Blackjack side bet probailities

Hi all,

Hoping to get some help with some probabilities. I'm looking at calculating the probabilities for the side bets for the European No Hole Card Blackjack game (4 decks used) and was hoping to have someone here cast an eye over my workings (I'll offer brief descriptions to make any mistakes/oversights evident):

3 of a kind suited - 0.000035
-> used the reasoning that there are 52 suited 3OAK combinations available over C(208,3)

3 of a kind unsuited - 0.000106
-> 208 unsuited combinations available - the 52 for 3OAK suited as above over C(208,3)

Straight flush - 0.000130
-> 12 possible sequence of 3 cards for each suit therefore 48 possible sequences in one desk => 192 possible sequences in 4 decks over C(208,3)

Now given the low probabilities of these leading me to assume I have done something wrong, I haven't tackled the side bets Straight, 3 card poker hand bar pairs, mixed, coloured, or suited as something seems amiss. Can anyone see anything that is clearly wrong or that I haven't factored in? Or do these probabilities look alright? If anyone needs any further information please let me know.

If anyone has any good reading material that they think would be helpful to determine these probabilities please let me know.

3 of a kind suited:

13 ranks 4 suits pick 3 of 4 of of the cards.
13*4* c(4,3)

Other 3 of a kinds

13 ranks pick 3 of 16 of the cards. - suited trips
13 *c(16,3) - 13*4*c(4,3)

Straight flush

12 sequences 4 suits. 4 cards of each rank
12*4*4^3

3 of a kind suited should be 0.0001407.
Your working is missing the fact that any of the four decks could be the missing card, meaning there are 208 combinations of suited 3ok in C(208,3).
3 of a kind unsuited 0.004780
Once again there are 13*C(16,3) -suited 3oK
Straight Flush I think there are 3072 combinations as each card can come from any of the 4 decks.

In all three calculations you have missed the fact that the included cards can come from any of the decks. A good way to think of this is to imagine that instead of playing with 4 identical decks, you have 1 red, 1 blue, 1 green, and 1 yellow deck, and then think about the possible combinations.
Hope this helps

Quote: miplet

3 of a kind suited:

13 ranks 4 suits pick 3 of 4 of of the cards.
13*4* c(4,3)

Other 3 of a kinds

13 ranks pick 3 of 16 of the cards. - suited trips
13 *c(16,3) - 13*4*c(4,3)

Straight flush

12 sequences 4 suits. 4 cards of each rank
12*4*4^3

Thanks for this, so is it safe to divide this by C(208,3) to get the probabilities?

Quote: someone

3 of a kind suited should be 0.0001407.
Your working is missing the fact that any of the four decks could be the missing card, meaning there are 208 combinations of suited 3ok in C(208,3).
3 of a kind unsuited 0.004780
Once again there are 13*C(16,3) -suited 3oK
Straight Flush I think there are 3072 combinations as each card can come from any of the 4 decks.

In all three calculations you have missed the fact that the included cards can come from any of the decks. A good way to think of this is to imagine that instead of playing with 4 identical decks, you have 1 red, 1 blue, 1 green, and 1 yellow deck, and then think about the possible combinations.
Hope this helps

Thanks for that I realised that I did 13 (cards) * C(4,4) (suit) * 4 (decks) to obtain the 52 combinations for 3oK suited. Revised my formula and got the same number of combinations now.

As for the unsuited I am slightly confused as to why we multiply by C(16,3)? Is it because we have 4 numbers of each number and 4 decks, so need to get the 3 from a combination of these 16? That's the only thing that I can think of.

Sorry for bombarding you with these questions - Wondering how my calculations were so wayward...:( and I'm just trying to wrap my head around this and Black probabilities has my brain overheated as of late.

Many thanks even if you don't get an opportunity to respond!