Since Sebastian knows what his opponent has and what cards are used, he knows that the only thing that would beat his friends hand is a straight or higher three of a kind. Theres no flush potential so everything else higher than a flush is eliminated. What Sebastian needs is a:
- king and a king, to make a higher three of a kind. There are three kings left and two of them can be his own cards. He doesn't know because he went in blind but it's possible but he has pocket kings.
- Two and a Two, to make a three of a kind kings, because two's are wild and can be any card. The highest face on the table is a king and the pocket two's can make a three of a kind.
- King and a Two, two's are wild and with the kings it can be a higher three of a kind.
- Ace and a Ten, that would make a straight because it'll be Ace, King, Queen, Jack, and Ten.
- Ace and a Two, the two can have a face value of a ten.
- Two and a Ten, the two can have a face of an Ace since it's a wild card.
- Ten and a Nine, that also makes a straight, king, queen, jack, ten, and nine.
- Ten and a Two, because the two can have a face value of a nine.
To get a king and a king or a Two and a Two:
Sebastian already knows that a king is used on the table so there are three kings left. Out of those 3 kings he only needs 2 and it doesn't matter the order in which he gets the kings because the outcomes still the same. So its only 3C2, because out of the three kings left he only needs two of them and suit doesn't matter. This also applies to pair of two's, same concept.
3C2 + 3C2 = 6
To get a King and a Two:
There are three kings left and he only needs one of them and since order doesn't matter it'll be 3C1. One of the Two's are used in the opponents hand so there are 3 Two's left and out of those 3 Two's he needs one of them which is also 3C1. Then they have to be multiplied.
(3C1)( 3C1)= 9
To get a Ace and a Ten:
There are four Aces left because they are not used and he only needs one of them so 4C1 and there are four Tens that are not used either and out of those four Tens he needs one of them so 4C1. This also applies to Nine and a Ten because the Nines are not used either so out of the four Nines he needs one.
(4C1)(4C1)(2)= 32
you multiply 16 by two because it's the same thing that happens to getting a Nine and a Ten.
To get an Ace and a Two:
There are four Aces left because they are not used and he only needs one of them so 4C1 out of the four Aces he needs one. For the Two there are three left because one of the Two's is in the opponents hand, so out of the 3 he only needs one, so 3C1. This also applies to getting a Ten and a Two, and a Nine and a Two.
(4C1)(3C1)(3)= 36
To find the favorable outcomes:
Add the numbers up because these are all the hands that he wants to get and that's what favorable outcomes are, the outcomes you want.
3C2 + 3C2 = 6
(3C1)( 3C1)= 9
(4C1)(4C1)(2)= 32
(4C1)(3C1)(3)= 36
total of 83
To find All possible outcomes:
Since ten cards were used (5 for the table, 3 for the burn cards, and 2 are the opponents cards) there are only 42 cards that remain. Out of those 42 cards he'll need two for his own hand, so it would be 42C2, which is 861.
To find the probability it would be favorable outcomes over all possible outcomes. And whatever that number is, in this case .0964, is the probability.
Sebastian already knows that a king is used on the table so there are three kings left. Out of those 3 kings he only needs 2 and it doesn't matter the order in which he gets the kings because the outcomes still the same. So its only 3C2, because out of the three kings left he only needs two of them and suit doesn't matter. This also applies to pair of two's, same concept.
3C2 + 3C2 = 6
To get a King and a Two:
There are three kings left and he only needs one of them and since order doesn't matter it'll be 3C1. One of the Two's are used in the opponents hand so there are 3 Two's left and out of those 3 Two's he needs one of them which is also 3C1. Then they have to be multiplied.
(3C1)( 3C1)= 9
To get a Ace and a Ten:
There are four Aces left because they are not used and he only needs one of them so 4C1 and there are four Tens that are not used either and out of those four Tens he needs one of them so 4C1. This also applies to Nine and a Ten because the Nines are not used either so out of the four Nines he needs one.
(4C1)(4C1)(2)= 32
you multiply 16 by two because it's the same thing that happens to getting a Nine and a Ten.
To get an Ace and a Two:
There are four Aces left because they are not used and he only needs one of them so 4C1 out of the four Aces he needs one. For the Two there are three left because one of the Two's is in the opponents hand, so out of the 3 he only needs one, so 3C1. This also applies to getting a Ten and a Two, and a Nine and a Two.
(4C1)(3C1)(3)= 36
To find the favorable outcomes:
Add the numbers up because these are all the hands that he wants to get and that's what favorable outcomes are, the outcomes you want.
3C2 + 3C2 = 6
(3C1)( 3C1)= 9
(4C1)(4C1)(2)= 32
(4C1)(3C1)(3)= 36
total of 83
To find All possible outcomes:
Since ten cards were used (5 for the table, 3 for the burn cards, and 2 are the opponents cards) there are only 42 cards that remain. Out of those 42 cards he'll need two for his own hand, so it would be 42C2, which is 861.
To find the probability it would be favorable outcomes over all possible outcomes. And whatever that number is, in this case .0964, is the probability.
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