**Over the past few years, we have become acquainted with slot games featuring variations on the stick’n’respin theme. In one of the most popular of those, a 3×5 machine, the wild symbol is present on Reels 2-4. When it lands in the window, it expands vertically to the entire visible part of the reel. The paylines are then evaluated (both left-to-right and right-to-left!) and the wild reel sticks during a free respin. If during a respin an additional wild symbol appears, it too will expand to the whole reel before line evaluation, and stick for yet another free respin. The process continues until no additional wild symbols appear. Thus, a player can get up to 4 spins with an increasing number of reels covered by wild symbols, all by staking and pressing the PLAY button only once.**

As a continuation of the previous post, it is briefly explained below how this can be solved by means of Markov chains. Each spin in a “stick’n’respin cycle” is started with each of Reel 2, Reel 3 and Reel 4 being either wild or not. We will therefore work with 9 states numbered from 0 to 8, where 0-7 represent spinning with one of the possible wild distributions (think binary!) and 8 represents “no more spins”. Knowing the probabilities that a wild symbol appears on a specific reel, we readily get the transition probabilities between the states. Always performing the initial spin in State 0, it turns out that we will surely end up in State 8 no later than prior to the 5th consecutive spin (which is thus not executed).

If we consider a 3-bit binary number, where the least significant bit represents Reel 2, then we get a natural enumeration of the states 0-7. If the “wild probabilities” are, say, 0.02, 0.05 and 0.03, for reels 2, 3 and 4, respectively, and we repeatedly multiply the resulting transition matrix from the left by an initial vector with probability 1 for State 0 and probabilities 0 for all other states, we end up at the following table.

It is then a matter of calculating the conditional averages given that some of Reels 2-4 are wild which, to a large extent, is a matter of copy and paste (once we have the average for State 0). The *n*th column in the above table contains the probabilities that the *n*th spin is executed in State *k*, where *k=8* means that the spin is not executed at all.

Another popular concept involving sticky wilds is to let all wild symbols popping up during a freespin session stick in their positions for the remainder of the freespins, thereby overriding any symbol that appears underneath. Assuming a possibly random but fixed number of freespins, we can solve this in a similar manner. Note that we may consider one payline at a time, and expand the average return linearly to all lines. If the wild symbol is present on all five reels, we have to deal with 32 states and need the conditional average for each of them. The corresponding probability table dictates how often the *n*th freespin is played with wild configuration *k*. These are then multiplied with the corresponding conditional averages and the sum of the products is the average total session win. Watch out if retriggers are possible!

A third variation on the same theme is to let the triggering scatter symbols turn into wilds and remain in their positions during an entire freespin session. It is standard that three or more scatter symbols are required (anywhere in the symbol window) in order to start the freespins. There are a number of differences compared to the previous concept, the most important ones being that we do not start at State 0, and different paylines are in general affected differently by the wild symbols. For simplicity, we assume that no additional wild symbols can appear during the freespins. Suppose that the game has 20 paylines. Then, given a triggering scatter/sticky wild combination, some of the paylines may not be affected at all while others have one or more wild symbols given.

Thus, in general, for each possible triggering combination we get one “line count” for each of the 32 possible wild configurations, adding up to 20. Again, we need the conditional averages for all possible wild configurations. Once those are in place, we weigh them with the line counts mentioned above to get the average total win per spin.

In order to complicate things, the latter two can be combined. It gets a little messier if we allow retriggers, since then we don’t know in advance how many freespins will be played. Also, we need to define whether or not a scatter symbol shall count should it appear underneath a sticky wild. All in all, however, it is usually possible to solve the game analytically with a bit of caution.

Hi, you say it gets a little trickier when introducing retriggers. Can you explain how you would solve this problem if you introduced a retrigger symbol on the reels please? For example 5 of a kind jacks on a winline gives an extra 4 spins. Thanks and great post!

Hello Mike, and thanks for your comment.

It is usually possible to calculate the total number of spins separately. Note, however, that an average number of spins is not good enough, in general we need the probabilities for a total of

nspins for any possiblen. These may or may not depend on the states, depending on the retriggering rules.In your example, with a non-decreasing number of sticky wild symbols and a line win combination as the trigger, it may happen after a few spins that a payline entirely has wild symbols on it. Does that mean we get 4 extra spins for every spin and stick in an infinite loop of freespins, or should 5 wilds count as something else? Do we get 8 extra spins for 5 of a kind jacks on two lines on the same spin? It might be a good idea to have triggers independent of the lines, but either way I think the above would cover this case as well (given that the rules are clear).

I’m glad you like the post!

Thank you for your detail reply. I must admit that seeing sticky wild calculations are possible without simulations has got me excited as I never thought if was possible.

In reply to your questions all wilds on a winline would only give a cash prize and not additional freespins. However if you did get 2 winlines with 5 jacks on the same spin that would indeed award 8 freespins.

The part I am failing to understand is the retriggers still. I can easily calculate the probability of getting 5 jacks from a single spin. However the probability of getting 5 jacks on the nth spin will be different because there are more wilds in play from previous spins.

So I have no idea what the table will look like which takes into account both sticky wilds and retriggers because although you say that retriggers can be calculated separately don’t they both affect the other?

Also I assume the maths only ever need to focus on one winline and then just multiply the final rtp up by the number of winlines to get the total rtp?

Thanks again for your reply. It has been a massive help

Mike, the probability of retriggering may or may not depend on the distribution of sticky wilds (I suppose it does in your case of 5 jacks). But the probability for a certain wild distribution after a certain number of spins is known once the Markov chain has been expanded, and the probability of retriggering on the

nth spin is then a weighted average. Assuming you work in Excel you are likely to need one tab per possible wild distribution anyway (32 if wilds can appear on 5 reels) and from each of them it should be possible to work out the triggering probability.Hi Leo,

Once again thanks for your reply.

I have been studying Markov chains with transition matrices and have successfully got sticky wilds calculations accurate. Minus retriggers.

You are correct I do have 32 sheets in excel, one for each wild combination. On each sheet I have also calculated the chance of getting a 5 jack retrigger.

Now assume I am giving 8 freespins then the Markov table has 8 entries for each of the 32 combination.

Now what I think I need to do is combine the 8 probabilities of each state in the chain and multiply it by that states retrigger chance. Then add all states retrigger chances together and you should end up with the correct number of spins.

So say 8 freespins averages 2 extra retriggers giving a total of 10 freespins. I assume I then need to extend the Markov chain from 8 to 10. This seems logical to me, however this could then change the retrigger calc from 2 extra to 3 extra. So from 8 to 11. So you’d then have to extend your Markov table by another 1 spin, which then repeats itself and never stops. How do you take this into account?

Similarly the number of retriggers may end up at 4.25 giving a total amount of freespins as 12.25 which cannot be represented in a Markov table. How does this work?

Thanks again for taking the time and helping me.

Kind regards,

Mike

Could you explain more what are the 9 states?

Hu Mike,

Can you kind enough let me know how to apply markov chain for sticky wild. Maybe you can get me some basic example. Thanks

Hello Kent,

State 0 means the initial state, where the reels did not spin yet.

State 8 means the reels have spun at least once, but there are no more respins.

States 1-7 mean one or more reel has a (sticky) wild on it and the other reels will spin again. 1, 2, 4 represent the reels 2, 3, 4 (these are the only reels that have wild symbols on them). If more than one of these reels are wild, the corresponding numbers are added to get the state.

Hope this helps!

Hi Leo,

Great blog, really informative.

I know this question has nothing to do with sticky wilds but wasn’t sure where else to ask.

If you have a set of reel bands, is it possible to calculate the payout percentage of these bands if they were used as “tumbling reels”, just like da Vinci’s diamonds.

Thanks

Hi Jack, thanks for your comment.

It’s quite straightforward to do tumbling reels. However, it is not practically possible to do them in your everyday Excel sheet. I write a piece of code to step through all combinations of initial stop positions. Note that given the reels, the sequence of tumbles is completely determined by the initial stop positions.

Good luck!

Hi Leo,

Thanks you for the awesome article. Im trying to recreate the results you show in the article, but I cant seem to be able to get the transition matrix right.

Can you help?

Thanks 🙂

Hi Tony, thanks for your comment.

What seems to be the problem with your transition matrix?

I use a column vector which I multiply from the left with the transition matrix, thus the entries of each column in the matrix need to sum up to 1. It is common to use row vectors which are multiplied from the right with the transition matrix, the latter of which thus needs to have rows adding up to 1. Could this be causing confusion?

Hi Leo,

Thanks you for reply, I multiply from the left with the transition matrix, which I calculate for all the possible start/end start that WILD can appear on reel 2, 3, 4. But I cant seem to get the result in your article.

Here is my calculation sheet: https://docs.google.com/spreadsheets/d/1XfpluQxZ8PgPdRd-8Arn1ge2p9Xvd7nC4JnT5o78Olk/edit#gid=0

It would be great if you can help point me to the right direction.

Hi Tony,

Note that the entries in each column in the matrix need to add up to 1.

Leftmost column is the probabilities to transfer from initial state to each of the possible states.

I use an additional state “no more spins” as the end state, so that a chain is a complete cycle of original spin possibly followed by a number of re-spins. The top left entry in the transition matrix should thus be zero, since no wilds on initial spin means that we transfer to the end state.

Second column contains the probabilities to transfer from state “wild on Reel 2”. There are four possible states to transfer to, namely “wild

on Reels 2+3”, “wild on Reels 2+4”, “wild on Reels 2+3+4” and “no more spins”. The probabilities are the products of the probabilities of wild/no wild on Reels 3 and 4. The remaining rows in this column are 0.

And so on…

I hope this helps!

Thanks you for the explain and sample. It helps a lot.

I have a quick question. Dont I have to multiply the probability of Wild for each reel with 3? since there are 3 rows on each reel?

Thanks

The example probabilities mentioned in the text include the “*3”. In general, you need the probabilities for a wild to appear anywhere in the symbol window, of course.

Let me know how it goes!