly like blockage's results are incorrect. I think that the house edge should be a lower-bound for the kelly, and blockage's result is under that. Although, it wouldn't be the first time he's proven me wrong.

Anyway, maths if for nerds. Just simulate it and it should be be easy to see

Yes, he got very lucky.

ranlo

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Quote from: RHavar on January 29, 2016, 07:07:44 PM

Quote from: dooglus on January 29, 2016, 06:40:13 PM

When I set up the payout table like that myself it tells me the max bet is around 9k bits, so I don't understand why this bet was allowed to go ahead.

That's definitely a huge underestimate. The max bet should be considerably larger.

Although, I'm quite surprised it would allow 0.4 BTC bet -- that seems a bit much to me. By any chance, did the owners of MoneyPot.com change the bankroll risk to a 10x kelly?

If anyone wants a worked example of figuring out the max bet for a normal kelly the maths is this:
http://math.stackexchange.com/questions/662104/kelly-criterion-with-more-than-two-outcomes

If no one else does it, i'll go through it for that bet and see what it should be

To answer this, none of the site's code has been touched since the takeover still. No alterations have been made anywhere in the site, nor has it been rebooted or anything. Current dev. is supposed to be getting this done asap (including things like the support emails), afaik.

blockage

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Quote from: dooglus on January 30, 2016, 12:09:50 AM

Edit: I should post the equation of the curve I'm plotting so you can check I didn't do anything stupid. I'm looking at the bet 'backwards', ie. from the house's point of view. I am considering the house bet size to be the most it can lose on the bet. So the 'b' for the jackpot is -1 (a profit of -1 times the bet size). 'b' for the 2nd highest win is -46/120 (we only lose 46/120ths as much as we were willing to risk because the highest two payouts are 121 and 47). Etc:

Code:

plot [0.352:0.354] \
      1/65536.0 * log(1 + x * -120/120) + \
     16/65536.0 * log(1 + x *  -46/120) + \
    120/65536.0 * log(1 + x *  -12/120) + \
    560/65536.0 * log(1 + x *   -4/120) + \
   1820/65536.0 * log(1 + x *   -2/120) + \
   4368/65536.0 * log(1 + x * -0.4/120) + \
   8008/65536.0 * log(1 + x *    0/120) + \
  11440/65536.0 * log(1 + x *  0.5/120) + \
  12870/65536.0 * log(1 + x *  0.7/120) + \
  11440/65536.0 * log(1 + x *  0.5/120) + \
   8008/65536.0 * log(1 + x *    0/120) + \
   4368/65536.0 * log(1 + x * -0.4/120) + \
   1820/65536.0 * log(1 + x *   -2/120) + \
    560/65536.0 * log(1 + x *   -4/120) + \
    120/65536.0 * log(1 + x *  -12/120) + \
     16/65536.0 * log(1 + x *  -46/120) + \
      1/65536.0 * log(1 + x * -120/120)   \
title "maximise this"

Why exactly do you divide by 120? The only difference to what I calculated is that factor in the 'b's. So my plot looks like

Code:

plot [0:0.006] [-1e-5:3e-5] \
      1/65536.0 * log(1 + x * -120) + \
     16/65536.0 * log(1 + x *  -46) + \
    120/65536.0 * log(1 + x *  -12) + \
    560/65536.0 * log(1 + x *   -4) + \
   1820/65536.0 * log(1 + x *   -2) + \
   4368/65536.0 * log(1 + x * -0.4) + \
   8008/65536.0 * log(1 + x *    0) + \
  11440/65536.0 * log(1 + x *  0.5) + \
  12870/65536.0 * log(1 + x *  0.7) + \
  11440/65536.0 * log(1 + x *  0.5) + \
   8008/65536.0 * log(1 + x *    0) + \
   4368/65536.0 * log(1 + x * -0.4) + \
   1820/65536.0 * log(1 + x *   -2) + \
    560/65536.0 * log(1 + x *   -4) + \
    120/65536.0 * log(1 + x *  -12) + \
     16/65536.0 * log(1 + x *  -46) + \
      1/65536.0 * log(1 + x * -120)   \
title "maximise this"

Quote from: dooglus on January 30, 2016, 12:09:50 AM

Edit2: I tried reading the Haskell code. (I don't know Haskell. I don't even know how many L's it has):

Code:

table =
  [ ( 65536     / (2^32) , 1 - 121.0 ), ( 749731840 / (2^32) , 1 - 0.5   )
[...]
fun x = sum [ p * log (1 + b * x) | (p,b) <- table ]

It looks like you're using newton's method to maximize the sum of the products - but you're using the payout multipliers from the player's point of view. You need to look at it from the other side. The house never wins 121x. The player's big wins are the house's big losses. We need to calculate each house profit or loss as a multiplier of the amount the house is risking.

No, it's from the house's point of view. Notice the 1 - 121.0 in the second component of the first entry? It takes the stake of 1 into account and subtracts the payout to get the profit (-120) for the house in that case. Sorry I should've normalized the table a bit more.

Edit: Oh I see what you're doing. You're normalizing the stake for the house. Let me ponder about that some more.

RHavar

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Quote from: Emerge on January 30, 2016, 12:34:20 AM

So in layman's terms, did the bet go through due to an error or was it simply just luck? A huge heap ton of luck?

The bet going through or not is something completely deterministic. As in, if it's "kelly compliant" it will go through. If it's not, it won't. So there's no luck involved. The greater question is, was the bet "kelly complaint" or not? I'm not really sure. Both Dooglus and I have independently calculated it, and are getting the same result (which would indicate it is). Blockage has calculated it, and believes the house is risking ~120x too much.

Honestly, I don't know which is right. I have very little confidence in the hacked up code I wrote and the results are rather unintuitive (e.g. in some cases, Dooglus found it's recommending the casino to risk 90% of it's bankroll for a plinko bet with < 1% edge). Perhaps Dooglus and I are making the same mistake and getting the same result. But I also feel strongly like blockage's results are incorrect. I think that the house edge should be a lower-bound for the kelly, and blockage's result is under that. Although, it wouldn't be the first time he's proven me wrong.

Anyway, maths if for nerds. Just simulate it and it should be be easy to see Tongue

Emerge

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So in layman's terms, did the bet go through due to an error or was it simply just luck? A huge heap ton of luck?

Topic: Moneypot just took a huge loss? - page 6. (Read 7588 times)