Toy Blast – It’s A Force Delete

It’s official – I was finally forced to delete Toy Blast from my phone. Some time ago I thought this was one of the cool games, but it seems the random number generator is rigged, just like every other game. Of course, I recognise this is only my opinion, which I can’t force onto others. All I can do is turn to page 7374 of the Big Book of Clichés and say “judge for yourself”.

The problem is if you attempt to design a game with rigged RNG, it’s probably too easy to design an experiment that can expose certain issues. The culprit is level 7374 which looks like this:

It is beyond the scope of this document to explain the rules of Toy Blast in detail. The essential points are:

To beat this level, one of the requirements is getting at least three slices of bread. This is difficult since there is only one toaster which cannot be accessed with ordinary cubes. Power-ups are a must.
The basic power-ups are rotors, TNT and Rubik’s Cube. Rubik’s Cubes are always a particular color (one face is a solid colour, other faces are random).
Power-ups are obtained by matching 5 or more ordinary cubes of the same colour in one move.
Combos can be obtained by getting two or more adjacent power-ups. Adjacent means touching horizontally or vertically, but not diagonally.
If the player activates a Rubik’s Cube + rotor combo then every ordinary cube with matching colour turns into a rotor that can be oriented horizontally or vertically.

Let us define a “coin-flip” as follows. The player activates a Rubik’s Cube + Rotor combo when exactly one of the four cells in the top row is a matching colour. The following four images show example of coin-flips. For instance, the first diagram has Y-B-G-B in the top row and a Yellow Rubik’s Cube next to a Rotor in column 6.

Note that if more than one cell is the matching colour, we do not count it as multiple coin flips. For instance, if the top row was Y-G-G-G and we activated a Green Rubik’s Cube + Rotor combo then I do not count that as three coin-flips. It’s not even worth one coin-flip. It’s ZERO coin-flips – the same number of points you get for playing a phoney in Scrabble.

The term “coin-flip” should be self-explanatory. If (in the first diagram) every yellow cube turns into a rotor then the yellow cube in the top row will collect one bread with 50% probability. Any other yellow cube on the board cannot collect bread, even if the player were allowed to call directions for every individual rotor. As every student of probability knows, all this assumes rotors are indeed horizontal or vertical with 50% probability.

I played level 7374 multiple times with the objective of getting as many coin flips as possible (not necessarily maximising my chances of beating the level). With enough skill and luck, it’s possible to get three coin flips in a single game. After sufficiently many games I got 50 coin flips and every time the rotor was vertical instead of horizontal. I did not hit the toaster once in 50 coin flips. Yes – Every Single Time.

On one occasion, I was down to my last move (see images above) and was destined to fail the level for multiple reasons. And the software still saw fit to give me the vertical rotor in the top row.

Just out of interest, I have already beaten level 7374 (I’m up to 7377) and managed to beat all levels without spending any coins (e.g. getting 5 extra moves or strategically changing the colour of a single cube). It’s hard to avoid the conclusion that Toy Blast is trying to punish players that are playing a little too well.

It is quite possible that something untoward happened to Peak Games (e.g. acquired by another company) when I wasn’t looking – but I do not intend to discuss this in detail. I recommend the reader google search “Peak Games” and “Zynga”.

And if you happen to be a game designer – don’t even think about it. You are not fooling anybody.

Let me repeat that for the 7374^th time:

You. Are. Not. Fooling. Anybody.

It goes without saying the same holds for Spider Solitaire. I believe we have reached the point if Joe Bloggs claims the game is rigged then the onus of proof should be on the game developers, not the player.

To be fair, I should point out that Toy Blast survived far longer than some other game apps on my phone. Some really awful games (both match-three and card games) basically “didn’t even try to hide it”.

Mathematics of Doubling in Match-Play

In this post I wish to discuss the mathematics of doubling in some detail.

Let us assume Hero and Villain are playing a “match-to-three”. That is, first player to win three VP is the overall match winner. Each individual game is worth 1 VP. The actual game being played is irrelevant: it could be Chess, Agricola, Snakes and Ladders, the royal game (with Hero moving the cards and Villain trash-talking at every sub-optimal play), or a well-known poker variant where both players start with three items of clothing. To make things interesting, assume the winning probability of an individual game is slightly less than 50%. To be specific, I will set the win rate to 40%. What are the chances of Hero winning the overall match?

This kind of question is most easily solved with dynamic programming. The boundary condition says that if someone already has three wins then the overall winning chances is either 0% or 100%. Next, we can compute the winning chances when the score is 2-2. We can then work our way backwards, eventually arriving at 31.7% winning chances for the Hero at 0-0.

You will notice that if the match score is X-X then the Villain prefers smaller numbers of X. Intuitively, smaller values of X means that it is less likely that Hero will find enough “random noise” to overcome Villain’s advantage in the long run. The following diagram summarises the probability of Hero winning the match at every possible match-score:

I should mention that Bart has already done his homework, and he knows how to compute the probabilities corresponding to each match score. If the winning chances of an individual game are 0.25 then he correctly computes the following:

0.049 chance of winning 5-point match, no doubling
0.104 chance of winning 3-point match (equivalent to 5-point match with 2VP per game)

I will leave this as an exercise for the reader.

Now let us give Hero the following handicap: Before each game, Hero can demand the game be played for 1 VP or 2 VP. Moreover, there is no bonus for winning the match with 4 VP instead of 3 VP.

This means, for instance, if Villain has 2 points then Hero will always play for 2 VP. Again, we can use dynamic programming to compute the winning chances. If you use Excel to perform the dynamic programming then you will need the function max(FOO,BAR) somewhere in your calculations. You will notice several things:

The boundary conditions now include either player having 3 or 4 VP.
The table only shows winning percentages, but not whether Hero should play for 1 VP or 2 VP.
The numbers look weird: The winning chances on the main digonal is no longer strictly increasing and the winning chances for 1-2 is the same as 1-1.

The latter is easily explained. Since Hero can choose to play for 2 VP, Villain gets no advantage from being 1-2 instead of 2-2. Slightly more interesting is a match-score of 1-1. If Hero plays for 2 VP, then the next game decides the match. If Hero plays for 1 VP then the worst case scenario is the match score becomes 1-2, in which case Hero can play for 2 VP and the next game decides the match. Therefore, it must be correct to play for 1 VP. It turns out Hero is close to breaking even thanks to his handicap.

It is not hard to get an Excel spreadsheet to crunch the numbers for different parameter values. We can, for instance, figure out what happens in a match to 13 points with Hero’s winning chances down to 30% for an individual game.

When we talk about longer matches, it is generally more convenient to think in terms of number of VP remaining instead of number of VP already scored. For instance, 19-18 in a 21-point match is equivalent to 5-4 in a 7-point match and it makes sense to describe both of them as “2-away 3-away”.

Redoubles

Now what happens if either side has the right to double the stakes, but there are no redoubles? This is a trivial case: it makes no sense to refuse a double. If one side declines to double, he can’t prevent the opponent from doing the same. Therefore, it must be correct for either side to double.

But what if redoubles were allowed? This means e.g. if Hero proposes to play for 2 VP then Villain can propose to play for 4 VP before the game starts. Then things may get weird. I will leave the analysis as an exercise for the reader.

Doubling During the Middle of a Game

Hitherto, we have assumed that doubling could only occur before the start of every game. In this case, each individual game can be thought of as a black box – the only relevant parameter is the winning chances of a single game. Moreover, it never makes sense to pass a double since the loser gets the same “pre-starting position” and the winner gets a free point.

But if doubling were allowed during a game then the “structure” of the game tree becomes important. For instance, two different games could have the same winning chances of 25% but a different structure. If Hero doubles judiciously then he can leverage the structure to improve his overall chances of winning the match (of course he can’t leverage the structure to improve his chances of winning a particular game – if Villain were hell-bent on winning the next game at all costs, then he would never pass a double).

To illustrate the concept of structure, assume we are interested in maximising the expected number of VP for a single game instead of a “match-to-N-points”. Consider the following “random-walk-game”: A happy star randomly moves to one of the twelve coloured leaf nodes, with each node occurring with probability 1/12. If the colour is Green (Red) then Hero wins (loses) one VP. If no doubling cube is used then basic math says the expected gain per game is 1/3 = 0.333 VP for both structures depicted in the left and right halves of the diagram below.

Now suppose that Hero has the right to double before the game and Villain can double when the happy star reaches one of the three “intermediate” nodes. On the left diagram, assume that Hero doubles (since there are more greens than reds). Villain should accept and then Hero can expect to win 0.667 VP. But on the right diagram Hero is only winning 0.5 VP if Villain uses correct doubling strategy. This is left as an exercise for the reader.

Therefore, structure is important: without the cube Hero wins the same expected VP in both games, but with the cube Hero prefers the first game. Another lesson is that ownership of the cube (i.e. exclusive right to make the next double) is worth some equity. As a general rule, if all other things are equal then whoever owns the cube prefers game states that require many moves before one side has a decisive advantage.

Hopefully this example should make it clear why a simple mathematical analysis breaks down when we consider real games with doubling decisions occurring during the game. Similar considerations obviously apply when computing optimal match strategy rather than expected VP in a single game.

Summary

In this post I show that optimal use of the doubling cube is a lot more complex than “always double (take) if our winning chances are at least 80% (20%)”. There are three caveats:

The elephant in the room is nobody knows how to reliable estimate the winning chances of a specific game state. Not even Spider GM can do this, unless there are very few cards unseen
80% and 20% turn out to be optimal parameters – only if we assume the winning chances change continuously rather than suddenly change (think Brownian motion instead of quantum leaps!).
The parameters 80% and 20% also assume we are playing for money (e.g. $1 = 1 VP, aim to maximise expected winnings). It doesn’t work in match-to-N-points, especially when we reach the pointy end with both sides close to victory.

If we can solve the elephant in the room, then this doubling strategy should be a good starting point – coupled with a few “common-sense tweaks” near the end of the match. For instance, you never double when you are 1 point away from winning the match etc. But one could argue it is precisely the elephant in the room that makes Spider Solitaire such a great game 😊

Fun Fact

If Hero has the exclusive right to make the next double then it is possible to construct a pathological game tree where one can change some green nodes to red while also changing Hero’s correct doubling decision from No-Double to Double. A well-known Backgammon example is the Jacoby Paradox.

Will 2022 Be Year of the Ninja Monkey?

“Twenty Twenty Two Gonna Be Great Year!” shrieks Ninja Monkey.

“How so?” asks the Wise Snail.

“Twenty Twenty Two – My Year! Year of Monkey!”

“You don’t have to jump up and down all the time,” gripes the Sand Griper. “It’s annoying.”

“Besides,” roars the Tiger, “you have no evidence to back up your claim. According to the Chinese Zodiac, it’s MY year. The formula for Tiger years is 12x + 6 for any whole number x. Substitute x=168 and you get 2022. Quod Erat Demonstrandum.”

“Tiger is right,” says the Elephant. “I may not be the sharpest tool in the box when it comes to Phil Hellmuth’s menagerie of poker animal types, but I can remember every sign of the Zodiac and which year corresponds to which animal.”

“Not so fast,” says the Ox. “Chinese New Year starts on February not January, so I still get to enjoy approximately 30 more days of fame.”

“To add insult to injury,” adds the rot13(fzneg nff), “the formula for Monkey years is 12x for any whole number x. That means you are the maximum possible distance in either direction from one of your good years – therefore 2022 is the worst possible year for the monkey”

“Monkey don’t care, Monkey don’t care! Monkey invent his own Zodiac!”

“But you can’t invent something out of the blue just because a few unpleasant facts got in the way of a really good story” says the Eagle.

Ninja Monkey presents his own version of events: there are exactly 337 animals in today’s meeting. Each animal represents a different species – if we conveniently ignore the fact Ninja Monkey brought his GF along. Ninja Monkey assigned himself the year 0, his GF the year 1 and the rest of the animals different years in no particular order. How convenient it was that 337 happened to be a prime divisor of 2022. Quod Erat Demonstrandum.

“The monkey raises a valid point,” says the Wise Snail. “If the Zodiac caters for only 12 animals, then the vast majority of us miss out altogether. Monkey’s suggestion is much fairer even if it is not based on accepted tradition”.

With nobody able to rebut the Wise Snail’s statement, everybody stews in awkward silence for a few minutes. Finally Bad Idea Bear #1 comes up with a resolution: The Eagle would deal ten random hands and Ninja Monkey would have to win at least one game without rot13(haqb) using his Improved Random Move Algorithm. If the Monkey could achieve this then the new Zodiac is in. Otherwise, the Monkey would have to reluctantly accept the Tiger’s version of events and wait another six years.

Nobody else has anything better to offer, and for once a suggestion from a Bad Idea Bear gets unanimous agreement. At least there is no BIB #2 around to come up with something even worse. So, without further ado Let The Games Begin!

Game 1

Game 2

Game 3

“I thought things had started well,” sighs the Wise Snail, who is Ninja Monkey’s best friend.

Game 4

“Some promising signs there,” sneers the Tiger. “Too bad in the end!”

<<Several hands later>>

“Okay fine”, says the Monkey as he slams the cards onto the ground after conceding the last hand. “Year of the Tiger it is!”

THE END

One for the Math Geeks

After the world’s most intense game of Spider SOLITAIRE played by Spider GM, two International Masters and three coloured blobs from Among Us, followed by some rather detailed post-mortem analysis I think now is a good time for some light relief with some rot13(zngurzngvpny znfgheongvba). More specifically, we answer the following question: if cards are dealt one at a time from a perfectly shuffled deck, how long do we have to wait until a full suit appears?

The relevance to Spider Solitaire players should be pretty clear: we have absolutely no control over what cards turns up. If, for example, there are no Queens in the first 20 cards then the chances of the next turnover being a Queen are exactly 8 in 84 – no amount of skilled or unskilled play can fight the basic laws of probability. All you can do is hope to mitigate the effects of not getting any Queens in the first 20 cards with skilful play. Similar considerations apply if you’re in the middle-game desperately waiting on the Three of Diamonds to complete a full suit.

If you insist on shifting the odds, we can commit the cardinal sin of Spider Solitaire by using rot13(haqb). Alternatively, we can somehow fail to obtain any turnovers in the tableau for the remainder of the game – in either case, the mere thought is too horrible to contemplate.

There is one minor complication: when dealing from the stock 10 cards appear simultaneously (which I mentioned more than once during the post-mortem of the Among-Us game). Here we will assume they appear sequentially. In the example below, a row of 10 cards has just been dealt. There are 61 cards visible, and by the time the Three of Diamonds is dealt we see every card in Diamonds. In other words, we needed 60 cards to obtain a full suit.

Clearly, the minimum number of cards required is 13 and the maximum is 96 (if we get e.g. no Kings until the last eight cards). During the Among Us game I thought we were fairly lucky to see a full suit after 60 cards. From experience I would expect we need more cards visible before the chances of seeing a full suit are 50%. However, I never bothered to quantify this. Fortunately, it is relatively straightforward to run a computer simulation. We might simulate X hands and find e.g. we need on average Y cards instead of 60 before there is an even chance of finding a complete suit. If Y cards corresponds to the Z^th percentile then we would know where we stand in terms of how lucky we were.

In this case, I ran 5000 iterations and found 60 or lower cards were needed to achieve a complete suit 1176 times. This corresponds to the 22^nd or 23^rd percentile. The mean and median is closer to 66.5 and 67.0 respectively.

As I suspected, we did get lucky with the Diamond suit, but nowhere near enough to justify improvising a rap song with the phrase “statistically significant” appearing once every ten seconds. To put this in Dungeons & Dragons terms, if this game of Spider Solitaire were a character, then we would have rolled better-than-average initial stats for complete-suits, but nobody in their right mind would accuse the dice of being rigged. Presumably we would have lousy initial stats for other abilities such as turnovers or shortages of particular ranks etc (I did have some misgivings about our winning chances during this hand therefore something had to be lousy), but unfortunately there is only so far one can go with the Dungeons & Dragons analogy.

Without the ability to remove Diamonds, I expect we would have been in more trouble than Ian Nepomniachtchi’s Bishop getting harassed by both Black rooks after capturing a poisoned pawn in Game 9 of the World Chess Championship since no other suit is close to completion. I’m not sure what’s the best way to test this conjecture via simulation, so if you have any clever ideas, please leave a comment 😊

Intermezzo

In Microsoft Window’s Spider Solitaire I found an unusual feature/bug: if a player completes a suit he receives a 100-point bonus but is not penalised 1 point for the move. For instance, suppose we have K-Q-J-T-9 of Hearts in one column and 8-7-6-5-4-3-2-A of Hearts in another and our score is 458. If we clear Hearts our score becomes 558, not 557. Another example: suppose we have no completed suits, a score of 400 and our target is to win with 1000+ points. The maximum number of moves we have is 208, not 200.

Note that clearing a suit is automatic, so we don’t have the option of completing the suit but refusing to move it do the foundations. In rare circumstances this option may be desirable, but giving an example is left as the proverbial exercise for the reader.

Here’s another fun question: what is the theoretical minimum number of moves required to beat Spider Solitaire if the cards fall perfectly?

To avoid accidentally revealing spoilers, I will insert the lyrics for one of my favourite anti-smoking songs.

Once a stupid smoker camped by a billabong

Under the influence of L.S.D.

And he sang and he smoked while his mates were drinking alcohol