# Blackjack

forkedknight
Posers and Puzzles 06 Jun '08 15:40
1. forkedknight
Defend the Universe
06 Jun '08 15:401 edit
in Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48

If you start with \$2048 and the minimum bet is \$1, what are your odds for doubling your bankroll using a strategy of:\
1) doubling your bet ( b[n] = 2 * b[n-1] )
2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Just to clarify if you aren't familiar with these strategies, you start with a \$1 bet. Each time you win, you reset your bet to \$1. Each time you lose, you follow one of the strategies above.
2. wolfgang59
Mr. Wolf
07 Jun '08 09:05
Originally posted by forkedknight
in Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48

If you start with \$2048 and the minimum bet is \$1, what are your odds for doubling your bankroll using a strategy of:\
1) doubling your bet ( b[n] = 2 * b[n-1] )
2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Jus ...[text shortened]... win, you reset your bet to \$1. Each time you lose, you follow one of the strategies above.
Option 2

When the odds are against you it is best to minimise the number of bets. The classic situation of having \$1,000 to play at roulette is to put the whole lot on red and walk out afterwards (win or lose).
3. 08 Jun '08 06:59
Originally posted by forkedknight
in Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48

If you start with \$2048 and the minimum bet is \$1, what are your odds for doubling your bankroll using a strategy of:\
1) doubling your bet ( b[n] = 2 * b[n-1] )
2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Jus ...[text shortened]... win, you reset your bet to \$1. Each time you lose, you follow one of the strategies above.
with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.
if this succceeds on the first or second attempt
a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose
which gives appx. 2050 cycles to double your money.
4. 08 Jun '08 08:50
Originally posted by preachingforjesus
with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.
if this succceeds on the first or second attempt
a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose
which gives appx. 2050 cycles to double your money.
As a math teacher I would not give full points for this answer. Why?
5. 08 Jun '08 13:50
not enough decimal places.
6. 08 Jun '08 16:11
Originally posted by doodinthemood
not enough decimal places.
ðŸ™‚
7. 08 Jun '08 17:03
Originally posted by FabianFnas
As a math teacher I would not give full points for this answer. Why?
well It might be because I didn't say:

the only way to not get the first dollar is to lose 10 times in succession which has a prbability (.52)^10 ~ 0.15%

now ~0.15% losing cycles makes for 1.0015*2048 total cycles
this gives ~2050 betting cycles with a ~99.85 success rate

you get 72.96% success in two hands soyou will average 1.x hands per cycle so it should come in at less than 4000 hands.
8. AThousandYoung
iViva la Hispanidad!
08 Jun '08 21:161 edit
Originally posted by FabianFnas
As a math teacher I would not give full points for this answer. Why?
then all you have to do is win 2048 more dollars than you lose
9. forkedknight
Defend the Universe
13 Jun '08 18:10
Originally posted by preachingforjesus
with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.
if this succceeds on the first or second attempt
a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose
which gives appx. 2050 cycles to double your money.
With method (1), you have enough money for 11 bets, as you will for the entire game.
Your chances of losing 11 in a row are:
0.52 ^ 11 = .00075168
which means your chances of winning a dollar before losing all your money is:
1 - .00075168 = 0.99924831 (I don't know where you got .9985444...)

In order to win \$2048, you will need to win exactly 2048 hands, which means you will have 2048 opportunities to lose 11 times in a row.

The oddes of _not_ losing 11 hands in a row on 2048 different cycles is:
.99924831 ^ 2048 = 0.214374
Therefore, there is a 21.4374% chance you will double your money before losing it.
10. forkedknight
Defend the Universe
13 Jun '08 18:211 edit
It's really case (2) that I'm more interested in (and comparing it with the first).

With method (2), unlike with method (1), you win more money per cycle by losing more hands. Also, you only have enough money for 10 bets. Your odds of winning x dollars in a cycle are:
x => odds
1 => .048
2 => .52 * .48 = .2304
3 => .52^2 * .48 = .129792
4 => .0674918
5 => .0350958
6 => .0182498
7 => .0094899
8 => .0049347
9 => .0025661
10 => .001334
11 => you lose

Therefore, there is not a set # of cycles you need to play (not a set # of wins).

How, then, do you find your odds of winning \$2048?