100 free spins no deposit, and the figure that needs four thousand turns

1

What a published return is a statement about

Of all the figures this site reads off a machine, the return to player is the one readers treat as a forecast and the one least able to behave like one. It is an average, worked over a number of turns large enough that the averaging has settled. Nothing on the studio’s page says how large, and the figure carries no band of uncertainty around it, so it arrives looking like a property of an afternoon when it is a property of a population.

The question this card asks is therefore arithmetical rather than sceptical. Given a figure the studio prints as 96.71 per cent the sheet, at claims 20 and 21, how many turns does it take before a reader’s own realised return lands near it often enough to be worth calling a description? Near enough is a choice and this card makes it the obvious one: within one house edge, which is 3.29 points, because that is the distance between the published figure and the point at which a reader stops losing money on average. A realised return further from the published one than the published one is from break-even has not described anything useful.

The award weighed below is a hundred turns for nothing paid in, winnings carrying thirty times before withdrawal a count and a multiple this card treats as typical rather than reading off anything, and a turn costing ten pence the reel’s own least stake, picked by this card. Both get named again at each use. A hundred is about the largest count out here that still reads as an afternoon rather than as a project, which is what makes it the right count to ask the question against.

2

The input the usual arithmetic needs and nobody prints

The ordinary way to answer takes three things: the band, the confidence, and the spread of one turn’s outcome. The first two are choices and are made above and below. The third is a property of the machine, and it is the quantity this entire network would reach for a volatility rating to supply.

No such rating exists. The word turns up on none of this studio’s thirty-six game pages, the session response a client receives has no field answering to it, and the studio’s announcements are silent as well game sheet, claim 104. Reviewers supply characterisations instead, and the three read for the sheet put the same machine in three different places, with one of the three disagreeing with itself from its table to its body claim 121, which prints all three and chooses none. A quantity with three values and no publisher cannot go into an arithmetic.

Deriving dispersion properly from what is published is a substantial piece of work and it is the business of the section on how long a balance survives, which is built around exactly that problem. This card needs much less. It needs a floor under the spread rather than the spread itself, because a floor under the spread gives a floor under the count of turns, and a floor under the count is the whole of what the question asks for.

3

A floor under the spread, built from two published figures

Two figures are published and both are primary. No round of this machine may pay above 2,100 times its stake, and how often a round pays precisely that is given in the session data, at 1 in 3,880,481 claims 27 and 29 of the sheet. That pair is enough on its own to put a floor under how widely outcomes scatter.

The reasoning is short. A spread is the average of the squared distances of outcomes from the mean. Every one of those squared distances is a positive quantity, so leaving all but one of them out can only make the total smaller. Keep the single outcome at the ceiling and drop the rest: the distance from the mean is a shade under 2,100 times the stake, the chance of it is one in 3,880,481, and the product of the chance and the square of the distance is 1.1354. The spread of one turn’s outcome cannot be less than that, so its square root, which is the figure the sample-size arithmetic actually uses, cannot fall under 1.0656 of one stake this card’s own arithmetic on claims 27 and 29.

Two things are worth noting about that figure before it is spent. Being a floor rather than an estimate is the first: the paytable carries other large outcomes and the feature round carries a multiplier climbing to ten, and all of them add to the spread while nothing subtracts from it. And it is already startling on its own terms. A spread of 1.0656 against a mean of 0.9671 means the typical squared departure of one turn is larger than the mean itself, which is a way of saying that one turn of this machine tells a reader nothing whatever, before anything has been computed about a hundred of them.

4

The count that follows, and why it is itself a floor

How loose the floor is can be judged from what went into it, which is one outcome and nothing else. Four high-paying rows stand in the same session data, their top entries reading 2000, then 1000, then 500 claim 49 of the sheet, and a money symbol may be worth as much as 2,000 times the stake, with a multiplier climbing to ten able to act on it claims 39 and 43. Every one of those is a large departure from a mean of 0.9671 and would push the floor upwards. None of their rates is published, so none of them can be counted, and the floor stays where one published pair of figures puts it.

Now the sample size. An average scatters less than a single turn does, by a factor that is the square root of how many turns went into it, so the count wanted to pull an average inside a stated band is the spread squared over the band squared, scaled by whatever the chosen confidence asks for. Take the band as one house edge, 3.29 points, and the confidence as nineteen times in twenty, which is the convention this card adopts and declares a band and a confidence chosen here, not read anywhere. The multiplier the convention carries is 1.96.

Put the floor from entry three through it and the answer is 4,030 turns. That figure is a floor twice over. Once because the spread it rests on is a floor, so the true count is larger. And once more for a reason worth stating plainly: the arithmetic above treats the average of many turns as settling into a symmetrical bell, which it does eventually and does slowly when the underlying outcomes are as one-sided as these. A machine whose spread is dominated by rare enormous payouts converges from below, so a count derived this way flatters the machine. Four thousand is the kindest answer the published figures permit.

At least 4,030 turns before a realised return sits within one house edge of the published figure nineteen times in twenty, which is forty times the count this award hands over

The band is the lever, and it is worth showing how hard it pulls, since the count moves with the square of it. Widen the band to twice the house edge, 6.58 points, and the count falls to about 1,008 turns, which an award of this shape with its own requirement would very nearly supply. Narrow it to half a house edge and the count climbs to about 16,119, which is eleven hours of continuous play and £1,612.00 of stake. Nothing about the machine changed between those three answers. Only the question did, which is the usual condition of a quantity nobody publishes a band for.

It is worth saying what the figure does not claim. It is not a prediction that 4,030 turns will produce 96.71 per cent; it is the count at which the published figure starts being a usable description rather than a distant one. And it is specific to the band chosen. Widen the band and the count falls fast, since the count moves with the square of the band: accept a realised return within twice the house edge and a quarter of the turns will do. The band this card chose is the one that makes the published figure mean something, which is why it was chosen.

5

How far a hundred turns is from four thousand

A hundred is about a fortieth of 4,030, and a fortieth is further away than it sounds, because the convergence runs on the square root. To halve the distance between a realised return and the published one takes four times the turns, so a hundred turns is not forty times too few in the sense of being nearly there. On the band this card chose, a hundred turns leaves the realised return scattered across a range about six times wider than the house edge itself.

The award does not stop at a hundred, though, and this is where the arithmetic gets interesting rather than merely discouraging. The thirty-times requirement makes a reader stake the winnings again, and staking is turns. Thirty times £9.67 of expected winnings is £290.10 of turnover, which at ten pence a round is 2,901 further turns. Add the hundred the award gave and a reader who clears the requirement has played 3,001 rounds without paying for any of them.

Three thousand and one against four thousand and thirty is a far better showing than a hundred against four thousand and thirty, and it is the only genuinely surprising thing on this card. An award of this size, taken with its condition rather than without it, gets a reader three-quarters of the way to the count at which the machine’s published figure begins to describe what happened to them. The missing piece is 1,029 turns, which at ten pence is £102.90 of a reader’s own money.

6

What reaching the count costs in money and in hours

Price the whole 4,030 and the cost is not the stake, which goes round and round, but the edge taken out of it on each pass. At ten pence a round, 4,030 turns lay £403.00 across the reel, and a house edge of 3.29 points takes £13.26 of that in expectation. That is the real price of finding out whether the published figure describes you: about thirteen pounds, paid for the privilege of a sample large enough to be worth looking at.

The time is the other half. A licensed reel is capped at 1,440 rounds an hour, that being the hour divided by the 2.5 seconds a game cycle may not come inside RTS 14D. At that pace 4,030 rounds take two hours and forty-eight minutes, without a single pause for thought, a cup of tea or a decision. A reader who stops to look at anything takes longer.

There is a tidier way of putting the money, and it is the one to carry away. Thirteen pounds and twenty-six pence is what the sample costs; thirteen pence is the award’s own expectation. A reader would have to be given about a hundred awards of this shape, and play every turn of every one of them, before the published figure started describing what had happened to them, and by then they would have paid out rather more than the hundred awards came to.

Set those two beside the award. A hundred turns is 250 seconds on the same clock, call it four minutes and ten seconds, and the award’s own expected close is thirteen pence. The sample costs thirteen pounds and nearly three hours; the award supplies four minutes of it and thirteen pence. The ratio between what the question costs to answer and what the award is worth is the plainest measure on this card of why nobody runs the experiment.

7

The requirement large enough to buy the sample

Entry five left a reader 1,029 turns short, and there is an obvious way to close the gap that turns out to be a trap worth laying bare. The requirement is what supplies the turns beyond the award’s own hundred, so a larger requirement supplies more of them. Solve for the one that supplies all of them: a requirement of about forty-one times on £9.67 of expected winnings generates enough turnover to carry a reader through the whole 4,030 rounds without a penny of their own going in.

Forty-one times is past break-even, which on this reel sits at 30.4. So the only requirement generous enough with turns to make the published return describable is one that takes the award’s expected close well below nothing on the way. At forty-one times the award closes at −£3.37 rather than at thirteen pence. The reader would have their sample, and the sample would be a sample of losing.

That is the shape of the thing, and it is not an accident of the figures chosen. Turns beyond the award’s own count are bought with turnover, turnover is bought by requirement, and requirement past the reciprocal of the house edge destroys more expectation than the award ever contained. A reader cannot get a large enough sample out of a free award without accepting terms that make the award worthless, and cannot get a worthwhile award without accepting a sample too small to tell them anything. The two desirable things sit on opposite sides of 30.4 times.

8

What the award comes to on two reels of the family

The award’s own arithmetic is quickly done. A hundred turns at ten pence lay £10.00 across the reel, of which a published 96.71 per cent gives £9.67 of it. The requirement then applies, and what survives is E = B × (1 − W × (1 − RTP) ∕ g). W here is the thirty; B is that £9.67; g is the share an offer counts a round of the clearing slot at, which is a whole round unless the terms reduce it counted whole, in the terms; and RTP is whatever percentage stands published against the reel the staking is taken to.

The two rows straddle something rather than differing by a little. Thirty times is four-tenths of a multiple inside break-even on the upper reel and more than four multiples outside it on the lower, so the same award is worth a positive thirteen pence on one and a negative £1.73 on the other. Sixty-four hundredths of one point has moved the answer by £1.86 and turned a gain into a loss, which is what a requirement sitting almost exactly on a break-even boundary does to a family with a dozen different published figures in it.

Which brings the two halves of this card together. A reader cannot tell from a hundred turns which of those two rows they are in, because a hundred turns is a fortieth of the sample that would settle it. They can tell by reading the published figure of the reel they are about to clear on, which takes a moment and settles it exactly. The published return is a poor description of a short session and a perfect description of which row of that table a reader has chosen.

9

The three things a hundred turns does settle

Three statements about an award of a hundred turns are exact, and they are exact because none of them is a statement about an outcome. The stake is £10.00 at the price supposed here, because a count multiplied by a price does not have a distribution. The time is 250 seconds at a licensed reel’s own fastest lawful tempo. And the expectation is thirteen pence on the reel weighed here, which is a true statement about an average of a great many awards of this shape and about nothing a single reader will see.

Two statements are not available and no page should supply them. What a reader’s own hundred turns will return is the first; the honest answer is that it is scattered across a range several times wider than the house edge, with a floor under that scatter and no published ceiling on it. The second is how close a hundred turns gets to the published figure, which this card has bounded from one side only: at least 4,030 turns are needed and the true figure is larger.

There is a smaller count where the question changes shape entirely rather than merely getting worse, and it has a card of its own at 5 free spins no deposit, where expectation stops being the right instrument at all and only the published ceiling has anything to say. In the other direction, the count that decides whether the machine’s feature round is reached at all runs into a figure nobody publishes, and that is settled at 40 free spins no deposit.

One thing could change all of this and it is not in a reader’s gift. A trigger frequency, a base-game hit rate or a dispersion figure published per title would turn every bound on this card into an answer, and the sheet records that the studio publishes none of the three. What is left to a reader is the one figure published per title, read before the choice of reel rather than after it, because that choice is settled exactly and in advance, where the outcome of a hundred turns is settled by nothing anybody can see.

So the finding of this card is a comparison rather than a number. The machine’s headline figure needs at least 4,030 turns before it describes anything a reader experiences, and the award on the table supplies a hundred, or three thousand and one if its own requirement is cleared. Thirteen pence is what the award is worth. Thirteen pounds and nearly three hours is what finding out would cost.