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 Zth 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 22nd or 23rd 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 😊