Chances of winning 2, 3 or more bets

Guys

I wondered whether anybody knew (or would care to calculate) the odds of (1) being dealt hand(s) that could be turned into 2, 3 or 4 hands by splitting and (2) what are the odds of winning all of those bets. To add another level of complexity what are the odds of (3) being able to get 2, 3 or 4 bets out by either splitting and/or doubling and (4) what are the odds of winning all of those bets.

To simplify (I think)

(1) Probabilities that the next hand dealt can be split (i) once, (ii) twice, and (iii) 3 times. So probabilities for each of those 3 different outcomes.
(2) Probabilities that each of those 3 outcomes will result in a win for all hands.
(3) Probabilities that on the next hand dealt you can get (i) 2 bets out by either splitting or doubling, (ii) that you can get 3 bets out by splitting, then either splitting or doubling one of your new hands, and finally (iii) that you can get 4 bets out by splitting, then either splitting or doubling both of your new hands.
(4) Probability that you win all your hands in (3)

I am working on my own answers but I reserve the right to wait and see what you guys come up with and change my results accordingly.

Cheers

Reachy

 

Strategy

You can always double, so you can always get 2 bets out on any starting hand, and 4 bets out on any starting pair. (So long as you have the BR.)

The probability of winning all the bets must be a function of the stratgey employed, too. If you absolutley must win them all, you'll be following a different strategy to that which you would otherwise employ. (Ken posted a while ago on this subject, refining Wong's no-bust approach.)

As a final complication, it's apparently very rare, but there are some games in which you get to see the first card to each of the split hands before you have to act on the first hand. Thus, if you need to get 3 bets out, you can make a more informed choice about which of the two hands to double.

 

Approximate

I was looking for ball-park figures really, just out of interest, as some teasers I've read recently have related to multiple bets.

Anyway, I've done some sums which may be of interest.

Prob. of being dealt a hand that can be split = 14.8%. That is, 1/13 chance of getting the same 2nd card as 1st card except tens where it is 4/13 (i.e. 9 x 1/13 x 1/13)+(4/13 x 4/13))

Prob. of winning both those hands = 4.4%. Borrowing from Wong CTS Table 4 there is a .30 probability of 2 hands beating the dealer :. 14.8% x 30% = 4.4%. Interestingly this number is similar to the prob. of getting a BJ.

More to follow...

Cheers

Reachy

Edit: Assumes infinite deck

 

I'm not sure how valid that .30 would be as an approximation. There must be a big difference between two random hands and two hands with a card in common. Presumably there is greater correlation between the results.

 

At ClubUBT those willing to split 10s out 4 times tend to win all hands. Time after time I have seen 7/8 cards of 10 or better. Calculate that!

BTW Reach a black eye on right side? Never forget about the left hook LOL

 

multi-bet winnings

Reachy, I never had the time to recover and work with my messed up bj files so this is not exact.
Off the top of my head I would say that DDD (desperate double down) is about 30% but if you include optimal splitting it goes up to almost 34%. DTB (desperate triple bet) wins about 4.5%, and DQB -quadruple bets succeed about 2% of the time. It would be nice to have this numbers redone properly.
It is a complicated task if done “manually”, but doing it would be an invaluable tool to any serious tournament player as it opens eyes to playing strategy. The frequency of winning multiple hands based on dealer busting may be a surprise to some. Did you know that winning split eights vs. dealer 9 or Ten comes in almost 90% of cases from the dealer’s bust? Or that it is better to have pair of not Tens if you need to win three bets?
Watch out for existing stats, chance of dealer having a bj need to be converted and can wreck havoc if missed. Also there are quite lengthy analyses what is the optimal play with multiple hands to be played concentrating on when to double or split if more, sometimes yet not fully known cards, will be dealt. Watch out for not obvious improvements like chance of splitting and especially doubling more than necessary hands if first hands were stiffs (17 is a stiff if you need to win).
Have fun.

PS
Your 4.5% for two doubles can’t be applied. If they were independent hands you would have to multiply .3 by .3 and then again by the chance of having two hands to double. I think it is over 25% chance to win triple bet if you get pair of Tens to start with and over 40% for other pairs, but zero with no pairs .

S. Yama