Final hand of a tournament 2 players left one advances Br 2 bets first bets 235 holds back 17.50 Br 1 bets 300 holds back 27.50 Br 2 is dealt a hard 19 and stands Br 1 is dealt a hard 16 Dealers up card is a 10 If Br 1 stands they have the dealer bust which is 24 % plus the dealer making a 20 which is an additional 36% plus the dealer 21 which is close to 4 % Can Br 1 do anything different than stand to increase their chances beyond 64 % to advance Joep2
Stand... If the BR1 has 16, the best play is stand. Stand is the best play even for BR1's hard 15 or hard 14. If anyone can prove that it is better to hit with hard 16, then it will be the answer for Ken's "THEORY QUESTION" in another thread.
In a one deck game having seen 5 cards 2 cards of BR 1 2 cards of BR 2 and the dealers up card That would leave 47 cards left in the deck of which the 12 cards comprising of 3 -4 -5 which makes BR 1 an instant winner. So the chance of seeing one of these 12 cards works out to be 25.53 % which is more than the dealer bust probability of 23 % .If they were to catch a 2 for 18 they now pick up the dealer making a 17 which should make hitting the 16 a better play what am I missing here ?. Joep2
Single deck Effects of removal in single deck are very strong. (You got us this time, but it would be nice of Joep3 in the future to specify full conditions beforehand.) So actually, if you hit and get an Ace, deuce, or trey the dealer busts 21.63%, 20.76% 20.83% respectively. Joep you were right that in this case hitting sixteen once is a better play. Chances of winning by hitting are 74.04% versus standing 63.72%. S. Yama Hold on a moment. Are you sure it wasn't a single deck Spanish 21 game?
How many decks So would it still be better to hit the 16 in a double deck game ? What amount of deck changes this sitution to stand as opposed to hit. Mr Yama where did you see Joep3 at ? Joep2
Have you not already answered your question? Hi As you know I am learning to make probabilty calculations for BJ (under the kind tutorship of various forum members - thanks guys!) so this is my stab at it. I went through all the dealer outcomes and player actions and concluded that standing is the best option. The next best action is to hit to at least 19 which gives you a probability of 56.5% of being successful. That was before I realised you were talking about a single deck. I haven't figured out the probabilities for single deck games yet.... I'd like to ask you a question about one of your statements though. You said however in your first post you said that standing wins if the dealer busts OR hits to 20 or 21. Can you ignore the 20 or 21 outcomes in this case? Cheers Reachy
Yes, ignore the 20 or 21! Reachy, by “instant winner” Joep2 means BR1 wins the table (as opposed to beating the dealer’s hand) regardless of the dealer outcome. He has the high, the low, and with 19, 20 or 21 he could not be swung by BR2.
I know that but... Thanks Jackaroo. I understand what Joep means by instant winner, what I was querying was whether you could compare the odds of hitting 3, 4 or 5 with just the odds of the dealer busting. The way I understand what Joep was trying to say was that the odds of locking up the win by hitting a 3, 4 or 5 are greater than the odds of the dealer busting if you stand, therefore you should hit. That was why I wondered whether you could now ignore the odds of the dealer hitting 20 or 21 if you stand because you win in that event also. I'm likely to have missunderstood what Joep was trying to say so apologies. Cheers Reachy
More info please As the question was originally stated, arlalik's answer was obviously correct since one has a right to make the most likely assumptions (6 deck shoe) if nothing is stated to the contrary. However, now we are looking at a single deck and being asked to give an answer which takes into account card removal but again forced to make assumptions some of which are: 1) Was there a burn card and if so do we know what it is? 2) What is the composition of BR1's hand: 7-9, 8-8, or 10-6?3) Other assumptions that my exist but I can't think of right now. After all these assumptions (and probably others) are made, the dealer bust probabilities must be recalculated especially since there is a greater proportion on low cards remaining which means the dealer bust probability is lower and the chances of dealer making a 20 or 21 is higher. Also, at least two 10's have been removed and possibly more. How does this affect the probable outcomes? My point is that this problem cannot be solved without more specific information and with that additional information, the answer must involve some very complex calculations. I am not prepared to make these calculations. Just food for thought.
strongly disagree Joep, as Reachy mentioned we cannot disregard the cases when dealer makes 20 or 21. In your example we have two simple formulas for Hit and Stand. STAND: BR1(adv) = D(bust) + D(20) + D(21) HIT: BR1(adv) =D(bust)*20/47 + D(17)*16/47 + D(18)*12/47 + D(19)*12/47 + + D(20) + D(21) Yama, please substitute the Dealer's probabilities in both Formulas to get 63.72% for Stand and 74.04% for Hit. arlalik & tirle
correction, still hit I rechecked what I've done last night and found one mistake. I used 27/47th for player advancing when the dealer bust but I should have used 20/47th. With this correction hitting is sill slightly better 64.55% to 63.72%. Here is the whole procedure. It is not necessary to be concerned with the burn card, as without knowing what it really was the probabilities of appearance of other cards remained the same. If BR1 ends up with seventeen he must get an Ace as the hit card. I removed three tens, one nine, one six, and one Ace from the deck and got the numbers for the dealer's bust, twenty, and twenty-one using deck comprising of 46 cards. Then, I removed a deuce (BR hits eighteen) instead of an Ace and recalculated the dealer’s seventeen, bust, twenty, and twenty-one. I used 4/47th chance (known cards) for calculating BR1’s chance of hitting and ending up with nineteen to twenty-one, the dealer’s outcomes in this case don’t matter, BR1 matches BR2. It doesn’t matter that much if we remove nine and seven, or two eights for the player’s sixteen as those cards are bust cards for BR1 hitting and they make pat hand for the dealer – both outcomed are detrimental for the BR1’s fate. It will make a slight difference if sixteen was made by a Ten and six (and that’s what I calculated), as without knowing the real cards chance for this was 4.8% versus the other combinations of sixteen 2.5%. Also, nineteen could be made by an Ace and eight, but Joep would tell us about it, wouldn’t he? Hope it somewhat clears the situtation. S. Yama
great calculators I wouldn’t have time to recalculate all those dealer’s outcomes manually. There is a great site provided by Norm Wattenberger of Casino Verite Blackjack fame. He also has made tons of other data available http://www.qfit.com/CVDPC.htm Dealers Probabilities Calculator or other blackjack calculators http://www.qfit.com/calcp.htm Happy number crunching, S. Yama
still disagree Yama, thank you for valuable info. Unfortunately we cannot use this table of probabilities, because it does not take into consideration that dealer's hole card cannot be an A (Dealer's BJ is NOT exluded). We'll return to this topic later.
This case was interesting enough to make me go back to it for a solid calculations. I not only removed an Ace or a deuce when BR1 made seventeen or eighteen, but also calculated separate dealer’s outcomes for BR1 busting with six through Ten. I assumed BR2 has a Ten and a nine. BR1 (T, 6) Standing - 63.72% Hitting 60.60% BR1 (9, 7) Standing – 67.28% Hitting 63.60% BR1 (8, 8) Standing – 67.08% Hitting 63.70% After all, it is better to stand on sixteen in a single deck game by more than 3%. S. Yama P.S. If the dealer has a blackjack BR1 advances regardless the play, so you can use the table.
Misunderstandings Tirle, it was I, not Joep, who said we could ignore dealer 20 or 21. I was responding to Reachy’s question in his post: I took this case to mean the one quoted, namely BR1 hitting and catching a 3, 4 or 5 where the specific dealer outcome is irrelevant--BR1 advances with all dealer outcomes.
Hit vs Stand with no BJ for Dealer We just finished calculations for 3 different BR1 cases(T-6, 9-7, 8-8) assuming Dealer T-up but no BJ. All the numbers are calculated with Ace,2,..Ten removed from the single deck also. BR1(T-6)....Stand = 60.36%, Hit = 57.73% BR1(9-7)....Stand = 64.23%, Hit = 59.45% BR1(8-8)....Stand = 64.01%, Hit = 60.34% Here are the numbers for Infinite deck. Please note that BR1's Hand combinations make no difference. BR1( 16 )...Stand = 63.79%, Hit = 58.93% And because the STAND is dominating in every case, that means the answer for Joep2's question for 2 deck game is STAND. tirle and arlalik.
Hole card cannot be an ace In post #13 tirle suggested about not using a table of possibilities because it did not take into consideration that the dealer's hole card cannot be an A. In this post I hope to clarify how probabilities change for the dealer and player when the dealer has a 10 (or A) for his upcard. In the problem where: ..........BnkRll........Bet...........Cards..........Action BR2.....252.5........235..........10 + 9..........Stand BR1.....327.5........300..........10 + 6.............? Dealer....................................A The Dealer's exact probabilities, for 1 deck with cards 10,9,10,6 removed from the deck (and the dealer 10 up card accounted for) before BR1 acts is as follows: Bust..............21121................23086 17...............12622................13796 18...............12256................13396 19...............10325................11285 20...............31416................34338 21...............03751................04100 BJ...............08511................00000 Before the dealer peeks at the hole card the probability of a dealer BJ is .08511. After peeking at the hole card and determines that there is no ace, the "A.Peek" column for dealer totals is then used. We can call this change in probabilities the "Peek Effect for the Dealer". Therefore if BR1 stands the probability of BR1 advancing should be: D(bust) + D(20) + D(21) = .23086 + .34338 + .04100 = .61524 = 61.52% Now if BR1 takes a single hit things get complicated for the probabilities of the hit card received. Take the probability of BR1 receiving an ace. It is NOT 4/47. there are 47 unseen cards, one is the dealer down card and 46 cards in the undelt pack. Since the dealer down card cannot be an ace then all 4 aces are in the 46 card undelt pack - therefore the probability of BR1 receiving an ace rises from 4/47 to 4/46. Consequently the probability of receiving a card of a different denomination than an ace slightly reduced (since it has a chance of being the dealer down card). Before the peek the probability of recceiving say a 2 as a hit card is (40)/47. After the peek the probability of receiving a 2 becomes (4)*42/(46*43). Since there are 43 unseen cards other than an ace then the probability of receiving a card other than ace is: (43)*42/(46*43). This simplifies to 42/46. Note: Prob(A) + Prob(other) = 4/46 + 42/46 = 1 as it should be. We can call this change in probabilities for player hit cards the "Peek Effect for the Player". From the the probability theory this change in probabilities for the dealer and player is called a "Conditional Probability" When then player takes a hit card things get much more complicated. The first term in BR1(adv) from post #10 was D(bust)*20/47. Here 20 represents all the cards in the unseen deck that do not bust BR1's 16. If an ace is the hit card then this term becomes: D(bustA)*4/46 + D(bust2)*(4*42)/(46*43) + D(bust3)*(4*42)/(46*43) + etc. where D(bustA) is the dealer's A.Peek new probability of busting when an ace is removed in addition to the cards removed earlier. Similarly the same for bust2 and bust3 etc. ................................BlueLight
