Moneypot just took a huge loss?

[klf]

Well, anybody who bets all maniac like that on games with a house edge is bound to throw it all back sooner or later...variance is a sour apple at times.  It's easy come and easy go in the minds of those who have the bug.  But, wow...it looked like investors were just cruising for awhile there...now a third of their investment gone! Ouch!  

In gambling site investments it is common and these are very high risky investments so no need to worry much. That's why we shouldn't put all our money in any one site to reduce our loses like this. Very rarely people win like this big amount from gambling and if he clever than he should withdraw his winnings.

[fullypak]

...Will take at least a few months to recover...

Maybe.

Or maybe he'll give it all back and then some in the next 20 minutes.

If you had won such a big amount of bitcoins, would you return it?
I don't think so 

Gamblers who are addicted often end up continuing to play, even after they win a large amount. Even though he was able to get 16 grand off MoneyPot, the temptation to play may arise again and he could potentially lose it all. Emotion definitely affects you hard.

That is true. After winning gamblers start thinking today they may be lucky so they can try some more time to win even bigger and mostly they will lose there winnings and only smart guys will withdraw either full or partial amount to enjoy there success.

[futurebit640]

...Will take at least a few months to recover...

Maybe.

Or maybe he'll give it all back and then some in the next 20 minutes.

If you had won such a big amount of bitcoins, would you return it?
I don't think so 

Gamblers who are addicted often end up continuing to play, even after they win a large amount. Even though he was able to get 16 grand off MoneyPot, the temptation to play may arise again and he could potentially lose it all. Emotion definitely affects you hard.

That is true. After winning gamblers start thinking today they may be lucky so they can try some more time to win even bigger and mostly they will lose there winnings and only smart guys will withdraw either full or partial amount to enjoy there success.

Once people addicted to gambling both winning and losing are problems for them because if they lose than they will try for winning and if they win and they will try for more profits so they don't have any intention to stop so gambling addiction is very bad for any one.

[blockage]

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

My (untested) calculations:

The kelly for that bet is a staggering 0.353028752 (aka the house should risk 35% of it's bankroll)?! wtf?

Seems rather unintuitive, and likely I made a mistake (and possibly twice, once when first writing the generalize kelly code).

Most of the work is in inverting the bet to be from the houses perspective, so here's what I came up with if it is of any help to anyone:

Well 35% seems rather large, but I believe that was within the range of the ~150-160 BTC bankroll, so bet #18530298 at least doesn't seem fishy that way. What's fishy is the 35%. So I went and calculated the kelly criterion for that  payout table. Here is some Haskell code:

Code:

table :: [(Double, Double)]
table =
  [ ( 65536     / (2^32) , 1 - 121.0 ), ( 749731840 / (2^32) , 1 - 0.5   )
  , ( 1048576   / (2^32) , 1 - 47.0  ), ( 524812288 / (2^32) , 1 - 1.0   )
  , ( 7864320   / (2^32) , 1 - 13.0  ), ( 286261248 / (2^32) , 1 - 1.4   )
  , ( 36700160  / (2^32) , 1 - 5.0   ), ( 119275520 / (2^32) , 1 - 3.0   )
  , ( 119275520 / (2^32) , 1 - 3.0   ), ( 36700160  / (2^32) , 1 - 5.0   )
  , ( 286261248 / (2^32) , 1 - 1.4   ), ( 7864320   / (2^32) , 1 - 13.0  )
  , ( 524812288 / (2^32) , 1 - 1.0   ), ( 1048576   / (2^32) , 1 - 47.0  )
  , ( 749731840 / (2^32) , 1 - 0.5   ), ( 65536     / (2^32) , 1 - 121.0 )
  , ( 843448320 / (2^32) , 1 - 0.3   )
  ]

ps = sum [ p | (p,_) <- table ] -- Should be 1.0

fun, der, der2 :: Double -> Double
fun x = sum [ p * log (1 + b * x) | (p,b) <- table ]
der x = sum [ p * b / (1 + b * x) | (p,b) <- table ]
der2 x = sum [ - b^2 * p / (1 + b * x)^2 | (p,b) <- table ]

newtons :: Double -> Double
newtons x = if abs ((x - x') / x) < 1e-10 then x' else newtons x'
  where x' = x - der x / der2 x

kelly :: String
kelly = show (100 * newtons 0) ++ "%"

Which tells me

Code:

λ> kelly
"0.29419062808971297%"
λ>

To me ~0.29% seems more realistic than 35.3028752%. So assuming the bet was a max bet (which it probably wasn't but was close) the house overbet by a factor of roughly 120x.

[EngiNerd]

Within Kelly? 

This is after the bankroll took big hit from 121x

Outside paid 29,900,000

The max bet fluctuates constantly right?


Wager: 9,997,748 bits

Profit: +999,774.80 bits

Investor Profit: -1000417.2563276 bits

App Dev Profit: 642.456327602647 bits

House Edge: 0.96%

https://www.moneypot.com/bets/18559115

Whoa, so it allowed a bet of 9 BTC?? How is that possible?

[RHavar]

Whoa, so it allowed a bet of 9 BTC?? How is that possible?

The restriction is on profit, not wagered. You could wager >1000 BTC if you wanted, as long as you weren't trying to win too much

To me ~0.29% seems more realistic than 35.3028752%. So assuming the bet was a max bet (which it probably wasn't but was close) the house overbet by a factor of roughly 120x. Good job!

While it's more realistic, I think it doesn't pass the sniff test either. I believe the absolute lower bound would be the house edge, which in this case is ~0.92%.

[actmyname]

It does seem a little bit sketchy to me, though. Perhaps some thorough investigation and people that are great at maths should go into the statistics of this.

[dooglus]

What's fishy is the 35%. So I went and calculated the kelly criterion for that  payout table. Here is some Haskell code:

To me ~0.29% seems more realistic than 35.3028752%. So assuming the bet was a max bet (which it probably wasn't but was close) the house overbet by a factor of roughly 120x. Good job!

Kelly tells us to maximise the sum of the probabilities times the log of (1 + b.x) where b is the profit divided by the stake, and x is the proportion of the bankroll to risk.

Plotting that function against x I see:

  

Find the x value where that peaks, and we have the proportion of the bankroll to risk. It looks like it peaks at around 0.3, meaning we risk 30%.

Let's zoom in:

  

So it peaks very close to 0.353, meaning we risk 35.3% of the bankroll.

Thanks Kelly!

Is your Haskell code remembering that we aren't risking the stake which is returned to the player? On the potential profit (payout - stake) is "risk" to the bankroll.

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"

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.

[Emerge]

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?

[RHavar]

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

[blockage]

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"

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.

[ranlo]

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.

[TwitchySeal]

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.

[blockage]

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.

[Game_Seller]

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.

[dooglus]

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]

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]

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.

[EngiNerd]

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.

[elm]

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