The reason I ask is because it is surprisingly unintuitive. Ignore the fact that an event is more or less likely than another and focus on permutations.

The number I get is 20. Nothing resembling any sort of factorials.

what about the permutations for 3 splits and exactly 2 doubles? Now I get 27

3 splits 3 doubles?

Is any one permutation of a scenario more likely than another?

With numbers so weird, I have to ask myself, am I right? What do you guys think

Quote:richodudegiven that in blackjack you are dealt a pair, how many ways can you split 3 times to 4 hands, and double down on exactly one of them?

The reason I ask is because it is surprisingly unintuitive. Ignore the fact that an event is more or less likely than another and focus on permutations.

The number I get is 20. Nothing resembling any sort of factorials.

what about the permutations for 3 splits and exactly 2 doubles? Now I get 27

3 splits 3 doubles?

Is any one permutation of a scenario more likely than another?

With numbers so weird, I have to ask myself, am I right? What do you guys think

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Here's what I get for 3 splits (4 hands):

0 doubles: 5 ways

1 double: 20 ways

2 doubles: 30 ways

3 doubles: 20 ways

4 doubles: 5 ways

We need to find how many ways the sequence of 2,2,2,2,9,9,A,A (3 splits 2 doubles) where at any point when reading from left to right, the numbers of 9s and As combined is not equal to or greater than the number of 2s, otherwise the rules of blackjack would be broken. The 8th card in the sequence us the exception to the rule where they must be equal parts. We almost must begin with 2 2s. that being said lets list the permutations

we know the first 2 cards must be 2 so its actually like finding the permutations of 2,2,9,A,A,A with the above rules still in places

2,2,2,2,9,9,A,A

2,2,2,2,9,A,9,A

2,2,2,2,9,A,A,9

2,2,2,2,A,9,9,A

2,2,2,2,A,9,A,9

2,2,2,2,A,A,9,9

2,2,2,9,2,9,A,A

2,2,2,9,2,A,9,A

2,2,2,9,2,A,A,9

2,2,2,A,2,9,9,A

2,2,2,A,2,9,A,9

2,2,2,A,2,A,9,9

2,2,2,9,9,2,A,A

2,2,2,9,A,2,9,A

2,2,2,9,A,2,A,9

2,2,2,A,9,2,9,A

2,2,2,A,9,2,A,9

2,2,2,A,A,2,9,9

2,2,9,2,9,2,A,A

2,2,9,2,A,2,9,A

2,2,9,2,A,2,A,9

2,2,9,2,2,9,A,A

2,2,9,2,2,A,9,A

2,2,9,2,2,A,A,9

2,2,A,2,9,2,A,9

2,2,A,2,9,2,9,A

2,2,A,2,A,2,9,9

2,2,A,2,2,9,9,A

2,2,A,2,2,9,A,9

2,2,A,2,2,A,9,9

equals 30. 3 more than 5am me got. how about 4 and 4?

2,2,2,2,9,9,9,9

2,2,2,9,2,9,9,9

2,2,2,9,9,2,9,9

2,2,9,2,9,2,9,9

2,2,9,2,2,9,9,9

equals 5. tired me got an extra one.

still though, is there any pattern to this? maybe someone could attempt the incredibly pointless problem of n splits and m double downs

Zero if the dealer has an Ace up. Are you asking about one deck? I suppose you are ignoring suits, too.Quote:richodudegiven that in blackjack you are dealt a pair, how many ways can you split 3 times to 4 hands, and double down on exactly one of them?

The reason I ask is because it is surprisingly unintuitive. Ignore the fact that an event is more or less likely than another and focus on permutations.

The number I get is 20. Nothing resembling any sort of factorials.

what about the permutations for 3 splits and exactly 2 doubles? Now I get 27

3 splits 3 doubles?

Is any one permutation of a scenario more likely than another?

With numbers so weird, I have to ask myself, am I right? What do you guys think

link to original post

Quote:MentalZero if the dealer has an Ace up. Are you asking about one deck? I suppose you are ignoring suits, too.

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given that the end result happened, how many ways could the cards have been ordered to make that result? It looks like ChesterDog got the rights answer