The result calculated by Don was for a new 6 deck shoe. It is also accurate for any random point with a 6 deck shoe. It does however assume that the streak will be completed within the same shoe.
For simulating streaks completed only within the same shoe (6 decks, 1.5 decks cut off) I get 6 tens in a row 1 out of 1498 times with a probability of .00066768
Thanks,
Phil
Interesting! Your result seemingly justifies what I have been saying all along. If the dealer (or player) jumps in at an earlier stage in a shoe, it’s more likely to have a long streak. At a later stage, it’s less likely to have such a streak. BTW, does the player plays basic strategy in your simulation?
Your error is in considering "undealt" cards remaining when you should be considering "unseen" cards.
There are 156 undealt cards, which are all unseen, but we can predict what they are.
Most likely, there are 48 Tens there, so the probability is 0.000676=1/1480;
Slightly less likely, there are 49 or 47 Tens there, so the probability is 0.000681=1/1469;
If there are 50 or 46 Tens there, the probability is 0.000695=1438;
…
If there are 90 or 6 Tens there, the probability is 0.017141=1/58;
...
If there are 96 or 0 Tens there, the probability is 0.002552=1/39.
Putting together, the probability of a long streak is the same as 0.000759 = 1/1317.
It is also accurate for any random point with a 6 deck shoe.
This part is still problematic for me. We consider this moment when the dealer jumps in right after the mid-shoe, so the most likely situation is a remaining deck of 48 Tens in 156 cards, so the probability becomes
48/156 x 47/155 x 46/154 x 45/153 x 44/152 x 43/151 = 0.000676 = 1 / 1480.
This result does not justify what you say above.
It would be helpful if you didn't offer probability calculations when you know absolutely nothing about how to do them. Instead of trying to learn from those who know better and are explaining things to you, you choose to be argumentative and repetitive. You are the embodiment of the word "incorrigible," which means incapable of being corrected. I've never once seen you say, Oh, I learned something, or Thank you for the explanation.
So, here's the deal: this thread is closed. If you post here again on this topic, I'm going to ask Al to suspend you for a month, and, and was the case on Norm's site, if you continue, the ban could become permanent. We are not your babysitters; we aren't here to please you. Grow up!
Don
The result calculated by Don was for a new 6 deck shoe. It is also accurate for any random point with a 6 deck shoe. It does however assume that the streak will be completed within the same shoe.
For simulating streaks completed only within the same shoe (6 decks, 1.5 decks cut off) I get 6 tens in a row 1 out of 1498 times with a probability of .00066768
Thanks,
Phil
Just to double check. You must only examine those streaks which had a chance to reach a length of 10. For example, if the dealer is dealt a ten up on the last round of the shoe, this does not count as a streak of only 1 when trying to match Don's result.
The result calculated by Don was for a new 6 deck shoe. It is also accurate for any random point with a 6 deck shoe. It does however assume that the streak will be completed within the same shoe.
For simulating streaks completed only within the same shoe (6 decks, 1.5 decks cut off) I get 6 tens in a row 1 out of 1498 times with a probability of .00066768
Thanks,
Phil
Just to double check. You must only examine those streaks which had a chance to reach a length of 10. For example, if the dealer is dealt a ten up on the last round of the shoe, this does not count as a streak of only 1 when trying to match Don's result.
Yes, I terminate any streak of 10s up at the end of the shoe. The next shoe starts with 0 tens up. It is streaks of 6 10s up in a row we are looking for.
Phil, I don't think I was clear. I'm interested in how your denominator is determined.
Don's result is the probability of encountering a single streak of 6 or more dealer tens in a row at the top of the shoe. To reproduce this by simulation, in strict terms, you would play rounds from the top of the shoe until one of a) the dealer is not dealt a ten or b) the dealer has been dealt 6 tens in a row. At that point you would record the result of one sample (yes or no), shuffle and repeat.
Now Don and I have both asserted that this result holds for any random point within the shoe where 6 rounds are guaranteed to be available. To verify this by simulation, you simply pick a random point in the shoe to begin rather than the top, making sure that there are enough cards remaining to complete at least 6 rounds, and do the same thing. Note that this is physically and mathematically equivalent to the dealer cutting the deck after shuffling.
Don has also asserted that the number of other players present and their actions have no bearing on this. To verify this by simulation, add additional players who play various strategies and repeat the two experiments above.
Your initial sim was a model of a player sitting at the table and playing shoe after shoe, allowing the streaks to span from one shoe to another. While the frequency of 6 or more dealer tens in a row under these conditions may also be a useful result, it is not quite the same as what Don calculated, although the resulting frequency should be quite close (and it was if I am not mistaken) .
Finally, I believe that you should be able to verify Don's result while playing full shoes but not allowing a streak to span two shoes, as you seem to have tried. In this case, you need to count all streaks of length 6 or more and to discard the data for initial dealer tens dealt within the final 5 rounds played in each shoe, since these rounds violate the condition of there being enough cards to complete 6 rounds.
Phil, I don't think I was clear. I'm interested in how your denominator is determined.
Don's result is the probability of encountering a single streak of 6 or more dealer tens in a row at the top of the shoe. To reproduce this by simulation, in strict terms, you would play rounds from the top of the shoe until one of a) the dealer is not dealt a ten or b) the dealer has been dealt 6 tens in a row. At that point you would record the result of one sample (yes or no), shuffle and repeat.
Now Don and I have both asserted that this result holds for any random point within the shoe where 6 rounds are guaranteed to be available. To verify this by simulation, you simply pick a random point in the shoe to begin rather than the top, making sure that there are enough cards remaining to complete at least 6 rounds, and do the same thing. Note that this is physically and mathematically equivalent to the dealer cutting the deck after shuffling.
Don has also asserted that the number of other players present and their actions have no bearing on this. To verify this by simulation, add additional players who play various strategies and repeat the two experiments above.
Your initial sim was a model of a player sitting at the table and playing shoe after shoe, allowing the streaks to span from one shoe to another. While the frequency of 6 or more dealer tens in a row under these conditions may also be a useful result, it is not quite the same as what Don calculated, although the resulting frequency should be quite close (and it was if I am not mistaken) .
Finally, I believe that you should be able to verify Don's result while playing full shoes but not allowing a streak to span two shoes, as you seem to have tried. In this case, you need to count all streaks of length 6 or more and to discard the data for initial dealer tens dealt within the final 5 rounds played in each shoe, since these rounds violate the condition of there being enough cards to complete 6 rounds.
My denominator is total number of rounds. My most recent numbers is for any occurrence of 6 in a row only within a shoe divided by total rounds.
With the method you outline above it sounds like approximately 1316 out of 1317 shoes would be shuffled away within the first 6 rounds. I understand if that’s how it’s calculated but seems very unusual for a simulation and would not happen in reality. It could likely be simulated though.
I my runs I was not solely counting 6 in a row off the top of each shoe and I am shuffling as normal with the cut card. I am resetting the 6 in a row count at the beginning of each shoe so that would discard the effect of 10s dealt in the final 5 rounds of previous shoe.
For the original question of this, I also often feel like the dealer always has a 10 up. I think it’s because I am really noticing it when it happens.
I don’t know if casinos or machines are cheating or not. But the people designing and programming the machines and making casino procedures do not have the welfare of the player in mind. The house is their priority. It might as well be cheating.
Phil
My denominator is total number of rounds. My most recent numbers is for any occurrence of 6 in a row only within a shoe divided by total rounds.
With the method you outline above it sounds like approximately 1316 out of 1317 shoes would be shuffled away within the first 6 rounds. I understand if that’s how it’s calculated but seems very unusual for a simulation and would not happen in reality. It could likely be simulated though.
I my runs I was not solely counting 6 in a row off the top of each shoe and I am shuffling as normal with the cut card. I am resetting the 6 in a row count at the beginning of each shoe so that would discard the effect of 10s dealt in the final 5 rounds of previous shoe.
Ok. So my point is that "total rounds" must not include rounds which were among the final 5 rounds of a shoe and which were not part of a streak of 6 or more.
And yes, the first simulation in my proposed list is purely theoretical, but if you are trying to verify Don's result, it is the model which matches the calculation.
To be clear, just for the record, my calculation is for a streak of six tens in a row off the top of the shoe, but no more. While the cumulative probability of such streaks of seven or more is basically negligible, it’s not identical to dealing six straight cards and then stopping.
Finally, everyone should understand that looking for such streaks of six while dealing out the whole shoe, for example, is a very different and somewhat more complicated calculation.
Saying I saw six straight dealer ten upcards at the start of the shoe is a lot more unlikely than saying the dealer dealt out the whole shoe and somewhere along the way, he got six straight upcards of ten. I hope everyone understands the difference.
Don
I have a correction: I am counting any streak of 6 10s or more and dividing by the total number of rounds. I am not counting any streaks of 6 or more that begin within the last 5 rounds of a shoe. (Before I said I was only counting streaks of 6 but am counting streaks of 6 or more)
When I was counting streaks of 6 or more through shoe boundaries I got 1 of 1317, when ending with shoe I get 1 out of 1498.
Phil
My denominator is total number of rounds. My most recent numbers is for any occurrence of 6 in a row only within a shoe divided by total rounds.
With the method you outline above it sounds like approximately 1316 out of 1317 shoes would be shuffled away within the first 6 rounds. I understand if that’s how it’s calculated but seems very unusual for a simulation and would not happen in reality. It could likely be simulated though.
I my runs I was not solely counting 6 in a row off the top of each shoe and I am shuffling as normal with the cut card. I am resetting the 6 in a row count at the beginning of each shoe so that would discard the effect of 10s dealt in the final 5 rounds of previous shoe.
Ok. So my point is that "total rounds" must not include rounds which were among the final 5 rounds of a shoe and which were not part of a streak of 6 or more.
And yes, the first simulation in my proposed list is purely theoretical, but if you are trying to verify Don's result, it is the model which matches the calculation.
I assume not counting the final 5 rounds is for when counting streaks of 6 or more anywhere in the shoe except tor the final 5 rounds?
Off the top of shoe I am seeing 6 dealer ten up cards in a row on the first 6 hands of the shoe 1 out of 1317 shoes with a probability of .0007595
I assume not counting the final 5 rounds is for when counting streaks of 6 or more anywhere in the shoe except tor the final 5 rounds?
No, it's because Don's calculation assumes that it is guaranteed that 6 rounds can be completed, which is not possible for the last 5 rounds.
I assume not counting the final 5 rounds is for when counting streaks of 6 or more anywhere in the shoe except tor the final 5 rounds?
No, it's because Don's calculation assumes that it is guaranteed that 6 rounds can be completed, which is not possible for the last 5 rounds.
Yes, same thing, no streaks starting in the last 5 rounds
Does anyone have any experience of a casion dealing themselves a consistently higher than average number of 10 face up cards, giving them a great advantage?
I'm not going to delve into the mathematical duel on this thread.
What I can you is that there is a dealer cheating method that can be used on hand shuffled games called the pick-up. The dealer will pick up the cards lo-hi, lo-hi and then stack them. When it comes time to shuffle they will use a false riffle to maintain the order of the cards.
You need to look out for a few things. 1. any variations in how they pick up the cards. They should be swooped up, so try to watch for any out of order payouts and card grabs or any interweaving.
Watch out for any changes in whether they expose the first or 2nd card in their own hand, as they may be peeking and then flipping over the higher card, to cause those with players with stiffs to bust more often.
Most casinos are not cheating, but the dealers themselves could be covering their tracks to use the same method to funnel money to a confederate, then filling back up the table with suckers money so they don't stand out to the pit.
Less likely is the casino using a prism shoe to peek on the cards in the shoe and deal seconds. This would be very risky for them if gaming commission ever found out, but the only ones who would ID this would be card counters who in most instances will just avoid that dealer/table.
Practical advice though, just change tables if the dealer is winning.
