Blackjack Game Problem
The two players have very different strategies. Player 1 decides to bet all of his money each hand. Since he will not be able to double or split, assume Player 1 will have a 45% chance of winning each hand (ignore blackjacks, surrender, etc.).
Player 2 is an experienced card counter. He bets 1/100 of his stack each hand and has a 1% advantage that always holds true (for every dollar he bets he gets $1.01 back).
Which player has a better chance of reaching $1,000,000 in the least number of hands?
Additionally, what dollar amount would both players reach (on average) in the same number of hands?
Player 1 is more difficult to figure out since there is no point at which he is assured to reach 1,000,000. The best we can do is figure out the average number of hands it would take. To reach 1,000,000 he needs to double up 10 times. The chance of this happening at 45% is about 1/3000. However, 3000 is not the average number of hands he will play since he will often win a number of hands in a row before losing and going back to zero.
I am almost certain that Player 1 has a better chance of reaching 1,000,000 before Player 2, although I am not sure how many hands on average he would play (my best guess is somewhere between 5000-10000). I would like to know if the Wizard has any insights into how one would solve for Player 1.
silly
Quote: mustangsallyThe 1 in ~3000 (2937) is only for the first 10 hands.
At 1542 hands there is a 25.010023% chance of hitting at least 1 - 10 win in a row streak.
3703 hands 50.008773% chance of hitting at least 1 - 10 win in a row streak (15.3% of at least 2 - 10 win in a row streaks)
5863 hands 66.667655% chance of hitting at least 1 - 10 win in a row streak (23.1% of at least 2 - 10 win in a row streaks)
12,279 hands 90.000052% of hitting at least 1 - 10 win in a row streak
(63.9% of at least 2 - 10 win in a row streaks) AND (37.4% of at least 3 - 10 win in a row streaks)
Average number of win streaks of 10 or longer in 69,000 hands. About 13.
I say player 1 can surprise everyone.
Those numbers all look right, how did you do the math on those?
I think a lot of people would be surprised that Player 1 has a much better chance of winning this challenge. I guess it goes to show that poor players can still have a better chance of winning big in the short term, but smart players always fair better long term.
silly
I get about 47.1116% (*2) as the expectation for playing when not having any money to double etc, but as you say quite a few of these will be ties. Even so that would only add a percentage of wasted hands that are just going to be replayed. It also ignores the effect of getting back over 1 for a winning BJ (where two winning BJ's mean you only need a run of 9).
I guess that it would be about one hand in 1857 that there's a run of ten. Even assuming he played the first hand, then sat out nine before starting again, that's still quicker than the grinder.
My gut feeling was the grinder only had 1000 to play with, whereas the risker has been taking lots of 1000's and eventually one of them will be lucky.
Quote: charliepatrickI agree with the about 69000 that the grinder (assuming the results are reasonably near average, which for that number they would be) eventually takes.
I get about 47.1116% (*2) as the expectation for playing when not having any money to double etc, but as you say quite a few of these will be ties. Even so that would only add a percentage of wasted hands that are just going to be replayed. It also ignores the effect of getting back over 1 for a winning BJ (where two winning BJ's mean you only need a run of 9).
I guess that it would be about one hand in 1857 that there's a run of ten. Even assuming he played the first hand, then sat out nine before starting again, that's still quicker than the grinder.
My gut feeling was the grinder only had 1000 to play with, whereas the risker has been taking lots of 1000's and eventually one of them will be lucky.
I would like to know the rule set in play for that statement. The OP did not specify a rule set for BJ. For example 6D S17 DA2 DAS LS 70%Pen has a lower HA, and Win Pct.than same offering w/o LS, and of course H17 is altogether different, and can be played w/o DAS or Soft DD. Location, location, location. Vegas has a variety of rules, here in N.E. US very few choices.
Also, Player 1's stategy and H.A is much different than Player 2's. For those with CVData, eliminate all DD and SPL, and run a sim to a billion. There are some very bad decisions there. Since Player 1 is not advantaged, presume Basic Strategy. If LS allowed the only Surrender change would be R 8-8 vs. X as one can't split.
Quote: iwannaiguanaTo reach 1,000,000 he needs to double up 10 times. The chance of this happening at 45% is about 1/3000. However, 3000 is not the average number of hands he will play since he will often win a number of hands in a row before losing and going back to zero.
It is not 3000 hands, it is 3000 trials. The trial is a number of hands played, until you lose a hand. With 0.45% to win, and ignoring pushes, the average length the trial is 1/0.55 = 1.82 hands. So, the average number of hands played until a 10-hand winning strike is
3000*1.82 = 5460, or, more precisely, 1/(0.55*0.45^10) = 5340 hands
Quote:I think a lot of people would be surprised that Player 1 has a much better chance of winning this challenge. I guess it goes to show that poor players can still have a better chance of winning big in the short term, but smart players always fair better long term.
It does not. If the second player used the same strategy, and bet all his money every time, he would only need 695 hands to reach the million. It is not about good vs bad so much as it is about "cautious" vs. "risky".
The punch line is pretty trivial thought - if you want to win (or lose) really fast and big, bet as much as you can on a single bet.
If you want to hit a certain target where there is no penalty losing (as in the original question), variance is your friend.
If the target is reasonably low (compared to house edge), a high-variance negative EV game will outperform a low-variance positive EV game.
Of course the simplest thing of increasing variance is higher bet size (which Player 1 does). Positive EV (which Player 2 does) will not help much.
Of course in real life the penalty of losing is - well - loss of money. Hence that strategy will not work (unless you don't care).
Very nice explanation.The result returned was for a win streak of 10 or more.(winning "strike"?... you be British?)Quote: weaselmanIt is not 3000 hands, it is 3000 trials. The trial is a number of hands played, until you lose a hand. With 0.45% to win, and ignoring pushes, the average length the trial is 1/0.55 = 1.82 hands. So, the average number of hands played until a 10-hand winning strike is
3000*1.82 = 5460, or, more precisely, 1/(0.55*0.45^10) = 5340 hands
One minor correction.
Since Player 1 would stop at the end of a win streak of 10 and not continue, your formula that includes a 'sum of an infinite geometric series' term is not needed for this example.
Instead one should use a formula that includes the 'sum of a geometric series' term.(all you high school students can look those two formulas up...a very good exercise) or just go to an Excellent Explanation HERE
I opt to use Feller's 'mean recurrence time' formula ((p^-n)-1)/(1-p) = 5337.824216 or 5338 for the mean or average number of trials.
Yes, Not much of a difference (2) but with very high values of p and small run lengths...Example p=0.99 and n=2 you can plug in the numbers and see what I mean.
Now, for you high school students still paying attention.
There are 4 or 5 more formulas that can be used. ( I often wonder why most high school students just want formulas and are really not interested in deriving formulas or equations as weaselman did above. Ahh yes, still teenagers.)Here without any proof. (p= prob of single trial success and n=length of run)
(here are 2 more just for giggles: (1/p^n)*((1-p^n)/(1-p)) or even (1-(p^n))/((1-p)*(p^n))
My point in bringing this up is in a run or streak or 'in a row' consecutive successes or failures the "average' or mean 'by itself' really does not give one a feel for the actual distribution. The average is not close to the peak of the distribution as in a normal distribution and one can see that since the median is less than the average.(extra credit)
yes, I have a high school daughter
Quote: weaselman
It does not. If the second player used the same strategy, and bet all his money every time, he would only need 695 hands to reach the million. It is not about good vs bad so much as it is about "cautious" vs. "risky".
The punch line is pretty trivial thought - if you want to win (or lose) really fast and big, bet as much as you can on a single bet.
I disagree. If Player 2 bet all his money every hand he would have essentially the same results as Player 1. Additionally, it is obvious that a card counter could not bet all of his money every hand; the vast majority of the time he must simply sit around waiting for a good count. This was simplified in the problem by something assuming he bets 1/100th of his stack. No professional card counter would ever be betting any significant portion of his bankroll.
I guess this could be better worded in terms of "good" or "bad" bankroll management. Player 1 (poor bankroll management) is much more likely to hit a big win (or loss) in the short term. Whereas if we were to ask who is better off after 100,000 hands it is almost always going to be Player 2 (good bankroll management).
This definitely applies in real casinos. The "big winners" that always brag are always the biggest losers as well. The people who do the best in the long term are the ones who quietly keep adding to their bank accounts.
Quote: iwannaiguanaI disagree. If Player 2 bet all his money every hand he would have essentially the same results as Player 1.
Not under the conditions you formulated. Player 1 will loose everything he has on about every other bet, while player 2 never loses at all. How is it "the same results"?
Quote:Additionally, it is obvious that a card counter could not bet all of his money every hand; the vast majority of the time he must simply sit around waiting for a good count.
Again, not in your problem, where he has a guaranteed 1% advantage all the time.
Unrealistic? Well ... how realistic is it that the other guy bets everything, loses and gets his original bankroll back constantly?
Quote:
This was simplified in the problem by something assuming he bets 1/100th of his stack. No professional card counter would ever be betting any significant portion of his bankroll.
If he is guaranteed to get his entire bankroll refunded in case of a loss, he'll be an idiot not to bet it all.
Quote:I guess this could be better worded in terms of "good" or "bad" bankroll management. Player 1 (poor bankroll management) is much more likely to hit a big win (or loss) in the short term.
There is no such thing as "good" or "bad" bankroll management by itself. You need to define the criteria of optimality first.
If the criterion is to maximize the amount won while minimizing the time played, then it is the second player, who uses "poor"/unoptimal bankroll management.
But such criteria only makes any sense when you cannot lose, which unfortunately almost never happens in real life.
Sometimes it does though. Sometimes a casino runs a promotion for example, that will reimburse your losses up to, say $500. How much would you bet in that case?
Quote:Whereas if we were to ask who is better off after 100,000 hands it is almost always going to be Player 2 (good bankroll management).
Well, not really. As you have just seen, in 100,000 hands the first player will have made about 20 million bucks on average, while the second one will be just getting past his first one.
Quote:This definitely applies in real casinos. The "big winners" that always brag are always the biggest losers as well. The people who do the best in the long term are the ones who quietly keep adding to their bank accounts.
You have to remember about the criteria before deciding who does "good" and who does "badly". If all you want is to be "better off long term", your best choice is not to go to the casino in the first place. (Even if you are an advantage player, there are quicker and better ways to make money). Most people go to casinos because they enjoy playing, and want to have a good time. If winning big fast is what makes them enjoy themselves, then slowly grinding by 1% of bankroll for the entire night is hardly a "good bankroll management" for them, it's just torture.
You are also mistaken in that you would always bet your entire bankroll under optimal strategy. After 100,000 hands the most likely outcome for Player 1 is that he still only has the 1,000 he started with. No matter how high he gets he always has a 55% chance of going broke the next hand. After 100,000 hands Player 2 will be assured to surpass 20 million.
If the goal was to optimize profitability after a set number of hands, and you could only choose between the two strategies, then you should use Player 1's strategy up to a certain point before switching to Player 2's strategy.
There is a generally accepted definition of bankroll management which is minimizes risk while maximizing long term profit. It can be applied from everything from stocks to poker. However, good bankroll management is not equivalent to amount of fun had. For this reason good bankroll management is not appropriate for casual gambler's and I agree that it does not make sense to follow it in most people's cases.
Quote: iwannaiguanaYou just missed the whole idea of the problem. Player 2 cannot bet his bankroll every hand because that would eliminate any advantage of counting and he would no longer have any advantage over the house.
So, you are saying that they are playing two different games then? One game has a constant expected value, and the other has EV, that depends on the bet amount? In that case, you can't compare the two strategies at all. They are both "good" - each player is playing the strategy, that is optimal for his respective game.
Quote:
If the goal was to optimize profitability after a set number of hands, and you could only choose between the two strategies, then you should use Player 1's strategy up to a certain point before switching to Player 2's strategy.
Well, maybe, except, you'd have to switch the games too. Using 2's strategy in 1's game is suboptimal, and vice versa.
Quote:There is a generally accepted definition of bankroll management which is minimizes risk while maximizing long term profit.
Under that definition, the optimal strategy for most people and most games is not to play at all.
The only exceptions are highly skilled VP players who can find the good paytable within reasonable distance from them, and, to a lesser extent, card counters with huge bankrolls, and a bunch of stupid pit bosses running one deck game with fantastic rules.
Quote:For this reason good bankroll management is not appropriate for casual gambler's and I agree that it does not make sense to follow it in most people's cases.
Exactly. Kelly's criterion (I think, that's what you are hinting about) is optimal for +EV games, which are a rarity in a casino. But even to the (second player's) game in your example it does not quite apply - first, because he can't lose, and second, because, even if he could, his losses would be reimbursed. The only case when he is really playing optimally is if, like you said, the advantage would decrease if he raised his bet.
All I was saying is that you would use a mix of the two strategies if your goal was to maximize profit after a period of time. I think Kelly's criterion would still apply to a similar situation since he is risking the money he has already won. If he has $10,000,000 he is obviously risking something since he will go back to $1000 if he loses. The only reason you can't use Kelly's criterion here is that I allowed for only to possible bet amounts.
Maybe a better question would be what strategy would you choose if you got to keep the money after x number of hands? At what point would you play it safe and switch from strategy 1 to strategy 2 in order to maximize your winnings?
Quote: iwannaiguanaThey're playing the same game with the rules I stated. The only two options are bet everything with 45% chance of winning or have the guaranteed win of 1% if they bet 1/100th of their bankroll. $1000 reloads are obviously not realistic in a casino, but are common in computer blackjack games such as the one on WoO (unfortunately counting is impossible on that one).
Reloads are unrealistic, guaranteed wins are unrealistic, and the dependency of the EV on the amount of bet is extremely unrealistic.
And they are obviously not playing the same game in any meaningful sense - just like a guy betting pass line is not playing the same game as the guy betting on twelve.
Quote:All I was saying is that you would use a mix of the two strategies if your goal was to maximize profit after a period of time.
And what I am saying is that you are wrong, the optimal strategy depends on the goal and on the game you are playing. If my goal was to maximize profit, provided that I cannot loose, I would only use the first strategy. If I can loose, but my game is +EV, then I would use the second strategy. If the game is -EV, then, the optimal strategy would be to not play at all.
Quote:I think Kelly's criterion would still apply to a similar situation since he is risking the money he has already won.
Yes. But betting 1% of the bankroll would not be optimal.
Quote:If he has $10,000,000 he is obviously risking something since he will go back to $1000 if he loses. The only reason you can't use Kelly's criterion here is that I allowed for only to possible bet amounts.
No, the reason you can't use it is that one game is -EV, and the other one is deterministic.
Quote:Maybe a better question would be what strategy would you choose if you got to keep the money after x number of hands? At what point would you play it safe and switch from strategy 1 to strategy 2 in order to maximize your winnings?
Well, if x=1, the expectation of the larger bet of N thousand is
E(1,N) = 0.45*E(0,2N)+0.55*E(0,1) = 0.45*2N + 0.55
With 2 hands remaining it is:
E(2,N) = 0.45*E(1,2N) + 0.55*E(1,1)
In general, E(k,N) = 0.45*max(E(k-1,2N),2N*1.0001^(k-1)) + 0.55*E(k-1,1)
Here is a spreadsheet, calculating this for x up to 100 hands, and bankrolls of up to a million bucks.
The bold values indicate cells where betting everything is beneficial compared to betting 1%.
Quote: weaselman
And what I am saying is that you are wrong, the optimal strategy depends on the goal and on the game you are playing. If my goal was to maximize profit, provided that I cannot loose, I would only use the first strategy. If I can loose, but my game is +EV, then I would use the second strategy. If the game is -EV, then, the optimal strategy would be to not play at all.
Yes, the optimal strategy depends on the goal and the game you are playing. And I agree with your math. However, you would be incorrect to only use the first strategy if you were trying to maximize profit. At the beginning the first strategy is obviously better. But once you reach a certain point it become better to use strategy two
Let's say you have $100,000,000 in this game. Do you really want to use strategy 1 knowing you will have a 55% chance of going back to $1000? No, in order to maximize expected profit you would use strategy 2, because the chances of you reaching $100,000,000 again are so small.
Quote: iwannaiguana
Let's say you have $100,000,000 in this game. Do you really want to use strategy 1 knowing you will have a 55% chance of going back to $1000? No, in order to maximize expected profit you would use strategy 2, because the chances of you reaching $100,000,000 again are so small.
Yes. I said, it was not optimal for the situation you described. Under some other set of conditions (like having won a hundred million bucks or something), it might very well be.
I would have to spend hours reviewing Blackjack Attack to come up with the expected number of hands player 2 would need to double his bankroll 10 times over, but it would be MUCH more than 6,000, by many orders of magnitude.
It goes to show that the advantage of milking the free reloads is much more powerful than that from card counting. Other posts have made this point. I just want to throw my support behind player 1.
silly
Quote: mustangsallyThe math is complicated to do without a spreadsheet or Markov chain.
Thanks to the Wizard he has solved these kind of problems in Excel. Maybe he forgot he did!
6,000 hands (45% win%) 67.5136340081363% chance of at least 1 run of 10 wins or longer (~30.8% chance of at least 2 runs of 10 wins or longer; ~10.3% chance of at least 3 runs of 10 wins or longer)
6,000 hands (47.5% win%)~84.20% chance of at least 1 run of 10 wins or longer
A nice online Matrix Algebra Tool can be found here. I know there are a few of them.
It is JavaScript so keep that in mind. (you can save and run from your desktop)
It has had no problems with results matching my larger matrix software.
And it will do fractions.
0.67513634 was returned for A^6000
A = [.55, .45, 0, 0, 0, 0, 0, 0, 0, 0, 0
.55, 0, .45, 0, 0, 0, 0, 0, 0, 0, 0
.55, 0, 0, .45, 0, 0, 0, 0, 0, 0, 0
.55, 0, 0, 0, .45, 0, 0, 0, 0, 0, 0
.55, 0, 0, 0, 0, .45, 0, 0, 0, 0, 0
.55, 0, 0, 0, 0, 0, .45, 0, 0, 0, 0
.55, 0, 0, 0, 0, 0, 0, .45, 0, 0, 0
.55, 0, 0, 0, 0, 0, 0, 0, .45, 0, 0
.55, 0, 0, 0, 0, 0, 0, 0, 0, .45, 0
.55, 0, 0, 0, 0, 0, 0, 0, 0, 0, .45
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
