Seven-Card Stud Probabilities

Example One

I have a four and an ace in the hole and a four on the door. What are my chances of pairing the ace in the next two cards? Assume no one else is showing an ace on the board, for that could cause you to rethink your position. When cards you are looking for appear on someone else's board, they are dead to you. They are referred to as "dead cards" for that reason. Only count "live cards" into the probabilities.

You want to have at least two pair by Fifth Street, one of them aces. Otherwise, it would be prudent to exit, particularly if the other players are betting anything. Recall that in stud you have to count all the cards you can see as dead, or out of the deck. So with five people playing, you can see seven cards (five door cards and your two hole cards). If you are third to bet, you will see as many as nine cards at the time your card is dealt (if no one quits), but this does not count for the calculation. Why? Because the calculation is made at the time of betting, not dealing. The deck has 45 cards now. If no one else has an ace, then your chances are 3 in 45 on this round, or 6.67%. Assuming no one else receives an ace and all are still in the game when the next round comes, the deck will still have three aces and will be down to 40 cards. The possibilities have improved to 7.5%. These are not independent probabilities, so do not add them. Instead, do the mental experiment of dealing out Fourth and Fifth Streets to all players from the 45 card deck (taking out five door cards and your two hole cards). The odds that one of the two cards dealt to you is an ace are combin(1,1)*combin(3,1)*combin(42;1) / combin(45;2) or 12.7%

Example Two

I have a high three card straight as an opening hand. What is the chance of making a four card straight on the next card? Remember that the odds of filling a straight go south quickly as the sequence bumps against the upper or lower limits of the cards. An ace-high straight has nowhere to go but down. A king-high straight is OK for the next round, but if you draw an ace, then you're still no better off than with an inside straight. So for starters, let's assume the "high" straight is 10-J-Q, to leave room for a two-card fill at either end. Using the same assumptions as before about the number of persons in the game, the deck will have 45 cards in it when betting time comes. There will be, let us assume, 8 cards that will solve the problem. If any of these cards appears on the board somewhere, be sure to knock down the count by as many dead cards as you see. If all 8 cards are live, the chances of getting to a straight of four cards on Fourth Street are 8 / 45 or 17.8%. The story is not really over with that answer, however, as the three cards are reasonably high, and drawing a match for a pair might still keep you in the game. Again, assuming there are no dead tens, jacks or queens on the board anywhere, the odds of pairing one of them up are 9 our of 45 or 20%. These outcomes are mutually exclusive and independent, so they can be added together. You have a 37.8% chance of being able to stay in the game.

Example Three

I started with a King in the hole. On Fifth Street, I got a second one. Can I hope for the third by the River card? Hope springs eternal. Of course you can hope for a third king, unless there is a dead king somewhere on the board. By Sixth Street you will be looking for one of two cards in a deck comprised of 52-17 cards. (The 17 are three rounds of 5 cards for door, Fourth and Fifth Streets, plus two in the hole.) This is 2/35 or 5.7%. If everyone stays in the game, on the last round the deck will be down to 30 cards after betting has concluded. That chance is now 2/30 or 6.67%. If both cards were dealt in a theoretical two card hand from a 35-card deck, and both kings were still live in the deck, the chance of pulling down one of them would be 11.1%.