Spider solitaire odds

My favorite opening position in 2 suit spider solitaire is when I get a king, queen and jack of one suit face up on the deal. I will often just keep hitting “new game” until I get there. It comes up fairly often it seems.

So what are the odds of it happening. How often should I expect to see that deal. It boils down to how often will the K-Q-J of one suit occur in the last 10 cards to be dealt.


I’ve no idea, but I’ve not played a 2-suit game that isn’t easily solved.

Normally questions like this get answered right away here. Perhaps, like me, everyone is too lazy to Google “two suit spider.” It sounds like you’re selecting 10 cards at random from a deck with S suits of K cards each. What are (S,K)? Not (4,13) I guess?

I don’t know Spider, but it looks like the question is “Pick 10 cards at random from a 52-card deck. What is the probability that there will be at least one set of KQJ of the same suit, among those ten?”

But my combinatorics-fu isn’t strong enough to answer that.

Type C(49,7) / C(52,10) into Wolframalpha.com to get the chance (0.00543) that you’ll get the KQJ of Hearts. Multiply that number by 4 to get (a number slightly more than) the chance of KQJ of at least one suit; about 2.17%. That’s not really “fairly often” so perhaps “2 suit spider” uses 2 suits instead of 4.

The expression C(49,7) / C(52,10) is more simply 1098/(525150). Exercise: Show directly that this is the answer to Chronos’ word problem.

It uses 4 sets of a black suit and 4 sets of a red suit. So ace to king of a single red suit, four times over. Same for the black. I don’t pay attention to what suit, to me it’s black and red. 104 cards total, 4 each of every card.


The number of distinct (and equally likely) ways to choose 10 cards from a deck of 104 is C(104,10). That’s about 26 trillion, but we needn’t worry about that — we’ll let Wolframalpha do the arithmetic for us.

The number of ways to choose exactly one of the 4 red K’s, along with exactly one each of the red Q’s and red J’s is 444. The number of ways to choose the remaining seven cards from the 92 cards that are not red face cards is C(92,7). Put this altogether to get
Enter that at the wolfram site, or just click this URL and see about 2.148%. But that is not the final answer OP seeks. With a symbolic math processor there are shortcuts, but I’ll just proceed laboriously.
The probabilities to get the red-picture goal with one of the pictures duplicated (or triplicated, or with two pictures duplicated, or with three duplicates, or with a duplicate and triplicate) are

which are 0.787%, 0.036%, 0.081%, 0.0023%, 0.0062%. Add these five cases together and get 3.06%. (We’ve not added in all the very rare cases — quadruplicate, two triplicates, etc.). Let’s guesstimate the grand total as 3.065%.

That’s just for red KQJ. Black KQJ will add another 3.065% BUT we must subtract the case where we get both red KQJ and black KQJ. For the case with 0, 1 or 2 duplicates these will be
which are 0.025%, 0.012%, 0.002%. The final guesstimated answer to OP’s question is 3.065% + 3.065% - 0.04% = 6.09%.

I scurried off to Monte Carlo, and had them shuffle a double-deck fifty million times. The result of that simulation, with 95% confidence, was 6.085% ± 0.003%

cool, so about 1 out every 16 deals. Thanks!


Two suits spider solitaire used to drive me nuts. Early on I set a personal record of 1205, I believe, in Windows. I must have caught a lot of luck. When I’d play I usually would try to win using the fewest moves and I could never touch that record. I even started to try to “cheat” by writing down what the face down and the not yet dealt cards were, devising a strategy, then restarting the game…just didn’t get too far with that.

My high is 1203. I also play the way you do. I’ve never been concerned about a high percentage of wins. Whenever I have a great game in the high 1190s I save it and replay it later. I don’t know if I have ever beaten my first run. But I’m not much of a strategist, never played chess or anything that required a lot of analysis.