Odds on player, dealer BJs
- Thread starter shadroch
- Start date
Easiest way of doing it
Probability of both you and dealer getting blackjack =
(probabllity of you having blackjack) *(probablity of dealer getting blackjack given you have blackjack)
Probability of your blackjack: There are two ways to get blakjack Ace 10 or 10 Ace; 16 10's 4 aces in a 52 card deck
= (4/52*16/51 + 16/52*4/51) = 32 / 663 =.0482
Probability dealer having blackjack given you have blackjack, same idea only a 10 and ace are missing in a 50 card deck.
=(3/50*15/49 +15/50 *3/49)= 9/ 245
Probabilty of both you and dealer having blackjack
= 32/663 *9/245 =.00177 or .177%
Probability of both you and dealer getting blackjack =
(probabllity of you having blackjack) *(probablity of dealer getting blackjack given you have blackjack)
Probability of your blackjack: There are two ways to get blakjack Ace 10 or 10 Ace; 16 10's 4 aces in a 52 card deck
= (4/52*16/51 + 16/52*4/51) = 32 / 663 =.0482
Probability dealer having blackjack given you have blackjack, same idea only a 10 and ace are missing in a 50 card deck.
=(3/50*15/49 +15/50 *3/49)= 9/ 245
Probabilty of both you and dealer having blackjack
= 32/663 *9/245 =.00177 or .177%
There are basic methods for counting problems such as these. The first is permutations and the second combinations. All card games use combinations because the order of the cards does not matter...ie. a 10,5 is the same value as 5,10.
As for the formula, nCr, simply means your choosing r from n elements. I'm assuming you know about factorials and the symbol ! that denotes it.
A direct formula of nCr looks like:
(n!) / (r!)(n-r)!
This equation can be used to solve any combination probability problems.
I'll do the example.
You already know there are 4Aces and 16Tens and a total of 52 cards.
You can break your question down into a 3 step problem.
1) How many total possible combinations of hands are there.
---Using nCr: a total of 52 cards are being used and you are choosing 2 from those 52. In other words, nCr = 52C2. Plug in n and r in the equation and you will get 1326 possible combinations.
2) The probability that you will receive a blackjack.
--Using the equation nCr: you have 4 Aces and you are choosing 1 from those 4. nCr = 4C1.
-- You also need to choose 1 Ten from the 16 in the deck. nCr = 16C1..again plug in.
Then multiply these 2 answers from step 2 and divide by Answer in step 1.
3) You now find the probability of the dealer having blackjack ALSO.
= 3C1 x 15C1 / 50C2.
Finally multiply the answer you get at the end of step 2 by the answer from step 3.
.048 x .0367 = .0017 is your probability that you and the dealer will both have a BJ.
As for the formula, nCr, simply means your choosing r from n elements. I'm assuming you know about factorials and the symbol ! that denotes it.
A direct formula of nCr looks like:
(n!) / (r!)(n-r)!
This equation can be used to solve any combination probability problems.
I'll do the example.
You already know there are 4Aces and 16Tens and a total of 52 cards.
You can break your question down into a 3 step problem.
1) How many total possible combinations of hands are there.
---Using nCr: a total of 52 cards are being used and you are choosing 2 from those 52. In other words, nCr = 52C2. Plug in n and r in the equation and you will get 1326 possible combinations.
2) The probability that you will receive a blackjack.
--Using the equation nCr: you have 4 Aces and you are choosing 1 from those 4. nCr = 4C1.
-- You also need to choose 1 Ten from the 16 in the deck. nCr = 16C1..again plug in.
Then multiply these 2 answers from step 2 and divide by Answer in step 1.
3) You now find the probability of the dealer having blackjack ALSO.
= 3C1 x 15C1 / 50C2.
Finally multiply the answer you get at the end of step 2 by the answer from step 3.
.048 x .0367 = .0017 is your probability that you and the dealer will both have a BJ.
callipygian
Well-Known Member
Actually, it's the same method, you just simplified the math in your head.
Starting with rollem411's equation ...
P = [C(4,1)*C(16,1)/C(52,2)]*[C(3,1)*C(15,1)/C(50,2)]
P = [4*16/(52*51/2)]*[3*15/(50*49/2)]
P = [4/52*16/51*2]*[3/50*15/49*2]
P = [4/52*16/51+4/51*16/52]*[3/50*15/49+3/49*15/50]
... you end up with your equation.
Okay, thanks for the formula. It's pretty much as I figured it was.
Let me take it a step further, because this is what I'm trying to ascertain.
In Laughlin, you can bet $1 bonus bet on the first hand of a new deck( SD). A BJ pays you 10-1,I believe. If both you and the dealer hit, it pays a progressive, which ranges from $50 to an occasinal $200plus. Is there a dollar figure at which this bet is worth making? Whats throwing me off is the smaller payoffs for only a player BJ. I'm not sure how to factor that payoff in.
Let me take it a step further, because this is what I'm trying to ascertain.
In Laughlin, you can bet $1 bonus bet on the first hand of a new deck( SD). A BJ pays you 10-1,I believe. If both you and the dealer hit, it pays a progressive, which ranges from $50 to an occasinal $200plus. Is there a dollar figure at which this bet is worth making? Whats throwing me off is the smaller payoffs for only a player BJ. I'm not sure how to factor that payoff in.
sagefr0g
Well-Known Member
serendipidous timing
ok guys i been trying to learn some of this stuff and could use a bit of an explaination of something i'm not understanding.
like i think i understand at least intuitively the reasoning that comes up with the equation:
nPn = n(n-1)(n-2)(n-3)... 3x2x1
and am ok with
nPn = n!
where i'm getting lost is when the author talks about;
if you have n objects, but want to make ordered arrays each consisting of r objects (r being less than n), then he calls that nPr or "the number of permutations of n things taken r at a time"
so ok i'm fine up to this point but then he goes on to say (here i get lost)
nPr = n(n-1)(n-2)(n-3)... (n-r +1)
he calls the right side of the equation essentily r factors.
and this is part of what i don't understand, i don't get the logic or justification of taking the nPn equation and taking out the 3x2x1 part and replacing it with (n-r +1)
then he goes on to say for the two equations:
nPn = n!
and
nPr = n!/(n-r)!
that we should be easily able to satisfy ourselves that (n-r)! in the lower part of the last of the last written fraction just kills off, by cancellaion, the unwanted tail of n! so as to make it properly stop with the r'th factor (n-r + 1)
and here i'm really lost as far as getting the logic
ok guys i been trying to learn some of this stuff and could use a bit of an explaination of something i'm not understanding.
like i think i understand at least intuitively the reasoning that comes up with the equation:
nPn = n(n-1)(n-2)(n-3)... 3x2x1
and am ok with
nPn = n!
where i'm getting lost is when the author talks about;
if you have n objects, but want to make ordered arrays each consisting of r objects (r being less than n), then he calls that nPr or "the number of permutations of n things taken r at a time"
so ok i'm fine up to this point but then he goes on to say (here i get lost)
nPr = n(n-1)(n-2)(n-3)... (n-r +1)
he calls the right side of the equation essentily r factors.
and this is part of what i don't understand, i don't get the logic or justification of taking the nPn equation and taking out the 3x2x1 part and replacing it with (n-r +1)
then he goes on to say for the two equations:
nPn = n!
and
nPr = n!/(n-r)!
that we should be easily able to satisfy ourselves that (n-r)! in the lower part of the last of the last written fraction just kills off, by cancellaion, the unwanted tail of n! so as to make it properly stop with the r'th factor (n-r + 1)
and here i'm really lost as far as getting the logic
Sonny
Well-Known Member
Sonny
Well-Known Member
so ok i'm fine up to this point but then he goes on to say (here i get lost)
nPr = n(n-1)(n-2)(n-3)... (n-r +1)
he calls the right side of the equation essentily r factors.
and this is part of what i don't understand, i don't get the logic or justification of taking the nPn equation and taking out the 3x2x1 part and replacing it with (n-r +1)
It’s just a way of writing the formula so that you can replace the n and r with any numbers you want to use. The (n-r+1) just tells you where to stop. You keep multiplying until you have done it r times. The reason we factor out the 4*3*2*1 is because we are only looking at the first 4 positions. We can ignore the results of the last 4 positions. The example below should make things a bit clearer.
then he goes on to say for the two equations:
nPn = n!
and
nPr = n!/(n-r)!
that we should be easily able to satisfy ourselves that (n-r)! in the lower part of the last of the last written fraction just kills off, by cancellaion, the unwanted tail of n! so as to make it properly stop with the r'th factor (n-r + 1)
and here i'm really lost as far as getting the logic
It might help if you write out the fraction to see how everything cancels out. Using the same numbers from above:
Code:
n = 8
r = 4
nPr = n!/(n-r)!
nPr = [U]8*7*6*5*4*3*2*1[/U]
4*3*2*1
nPr = 8*7*6*5
nPr = n*(n-1)*(n-2)*(n-3)The bottom numbers cancel out the top ones in order to eliminate the unwanted end position results. As an example, let’s look at a horse race with 7 horses. How many different ways can the race end assuming there are no ties? Well, any of the 7 horses could finish first. That leaves 6 horses that could finish second, 5 horses that could finish third, 4 horses that could finish fourth, etc. The total number of possibilities is:
Code:
n = 7 horses
nPn = n!
nPn = 7*6*5*4*3*2*1But what if we are only concerned with the horses that finish in the money? If we only look at the first, second and third horses then we still get 7 horses for first, 6 for second and 5 for third, and then we’re done. We can ignore the horses that come in fourth, fifth and the rest of the positions. Using the formula:
Code:
n = 7 horses
r = 3 positions
nPr = n!/(n-r)!
nPr = [U]7*6*5*4*3*2*1[/U]
4*3*2*1
nPr = 7*6*5-Sonny-
Last edited:
callipygian
Well-Known Member
Yes. I'm not sure why Sonny sent you a PM instead of posting it; there's a fairly straightforward way to calculate this.
To clarify, though, when you hit the progressive jackpot, you do not also get paid 10:1 for getting a blackjack, right? That is, if the progressive jackpot is $100, then you get paid $100 for a $1 bet, not $110, right?
I sent you a PM.
Is there a reason you didn't post it? Because I think posting it and how you arrived at that figure would be instructive as a general rule for estimating edges.
Sonny
Well-Known Member
I just wanted to preserve any potential opportunity until Shadroch gets there. I'd hate to burn it out before he got a whack at it.
-Sonny-
callipygian
Well-Known Member
Okay. GL shadroch!
On a semi-tangential note, how does the progressive jackpot get built up? By time or by the number of people who put a bet down? Because if the jackpot goes up by the number of people who put a bet down, you might be justified in making the bet at a lower figure at a relatively dead time in the casino (in which case you might be the only one making the bet).
Sonny,
I appreciate your thoughts, but feel free to post your answer here.
The game rarely gets into the positive range. I'm not sure I've ever seen it myself. People tend to look at the meters when choosing a game so the higher the meter, the more crowded the table.
The jackpot progresses by the amount of people betting, but it is a fraction of the bet. Something along the line of three people bet a dollar and it goes up a dime, more or less. On top of that, The Riverside now restricts doubling to 10/11, so its not a great game to begin with. Edgewater has a single 3-2 SD,DOA game that offers this, or did two months ago.
I appreciate your thoughts, but feel free to post your answer here.
The game rarely gets into the positive range. I'm not sure I've ever seen it myself. People tend to look at the meters when choosing a game so the higher the meter, the more crowded the table.
The jackpot progresses by the amount of people betting, but it is a fraction of the bet. Something along the line of three people bet a dollar and it goes up a dime, more or less. On top of that, The Riverside now restricts doubling to 10/11, so its not a great game to begin with. Edgewater has a single 3-2 SD,DOA game that offers this, or did two months ago.
Sonny
Well-Known Member
Alrighty. Below is a revised version of my PM to Shad (I adjusted the bankroll requirements and added some info):
The bet barely becomes positive when the progressive payout reaches $265 with an EV of 0.1% and a SD of 11.4 units. The point at which you would place a bet will depend on your bankroll size and your tolerance for risk. For example, a full Kelly bettor with a $9k bankroll would take the bet at a progressive payout of $273 or more and win about $1.50 per bet. A half Kelly bettor with the same bankroll would take the bet at a payout of $283 and win about $3.30 per bet. With a $5k bankroll a full Kelly bet would be at $281 for en EV of about $3 per bet, and a half Kelly bet would be at $303 with an EV of about $6.83 per bet.
As always, corrections are welcome.
-Sonny-
callipygian
Well-Known Member
I actually got $275, not $265, for the base EV.
P(player BJ)*P(dealer noBJ)*10+P(playerBJ)*P(dealer BJ)*X-P(no player BJ) = 0
(128/2652)*(1-90/2450)*10+(128/2652)*(90/2450)*X-(2524/2652) = 0
X = 274.6
[Side note: Sonny, did you take into account that dealer can't have a BJ if player has BJ and wins 10? If you omit P(dealer noBJ) from the first term you get X = 264.6]
I have no idea how Sonny calculated the risk-adjusted bets, but here are the EV's for progressive jackpot values from $274 to $306.
Edit: SD's too
P(player BJ)*P(dealer noBJ)*10+P(playerBJ)*P(dealer BJ)*X-P(no player BJ) = 0
(128/2652)*(1-90/2450)*10+(128/2652)*(90/2450)*X-(2524/2652) = 0
X = 274.6
[Side note: Sonny, did you take into account that dealer can't have a BJ if player has BJ and wins 10? If you omit P(dealer noBJ) from the first term you get X = 264.6]
I have no idea how Sonny calculated the risk-adjusted bets, but here are the EV's for progressive jackpot values from $274 to $306.
Edit: SD's too
Code:
Pot EV SD
274 -0.001 11.8
275 0.0008 11.8
276 0.0025 11.9
277 0.0043 11.9
278 0.0061 11.9
279 0.0079 12
280 0.0096 12
281 0.0114 12.1
282 0.0132 12.1
283 0.015 12.1
284 0.0167 12.2
285 0.0185 12.2
286 0.0203 12.3
287 0.022 12.3
288 0.0238 12.4
289 0.0256 12.4
290 0.0274 12.4
291 0.0291 12.5
292 0.0309 12.5
293 0.0327 12.6
294 0.0345 12.6
295 0.0362 12.6
296 0.038 12.7
297 0.0398 12.7
298 0.0415 12.8
299 0.0433 12.8
300 0.0451 12.9
301 0.0469 12.9
302 0.0486 12.9
303 0.0504 13
304 0.0522 13
305 0.054 13.1
306 0.0557 13.1
Last edited:
Sonny
Well-Known Member
Yup, that’s what I did alright.
I used Bet = Bankroll * Advantage / Variance. So when the jackpot is $283 and the player’s bankroll is $9k the bet would be $9000 * 0.015 / 148.8 = $0.907, not quite ready for a full Kelly $1 bet. But when the jackpot is $284 the bet would be $9000 * 0.0167 / 148.8 = $1.009. Now it’s time for that side bet. Because of the very high variance of this bet we have to wait until the bet turns favorable enough to justify making the bet. It requires a huge bankroll in terms of units, but since the unit size is only $1 it can occasionally be manageable.
-Sonny-
callipygian
Well-Known Member
Okay. So then here's the chart for bankroll required at full Kelly betting.
Code:
Pot EV Var Bankroll
274 -0.001 138.7 0
275 0.0008 139.7 181518
276 0.0025 140.7 55323
277 0.0043 141.6 32821
278 0.0061 142.6 23425
279 0.0079 143.6 18268
280 0.0096 144.6 15009
281 0.0114 145.6 12763
282 0.0132 146.6 11122
283 0.015 147.6 9870
284 0.0167 148.6 8884
285 0.0185 149.6 8087
286 0.0203 150.6 7430
287 0.022 151.6 6879
288 0.0238 152.7 6409
289 0.0256 153.7 6005
290 0.0274 154.7 5654
291 0.0291 155.7 5345
292 0.0309 156.8 5072
293 0.0327 157.8 4828
294 0.0345 158.9 4610
295 0.0362 159.9 4413
296 0.038 160.9 4235
297 0.0398 162 4073
298 0.0415 163.1 3924
299 0.0433 164.1 3788
300 0.0451 165.2 3663
301 0.0469 166.2 3547
302 0.0486 167.3 3440
303 0.0504 168.4 3340
304 0.0522 169.5 3247
305 0.054 170.5 3160
306 0.0557 171.6 3079
307 0.0575 172.7 3003
308 0.0593 173.8 2932
309 0.0611 174.9 2865
310 0.0628 176 2801
311 0.0646 177.1 2741
312 0.0664 178.2 2685
313 0.0681 179.3 2631
314 0.0699 180.4 2580
315 0.0717 181.5 2532very nice. now a question i have always had on sidebets when determining bankroll needed.
since we play these sidebets (including this, lucky ladies, etc) simultaneously with our blackjack hands, ive always maintained we cannot just size our bets for these sidebets off our regular BJ bankroll independently. it seems to me that if playing both regular BJ and these sidebets from the same bankroll, we need to account for the extra variance of both the sidebet and regular counting to determine how much we should bet on any given round in total between the blackjack hand and the sidebet.
now the difference might be so small that it doesnt make a difference, but im not so sure about that, particularly if you are about to put a full 1 unit bet down on a sidebet. obviously with this sidebet being discussed, the unit size might be significantly smaller than your regular counting unit (like say $1 vs $25 counting unit). but what about with lucky ladies? many people use a $25 blackjack unit and then will also put out a full $25 bet on lucky ladies when the time is right. thats a full extra unit now on the table...
anyone have thoughts on optimal sizing of sidebet amounts when playing both a box (or two or three..) and the sidebets simultaneously off the same bankroll?