I stand corrected, four is also a match.
In the OP, both decks are shuffled together. The cards are then dealt in one long row. If two of the same cards are next to each other, we have a match.
In Montmort, the two decks are side by side. Cards are simulataneously dealt off the top of each deck, one by one. The cards are not placed in one long row as in the OP. There is a match in Montmort only if on any particular deal, the two cards are the same.
If, however, we try to simulate the OP using Montmort, we could pretend the cards are being placed in one long row as they’re dealt from the two decks. The two cards from the first deal start the row, followed by the next two cards, and so on. This, however creates more possibilities for matches–the second card from the first deal could match the first card from the second deal. For example, the first deal from the two decks could be 2H and 8S; the second deal 8S and 5C. No matches for Montfort, but when they’re laid in a row:
2H 8S 8S 5C
A match for the OP.
No, MC’s question is different. For any card, there are only 72 cards in the two decks that don’t match it in either rank or suit, so the probability is (72/103)^103 = virtually zero.
Unfortunately, neither the OP’s problem nor MC’s problem can be done that simply.
If I have my two decks shuffled together and deal my first card, it’s true that the probability of no match when I deal the second card is 72/103. However, to say that the probability of no match in the entire row is (72/103)^103 is to assume that the probability of matching one particular pair of cards is independent of matching another particular pair of cards. For example, say the first two cards you deal are a match, say both clubs. What’s the probability of a match on cards two and three? Slightly less than what it was to start with, since there are now two clubs out of the deck, leaving fewer potential matches on the third card. You would have to use conditional probability with this approach, and I believe that would get very messy very quickly.
Yeah, I would have realised that if I had actually read the whole thread before diving in…
Wow, thanks everyone. I didn’t expect this to get “controversial” :). As I expected, the method of calculating the probability is not simple, but it’s also nice to see that the actual probability is not too far off from what I’d guessed from experience.
Wow. When I read this thread and saw the ref. to the Montmort problem I thought “But it’s a different problem!” But then I did the math myself and what do you know. It is right. Again, wow.
This kind of analysis comes up in Hashing functions in Computer Science. What are the chances of two values being “hashed” to the same address. So that 1-1/e business is quite familiar.
ftg, they are different problems, but the probabilities do end up being very close, which kind of surprised me, too.
The probability of a match in the Montfort problem 63.21205588%, but the probability of a match in this problem is 63.39019684%.
My imperical test came up with about 63.4%.