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Here's what I don't get. I've played a lot of Hearts for many years - thousands of games I'm sure. The number of times I've played a low heart, and the trick has gone 2H, 3H, 4H, 5H seems way out of probable reality. But the number of times I've seen all four players play the any of the same value card (e.g. 6C, 6D, 6H, 6S) is once. Why is the probability of 2H, 3H, 4H, 5H any lower than any 4 same value cards? In my naiveté it seems that in both cases, the probability calls for 4 players to have a specific card at the same time. Can anyone explain it to me like a 4 year old?

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xkcd.com/1364 –  TimLymington Jun 30 at 22:23
    
The fact that you see 2H, 3H, 4H, 5H seems surprising. If the player stuck with the 5 is trying not to accumulate hearts, I'd expect that player to slap down the largest heart card he has. If he's trying to shoot the moon, then the 5 makes sense. –  Ellesedil Jul 1 at 0:42
    
Sorry for not being clearer - the cards most times do not appear in that order but more likely 5H, 4H, 3H, 2H or some other combination started by the 4H or 5H. Thanks for your comments –  Ricksx Jul 8 at 16:00
    
I'm pretty sure most 4 year olds don't understand probability either. –  bwarner Oct 14 at 12:46

2 Answers 2

The reason H2,H3,H4,H5 is more likely than H6,S6,C6,D6 is simply the rules of the game. If a heart is led, it is mandatory to play a heart if possible, so most tricks contain four of the same suit, and a trick with one of each suit is extremely rare. When you add in the requirement for all four to be of the same rank, your second example is vanishingly improbable. This boils down to the fact that a card game is not a random selection of available cards (except when my partner is choosing an opening lead, of course).

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There's also the game theory argument that people want to avoid hearts and will be playing their lowest to try and avoid the points, whereas in a situation where they can discard a card off suit, it's much more likely that they might try and discard an off-suite card of high rank. I wouldn't be surprised if all 10s or all Jacks came out in one trick, but other values seem much more unlikely. –  Hao Ye Jul 1 at 0:32

The chances of each player getting exactly one card of a set, be that set {2H, 3H, 4H, 5H} or {6D, 6C, 6H, 6S} is the same. However, 6 was just an example; the chance of each player having a card with the same value but a different suit is actually bigger then the lowest four Hearts being distributed like that. So the starting position is actually in favour of four cards of the same value being played.

But players don't choose cards at random. First of all, they have to follow suit if possible, so for none of the players to be able to follow suit would mean that it's later in the game, or that the opening player has all cards of that suit.

And that's just following the rules. But there's strategy and tactics to consider as well. Unless a player is trying to shoot the moon, they will typically play the lowest Hearts possible. So a player may play a 4H or 5H, hoping that one of the others will have to play a 6H or 7H at least. A player following suit with a 5H on a 4H may hope for the same. If the other players don't want the trick, the only thing they can do is follow with the 2H and the 3H, sticking the trick on the player with 5H.
There is no strategic or tactical advantage to following value instead of suit, so it really comes down to coincidence — although I can imagine players following value just for fun, after the first two cards have been played.

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