TS is correct. It's hard to argue against you when you only say "No" and fail to back it up with any kind of argument about why increased volume wouldn't lead to decreased variance.


You seem to be misunderstanding. During the ~29,999 losing bets the bankroll grows by a factor of more than 100, so when the lucky bet wins 99% of it the bankroll is still bigger than at the start.

The current setting optimises the rate of growth of the logarithm of the bankroll, and so is set 'correctly', especially given the terms set out in the FAQ about allowing full-kelly bets.

I expect the people running the site will overreact to this lucky win and reduce the maximum bet. In doing so they will be reducing their expected profit, and also their variance. I did the same at Just-Dice - it's hard not to overreact in the face of suffering a big loss like this.

I didn't hear about clamtopia before.

The link to just-dice is missing a hyphen.

That could be anything in that package.

Where's the beans?

Withdrawals don't change profits. Wins and losses change profits.

The win already happened, and so is already reflected in the number you see.

Is that something I can get from shapeshift directly? Or something I need to implement myself?

Can you show me other sites which have what you're asking for?

I don't know whether OP won, but someone did.

Here's a list of all 114 of the winner's bets. Each line ends with a reference to one of 11 different payout rows he used. They are listed at the end:


These are the 11 different payout rows played against. A-Q are the 17 holes from left to right. Amounts are in satoshis. mm means millions, k means thousands. row-10 pays 4840mm satoshis for an outside hole; that's 48.4 BTC:

Your grammar is weird, but to be clear I do not and never have owned any part of moneypot.


Don't feed the troll.

It's clear nobody understood the risks, not even the guy who coded moneypot...


That is interesting. It took me a while to understand your point, so I guess others may not understand it too.

You noticed than none of the payouts in the array are zero, so if you bet 1 BTC you can't lose the whole 1 BTC no matter which payout you get. And so in a sense you aren't really risking 1 BTC. The losest payout is 0.3x, so the worst that can happen is that you lose 0.7 BTC. So you recalculated the multipliers such that you get the same result from betting 0.7 BTC with your new payouts as you would have got betting 1 BTC with the original payouts. Then there's a zero in the array, and you're truly risking the whole amount you bet.

Calculate the new multipliers as (oldMult - minOldMult) / (1 - minOldMult)
So the 121 jackpot becomes (121 - 0.3) / (1 - 0.3) = 172.4285


Right. The house edge of the original row of payouts is 0.9195%, and after 'normalizing' it, it goes up to 1.3136%. You're losing the same, but betting smaller, and so are losing a bigger percentage of your stake.


Yes. Before normalization the house edge of this line:
    [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];
is 0.9997%.

After normalization:
    [2.7083, 2.5234, 1.9583, 1.4036, 0.0, 0.5859, 0.7265, 0.7864, 0.9010,
     0.7864, 0.7265, 0.5859, 0.0, 1.4036, 1.9583, 2.5234, 2.7083]
it is 26.0342%.

'Kelly' tells us the same before and after normalization: risk 99.97% of the bankroll. This doesn't seem overly unreasonable. Who is going to play a 26% house edge game at large enough stakes to stand a chance of winning the whole bankroll?

To gain an intuitive understanding of why Kelly tells us to risk more than 26% of our bankroll on a 26% house edge bet, it helps to see the plinko bet as 17 separate bets. For the low payout bets where the ball lands towards the middle we would want to set quite a high max bet, since the payout is a small multiplier of the stake, but for the high payout bets on the outside edges we would want to set a low max bet. We can only set a single max bet, and so we need to find some way of balancing the competing forces. It turns out we maximise the log of the bankroll by giving more weight to the max bet implied by the more common (inner, low payout) outcomes.

As the wikipedia page says:




And so you should never play a plinko game that doesn't have a 0x somewhere on it?

You just need to stop your custom code, edit it to make whatever changes you want, then start it again.

I think you're looking at it wrong.

The blockchain forked when the transaction creating the 184 billion BTC was mined. "Correct" clients will have rejected that block and started building their own correct fork, while Bitcoin Core went off on an invalid chain.

Fixing the Core code and forcing it back onto the correct chain resolved the fork that was caused by the bug; it didn't create a fork.

It's possible that nobody found a block using anything but code derived from Core, and so the correct branch didn't get to diverge from the incorrect branch, but that's beside the point.

I hadn't been on your site for a week or so, and thought I'd maybe play a sit&go.

When I looked, there were only 4 games available:

 

Is that right? The buy-in for a 6-max game is either 0.1 or 100. That's $0.04 or $40. Isn't that too much of a gap? It seems kind of silly spending an hour on a 4 cent game, but $40 is probably more than most casual players would want to risk on a single tournament.

And the only heads-up game available is for 50 mBTC ($20). Last time I was on the site I played a 5 mBTC ($2) heads-up sit&go, but today I couldn't find it. I would suggest offering a much wider range of games if you want people to play.

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.


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.


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.

There's no affiliate program because the house edge is only 1% and the commission on investments is only 10%.

I could increase the commission, and use the extra money to pay affiliates, but I don't really want to. It's also hard to prevent people from abusing affiliate programs by referring themselves.

Auto-betting is just silly. The game has a positive house edge. If you bet automatically you will lose and not even get to enjoy playing.

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.

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).

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:


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


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.

I just checked his bet history. He made almost all his profit from a single bet:

  https://www.moneypot.com/bets/18530298

The payout table was set to:

  121 47 13 5 3 1.4 1.0 0.5 0.3 0.5 1.0 1.4 3 5 13 47 121

and he hit the 121x

He bet 400k bits and so made a profit of 48 BTC.

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.

 

This is in agreement with the moneypot FAQ, which says:

 

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; maybe it says we should risk this much. Could someone look into it and see?

OK, that's fine. I'm looking forward to it.


Betting on red and black simultaneously is a great bet for you guys. There is zero chance of the player making a profit, and a ~3% chance of them losing. I would positively encourage players to make that kind of bet against me all day long.


I don't understand the variables. What is meta.gameOdds? What does odds of 0.5 represent? Is that a 50% chance of winning? So any bet with more than a 50% chance of winning doesn't count in full towards the wagering requirement? What is meta.gameOdds for blackjack? Does it depend on what cards were dealt? I'm guessing the players win less than 50% of blackjack hands, the > 0.5 condition doesn't trigger. I don't see an 'else' - what happens for games with gameOdds <= 0.5?

I think you need to make it clear that players may need to wager much more than 888x the bonus amount, depending on how they bet. Currently your bonus terms are misleading, saying that you "only" need to wager 888x.

Bump.

Why are you avoiding answering this?

You've understood correctly, but it isn't wrong.

For (2), if you happen to pick exactly a block creation time, use the time between that block and the one before it. No need to disregard your selection.

Let's make it even simpler:

1. Pick 10k random numbers between 0 and 10k. Sort them in order.
2. Pick random numbers in the same range. Find where they fit in the sorted list above. Find the size of the gap they fit into.
3. Average those gap sizes.
4. Get 2.00

Astonishing, isn't it? 10k points in a 10k range are an average of 1 unit apart from each other. But randomly pick points in that range and find that the average size of the gaps you land in is actually 2!

Some code:


Some results:


See how in the results we are tending to find the bigger gaps (5.12, 3.73, ...) and not the smaller ones. It's because the smaller ones are smaller, and so less likely to be randomly picked.

For the record, it's the end of next week, and there's a 200 BTC buy wall at twice the price quoted by Mr. Vulture:



I'm not sure I've ever seen a less accurate prediction.