cavalier
03-11-2003, 11:41 PM
You are rolled up in 7-stud playing heads up. Your opponent doesn't show your quad card.
What are the odds that your last 4 cards will give you a full house or quads?
Step 1 : Making quads
There are 48 cards left in the deck.
Those 48 cards can yield (48,4) = 194,580 four-card hands.
In order for you to turn your trips into quads, one of the cards in the hand must to be the same rank as your starting trips. The other 3 can be anything. There are (47,3) = 16,215 possible combinations for the other three cards.
Therefore, the probability of getting the last one of your rolled-up trips is
16,215 / 194,580 = .0833 or about 1 in 12 hands.
Step 2 : Filling rolled-up trips with two or three of your opponents rank.
* Do not allow for a second pair in the other 2 cards.
* Do not allow for quading your rolled-up trips.
There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands. Since there are 11 other ranks that we don't want to pair, we have to remove the 6 pairs each rank can create. This leaves us 2970-(11*6) = 2904 hands.
2,904 / 194,580 = .0149 or 1 in 67.0
Step 3 : Improving without making your original quads and without using your opponents rank.
In order for your last four cards to include a pair that isn't the rank your opponent shows on 3rd, you would have 44 (4*11 ranks) cards to improve with ( we exclude his rank and your quad card ).
Those 44 cards can yield (44,4) = 135,751 four-card hands.
We have 11 ranks we can possibly pair. A pair can be made (4,2)=6 ways for each rank. The other two cards can be arranged (42,2)=861 ways. This gives us 11*6*861=56,826 hands in which we can improve by getting a pair, trips or quads of a rank other than the one we or our opponent started with.
56,826 / 135,751 = .4186 or 1 in 2.4
Finally, the chances of improving to a full house or quads when rolled-up in a heads up game with an opponent showing a rank other than your trips is
.0833 + .0149 + .4186 = .5168 or 51.7% of the time.
Note that his showing your quad card on 3rd takes away about 16% of your chances, leaving you a 43.35% chance of getting a full house, or 'other quads'.
If anything is wrong, or I overlooked something... please post!
What are the odds that your last 4 cards will give you a full house or quads?
Step 1 : Making quads
There are 48 cards left in the deck.
Those 48 cards can yield (48,4) = 194,580 four-card hands.
In order for you to turn your trips into quads, one of the cards in the hand must to be the same rank as your starting trips. The other 3 can be anything. There are (47,3) = 16,215 possible combinations for the other three cards.
Therefore, the probability of getting the last one of your rolled-up trips is
16,215 / 194,580 = .0833 or about 1 in 12 hands.
Step 2 : Filling rolled-up trips with two or three of your opponents rank.
* Do not allow for a second pair in the other 2 cards.
* Do not allow for quading your rolled-up trips.
There are 3 ways to make a pair out of the 3 remaining cards of his rank. Also, there are (45,2)=990 two-card combinations for the other 2 cards (45 since we leave out the one that makes your quads). This gives (3*990)=2970 possible hands. Since there are 11 other ranks that we don't want to pair, we have to remove the 6 pairs each rank can create. This leaves us 2970-(11*6) = 2904 hands.
2,904 / 194,580 = .0149 or 1 in 67.0
Step 3 : Improving without making your original quads and without using your opponents rank.
In order for your last four cards to include a pair that isn't the rank your opponent shows on 3rd, you would have 44 (4*11 ranks) cards to improve with ( we exclude his rank and your quad card ).
Those 44 cards can yield (44,4) = 135,751 four-card hands.
We have 11 ranks we can possibly pair. A pair can be made (4,2)=6 ways for each rank. The other two cards can be arranged (42,2)=861 ways. This gives us 11*6*861=56,826 hands in which we can improve by getting a pair, trips or quads of a rank other than the one we or our opponent started with.
56,826 / 135,751 = .4186 or 1 in 2.4
Finally, the chances of improving to a full house or quads when rolled-up in a heads up game with an opponent showing a rank other than your trips is
.0833 + .0149 + .4186 = .5168 or 51.7% of the time.
Note that his showing your quad card on 3rd takes away about 16% of your chances, leaving you a 43.35% chance of getting a full house, or 'other quads'.
If anything is wrong, or I overlooked something... please post!