Ok math people, I have a few questions and I KNOW there are great minds here who can answer them, but more importantly tell me how they calculated their answers so I don’t have to do this in the future. Ok, here goes: Assuming an infinite deck of cards and over 100 hands, what percentage of hands would one expect the DEALER to get the following: 1. Ace showing 2. Blackjack 3. Drawing to 20 or 21 4. Busting
part 1 Hi FGK The first 2 are easy. 1) Assuming an infinite deck the chances of any specific card coming out at any point is 7.69% or 30.78% for 10 value card. Therefore the chance of the dealer having an Ace showing is 7.69% or 1 in 13. 2) Probability of BJ is just over 4.73% or approx. 1 in 21 I've just calculated dealer bust probability by taking the dealer outcome charts and multiplying the probability of bust for each exposed dealer card by the probability of that exposed card occurring and then adding them together. I got... 3) Chance of dealer bust assuming infinite deck and H17 = 29.56%. S17 = 29.1% Number 4 needs some work/research, I'm sure wizard of odds has th numbers, I'll check... Cheers Reachy
Peasy Just realised that I could figure out No. 4 the same way as No. 3. Probability of dealer hitting 20 or 21 is 26.9% for S17 and 27.4% for H17. Cheers Reachy
dealer outcomes and theri distribution We could try to calculate it ourselves but why not use other’s toil. One of the easiest to access should be Don Shlesinger’s “Blackjack Attack” (I and II). In Chapter 4 he has, perhaps useful to you, table of final hand probabilities for dealer (and corresponding player’s results). Note that there is an interference of dealer’s blackjack with finishing players’ hands, and also whenever all the players bust normal procedure is to not finish the dealer hand. This is for standard Strip rules, six decks. Dealer outcomes in %: <17 – 0 17 -- 14.52 18 -- 13.93 19 -- 13.35 20 -- 17.96 21 (non-BJ) – 7.29 BJ -- 4.75 Bust -- 28.2 Somehow I have a feeling that there is more to you question and that it has to do with the distribution of those outcomes. Knowing the chances from the table above take one of many binomial distribution calculators available on the internet and set the “p” as the dealer probability, N as the number of trials, and nO as how many times it will occur, then you can check the probability of it happening as exactly that many times or that many times and more, or that many times or less. For example, I observed that the dealer had finished with nineteen and twenty 50 times in 200 rounds. I would take .1335 + .1796 = .3131 as “p”, 200 as N, nO as 50. I would get the chance for the dealer finishing with 19 or 20 in 200 rounds exactly 50 times is 0.94% (which most likely does not mean much by itself). But it would tell me the chance for the dealer’s totals 19 and 20 occurring 50 times or less was only 3.04% and happens on average only once in 33 trials. S. Yama
testing distributions if you want to calculate the odds against a particular number of hands showing, to determine if it is an improbable result - use the nO=> or nO<= calculations - so you want to know the number of hands of that many or more, or, that many or less -
I would guess you would need to create tables for yourself. Use Norm's site: http://www.qfit.com/CVDPC.htm This is Casino Verite Dealer Probability Calculator where you can set deck(s) and rules in any way you want. Then you would need to total the results for all dealers upcards, just as you did for the problem earlier on. (But) Of course if that's what you were asking about. I am working on my mind-reading abilieties combined with not-so-heavy math. S. Yama
So while everyone has their math hats on... According to the popular ploppy strategy of assuming the dealer has a 10 in the hole, how often is that actually correct? I'm guessing 16/52=30.77%, so it's true less than a third of the time!
Correct... ...assuming either an infinite deck or first deal off a fresh shoe. The actual odds will change depending on how many T's have been seen. Cheers Reachy
OH Monkey wrench! Any cards coming off the shoe or from the so called infinite deck will impact the outcomes of all hand totals. For example, in FGK's question what is the % chance dealer will show an Ace, its easy to estimate that on pure probability, but BJ isnt a game of pure probability it has a huge variance situation to situation. What if, in FGs case, 7 players took 5 Aces? The dealers probability of showing an Ace went way down. The infinite deck is a conundrum all of its own- Where are the cards replaced and what is the shuffling mechanism? BJ is a world guided by theory of events averaged over millions of hands, but the problem is you never know where you will sit during a given hand or playing session on that theoretical law called basic strategy. Your aces , Reach, off a fresh shoe or infinite deck- which are related -mathematically- but who knows where your immediate situation will fall? Every card has an impact on each situation and scenario. I wish someone would define the "infinite deck" and the standard card replacement technique. I did find this page that is interesting but still leaves the infinite deck as an impotent device to manage each count scenario. I know you got to start somewhere of course, but BJ from the players standpoint is choppy and not continuos as your ruin or reward removes the player from the infinite situation. http://www.pokeryoda.com/linear-proximation-to-the-infinite-deck-blackjack-function.htm
that geezers cat! You are correct Barney. I'm no maths or stats whiz by any stretch of the imagination but I do understand the limitations of using the infinite deck in probability calculations for BJ. Actually FGKs questions are impossible to answer accurately unless we know the specific situation he is talking about i.e. how many players, what cards they have, what cards have gone before, etc, etc. But does that mean that the information that his questions have generated is useless? Of course not. It's a best guess and depending on the situation the actual probabilities will tend to the figures mentioned; it gives us guidelines so that we aren't just making wild guesses. As far a Bet21 is concerned, and lets face it, that's where I get all my play and FGK gets most of his, we are almost playing with an infinite deck anyway since they shuffle after every hand. Yes there will be an effect on the probability of the dealer having a 10 under if there are several already showing but it's not as great as you might think. Consider this: Bet 21 with 7 players all being dealt 2 Tens each with the dealer showing a Ten up; what is the chance of her also having a Ten under? So that's 15 Tens already dealt, what's the chance that the dealer's undercard is also a Ten? You might think it's quite small but it's not much less than the 30.78% already mentioned. I'll let you (or whoever wants to) calculate the figure and see whether it correlates with mine. As for the definition of an "infinite deck", isn't it fairly simple? It's a theoretical shoe with an infinite number of standard playing cards where the effect of card removal is zero. In practical terms you could have a real world infinite deck if you took a single deck of cards and as you dealt out each card you replaced it with exactly the same card back into the deck and then shuffled it before you dealt another card. Of course that would be incredibly time consuming and pointless but you get my point? Cheers Reachy
Using ((n*16)-15)/(n*52) where n= number of decks... A 6 deck shoe less 15 T's = 25.96% chance of dealer T in hole An 8 deck shoe less 15 T's = 27.16% chance of dealer T in hole A 100 deck shoe less 15 T's = 30.48% chance of dealer T in hole
Nearly Masteff You missed out "-15" from the right side of the equation i.e. ((n*16)-15)/((n*52)-15) Cheers Reachy
Fyi I realise that the value of the following information may be questionable but I think it's reasonably interesting. What the tables show is the effect of card removal at Bet21 on any one hand. The first table shows the numbers for all card values other than 10; the second one for all 10 value cards. The Y axis corresponds to the number of players in the hand and the X axis corresponds to the number cards of any one value showing after the initial deal. So if you wanted to know the probability of a 7 appearing if there are 5 players and 3 7's are already showing you would use the first table and get the figure 6.98%. If you wanted to know what your chances of getting dealt a 10 when there are 3 of you and there is only 1 10 is already out you would use the second table adn get 31.15%. You could of course extend the tables as the cards get dealt out but I 'll leave that to anybody who may have the inclination! Code: No. of Cards Showing (Non-10 Value Cards) Plrs 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 7.82% 7.49% 7.17% 6.84% 6.51% 6.19% 3 7.87% 7.54% 7.21% 6.89% 6.56% 6.23% 5.90% 5.57% 4 7.92% 7.59% 7.26% 6.93% 6.60% 6.27% 5.94% 5.61% 5.28% 4.95% 5 7.97% 7.64% 7.31% 6.98% 6.64% 6.31% 5.98% 5.65% 5.32% 4.98% 4.65% 4.32% 6 8.03% 7.69% 7.36% 7.02% 6.69% 6.35% 6.02% 5.69% 5.35% 5.02% 4.68% 4.35% 4.01% 3.68% 7 8.08% 7.74% 7.41% 7.07% 6.73% 6.40% 6.06% 5.72% 5.39% 5.05% 4.71% 4.38% 4.04% 3.70% 3.37% 3.03% No. of Cards Showing (10 Value Cards) Plrs 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 2 31.27% 30.94% 30.62% 30.29% 29.97% 29.64% 3 31.48% 31.15% 30.82% 30.49% 30.16% 29.84% 29.51% 29.18% 4 31.68% 31.35% 31.02% 30.69% 30.36% 30.03% 29.70% 29.37% 29.04% 28.71% 5 31.89% 31.56% 31.23% 30.90% 30.56% 30.23% 29.90% 29.57% 29.24% 28.90% 28.57% 28.24% 6 32.11% 31.77% 31.44% 31.10% 30.77% 30.43% 30.10% 29.77% 29.43% 29.10% 28.76% 28.43% 28.09% 27.76% 7 32.32% 31.99% 31.65% 31.31% 30.98% 30.64% 30.30% 29.97% 29.63% 29.29% 28.96% 28.62% 28.28% 27.95% 27.61% 27.27% Cheers Reachy
