$10,000 for free?
Mr Burns has taken his mental pills and as a result offers you these crazy ideas for your Christmas Bonus.
1) $10,000 cash, no questions asked.
2) $5,000 cash and 10 spins on a single zero roulette wheel with 1 extra 18 and 1 extra 19 on the wheel, but you can only place outside bets.
3) $5,000 cash and 20 hands of six-deck blackjack (standard rules) where blackjack pays 10-1 (including splits)
4) $5,000 cash and 20 hands of three card poker, dealer exposes one card.
Which offer would you take?
Quote: WizardofEnglandOk, my attempt at a question for the Wizard,
Mr Burns has taken his mental pills and as a result offers you these crazy ideas for your Christmas Bonus.
1) $10,000 cash, no questions asked.
2) $5,000 cash and 10 spins on a single zero roulette wheel with 1 extra 18 and 1 extra 19 on the wheel, but you can only place outside bets.
3) $5,000 cash and 20 hands of six-deck blackjack (standard rules) where blackjack pays 10-1 (including splits)
4) $5,000 cash and 20 hands of three card poker, dealer exposes one card.
Which offer would you take?
Depends on the table limits and my bankroll. If I can bring more money to the table and bet big, #3 is the best bet. Blackjack pays 10-1 makes the game over 40% +EV, and if you include splits then you'd also split any pair of 10s and have a decent shot at hitting at least one blackjack in your 20 hands. If I'm stuck with the initial $5000, none of the other options have an EV of > 100% (and with no variance, to boot) so I pick #1.
Quote: MathExtremistDepends on the table limits and my bankroll. If I can bring more money to the table and bet big, #3 is the best bet. Blackjack pays 10-1 makes the game over 40% +EV, and if you include splits then you'd also split any pair of 10s and have a decent shot at hitting at least one blackjack in your 20 hands. If I'm stuck with the initial $5000, none of the other options have an EV of > 100% (and with no variance, to boot) so I pick #1.
One card poker has a player advantage of 3.48% on the 5/4/1 paytable.
Option 2 has a player advange, but I've never seen a roulette wheel with 39 numbers.
Quote: MathExtremistDepends on the table limits and my bankroll. If I can bring more money to the table and bet big, #3 is the best bet. Blackjack pays 10-1 makes the game over 40% +EV, and if you include splits then you'd also split any pair of 10s and have a decent shot at hitting at least one blackjack in your 20 hands. If I'm stuck with the initial $5000, none of the other options have an EV of > 100% (and with no variance, to boot) so I pick #1.
The limits would be $10,00 max per bet, but you can't bring your own money.
Quote: mipletOne card poker has a player advantage of 3.48% on the 5/4/1 paytable.
Option 2 has a player advange, but I've never seen a roulette wheel with 39 numbers.
It is only a theoretical wheel...... ;-)
Quote: WizardofEngland
1) $10,000 cash, no questions asked.
2) $5,000 cash and 10 spins on a single zero roulette wheel with 1 extra 18 and 1 extra 19 on the wheel, but you can only place outside bets.
3) $5,000 cash and 20 hands of six-deck blackjack (standard rules) where blackjack pays 10-1 (including splits)
4) $5,000 cash and 20 hands of three card poker, dealer exposes one card.
Which offer would you take?
First, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
Quote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
what would you make the payout on blackjack be, so it falls somewhere around the 2.5-3.5% shown on the other bets?
how does element of risk effect these bets?
Quote: WizardofEnglandQuote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
what would you make the payout on blackjack be, so it falls somewhere around the 2.5-3.5% shown on the other bets?
how does element of risk effect these bets?
if X is the number of additional units that a BJ would need to pay,and you want the game to return 3.35% (.65% HA-4% PA adjustment= 3.35% PA), then X*.0453=4%, so X=3%/.0453, so X=.883 units. So 2.38:1 would result in a 3.35% player advantage. Maybe call it 2.5:1?
Quote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
You would take option 3 over option 1? I think that if you sat down with $5000 and played 20 hands of a game with a 3.48% PA, you'd have a $9911 expected bankroll after your session.
Quote: rdw4potusQuote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
You would take option 3 over option 1? I think that if you sat down with $5000 and played 20 hands of a game with a 3.48% PA, you'd have a $9911 expected bankroll after your session.
Hmm, this is getting interesting. so option 1 is best? even with 10-1 payout?
Quote: WizardofEnglandQuote: rdw4potusQuote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
The probability of a winning blackjack is 4.53%. So 10-1 on a blackjack is worth an extra 8.5*0.0453 = 38.5%! I won't fuss with the rule about after splitting. Let's assume the normal house edge is 0.5%, so (3) is worth 38%.
(4) As I show in my page on flashing Three Card Poker dealers, the advantage seeing one card is 3.48%.
So I would go with option (3) for sure.
You would take option 3 over option 1? I think that if you sat down with $5000 and played 20 hands of a game with a 3.48% PA, you'd have a $9911 expected bankroll after your session.
Hmm, this is getting interesting. so option 1 is best? even with 10-1 payout?
Well, ok. #3 as written is by far the best option, #3 as adjusted to fall in line with the other options is slightly inferior to option 1.
*edit* Also, these nesting quotes are awesome. Thanks, JB.
I would also go with option 3, but the Roulette table has a bet that's better than Red: Both 18 and 19 are in the middle dozen. Since you'll win 28 (14*2) for each 39 bet, and lose 25 out of 39, there's a 3/39 = 7.69% advantage when betting the middle dozen.Quote: WizardFirst, it should be noted that 18 and 19 are both red, making the probability of a winning red bet 20/39.
So the advantage on (2) would be 1/39 = 2.56%.
Yeah, except if that's your attitude, you might as well cancel your plans to spend Christmas in Sin City....Quote: WizardofEnglandHmm, this is getting interesting. so option 1 is best? even with 10-1 payout?
It's also interesting to seriously consider this question in the light of EV vs. utility. Sure, by taking the $5000+ +EV bets, your overall expected value is greater than $10,000, but is it worth the (considerable) risk of winding up with much less than $10,000, despite your advantage?
People make this decision all the time, but mostly in the reverse context, for example when they buy fire insurance. It is obviously -EV to buy it, but they are risk-averse (probably, rationally so). When Mr. Burns hands you the ten grand, and that will pay for that $9000 Lasik eye surgery you've wanted for years, you'd be crazy to risk that for a relatively small additional EV.
Risk-aversion and selecting a lower than maximal EV is an ongoing dynamic on "Deal or No Deal", where just about every deal offered by the banker is -EV.
Quote: DJTeddyBearYeah, except if that's your attitude, you might as well cancel your plans to spend Christmas in Sin City....
Was only a hypothetical question.....;-)
If you lose early, you don't get a chance to bet the full number of rounds. Thus no +EV.
If you win early, you can bet the full rounds, but top bet is limited to 10k
To answer the question need do more math, or sims.
Quote: newbie49>>>It's also interesting to seriously consider this question in the light of EV vs. utility. Sure, by taking the $5000+ +EV bets, your overall expected value is greater than $10,000, but is it worth the (considerable) risk of winding up with much less than $10,000, despite your advantage?
If you lose early, you don't get a chance to bet the full number of rounds. Thus no +EV.
If you win early, you can bet the full rounds, but top bet is limited to 10k
To answer the question need do more math, or sims.
This is untrue. "Positive expectation" means that a bet is advantageous. Under the conditions you stated, all such bets would be advantageous to the player. Whether the individual bets are won or lost is irrelevant.
Quote: DJTeddyBearI would also go with option 3, but the Roulette table has a bet that's better than Red: Both 18 and 19 are in the middle dozen. Since you'll win 28 (14*2) for each 39 bet, and lose 25 out of 39, there's a 3/39 = 7.69% advantage when betting the middle dozen.
Good catch! I should have thought of that.
How hard is that?
Which option is best?
Does risk come into it?
How likely would you be to get more than $10k
How likely would it be that lose the whole $5k?
Would the kelly bet apply here?
Quote: WizardofEnglandWould the kelly bet apply here?
If you were limited by the $5000 bankroll, certainly.
hmm, I think you mis-understood me. Sure every bet you make has + EV, even if you end up losing the bet.
I think you are responding to my comment "Thus no +EV." I meant no +EV for the bets you do not make.
For example, you bet as much cash you have, starting from the initial 5000, then capped at 10,000. And suppose you lose everything after 5 rounds, then you don't make any positive EV on your 6th to 20th bet because you did not bet them.
Suppose you start with 10,000 cash and the bet is capped at 10,000. And any bet you make has 10% edge. Then for 20 rounds, you can calculate the profit as 20 x 10,000 x 10%. But it is not accurate, because in some cases you will run out of money, and you don't get to bet the full 20 rounds.
