# Ask the Wizard #167

Annette from Boise

You are right regarding the 1/2 and 2/3 probabilities. If you buy t tickets your probability of winning is t/(68+t). So for a 90% probability, solve for t as follows.

0.9 = t/(68+t)

0.9*(68+t) = t

61.2 = 0.1t

t = 612, or $51,000

For 95%...

0.95= t/(68+t)

0.95(68+t) = t

64.6 = 0.05t

t = 1292, or $107,666.67

Assuming the car is worth $27,000 to you, you should quit buying tickets as soon as the next ticket sold does not increase your probability of winning enough to warrant the price.

For a ticket to be worth the price it should increase your probability of winning by p, where...

27000*p=(500/6)

p=0.003086

Let's say t is the number of tickets your purchased where you are indifferent to buying one more ticket.

[(t+1)/(t+68+1)] − [t/(t+68)] = 0.003086

[(t+1)/(t+69)] − [t/(t+68)] = 0.003086

[((t+1)*(t+68))/((t+69)*(t+68))] − [(t*(t+69))/((t+68)*(t+69))] = 0.003086

[((t^{2} +69t+68)/((t+69)*(t+68))] − [(t^{2}+69t)/((t+68)*(t+69))] = 0.003086

68/((t+68)*(t+69)) = 0.003086

((t+68)*(t+69)) = 220.32

t^{2}+137t+4692 = 22032

t^{2} +137t - 17340=0

t=(-137+/-(137^{2}-4*1*-17340)^{2})/2

t = 79.9326

Let's test this by plugging in some values for tickets purchased, assuming the player can always buy tickets at $500/6 = $83.33 each.

At 79 tickets your cost is 79*(500/6) = $6,583.33, your probability of winning is 79/(79+68) = 53.74%, your expected return is $27,000*0.5374 = $14,510.20, and your expected profit is $14,510.20 - $6,583.33 = $7,926.87.

At 80 tickets your cost is 80*(500/6) = $6,666.67, your probability of winning is 80/(80+68) = 54.04%, your expected return is $27,000*0.5405 = $14,594.59, and your expected profit is $14,594.59 - $6,666.67 = $7,927.92

At 81 tickets your cost is 81*(500/6) = $6,750.00, your probability of winning is 81/(81+68) = 54.36%, your expected return is $27,000*0.5436 = $14,677.85, and your expected profit is $14,594.59 - $6,750.00 = $7,927.85.

So we can see that the maximum expected win peaks at 80 tickets.

John C from Crestwood

Thanks. Almost every casino gift shop sells basic strategy cards, but for some reason they don’t indicate when to surrender. There are not many situations to surrender, but those situations happen often, so I think they are worth memorizing. In a six-deck game surrender is worth 0.07% if the dealer stands on soft 17, and 0.09% if he hits it.

Gordon G. from Paramaribo

Thanks. The probability of getting at least one joker in two cards dealt from a 314-card shoe is 1-(combin(312,2)/combin(314,2)) = 1.27%. So the probability is 1.27% of turning an average hand into an automatic winner. If we assume an average hand has an expected value of −0.005 then the value of the first rule is 0.0127*(1-(-0.005)) = 1.28%. You can see from my blackjack section that the five-card Charlie rule is worth 1.46%. Assuming the cards are shuffled after every hand, or you are forced to flat bet, then given the choice as a player I would pick the five-card Charlie rule. However, the joker rule would be very easy to exploit further. The greater the ratio of jokers to cards in the shoe the more you should bet. With at least 50% deck penetration, this should easily be enough to make it the better rule.

Steven from Cary

I see Bodog follows the 1-2-3-4-4-9-15-25-200-800 pay table, which is known as “Ugly Ducks” and returns 99.42% with optimal strategy. Although I don’t indicate an Ugly Ducks strategy on my site, my Not so Ugly Ducks strategy should do quite nicely, and is indeed accurate for this play. To answer your question, the strategy depends only on the pay table. The number of hands makes no difference. What is right for one hand is right for 100 hands. On average you should get about 34 coins back playing the inside straight and 32 coins back tossing everything. However, actual results will vary. I would say you’ve just been unlucky with the inside straights if you have only been getting back 25 on average.

"Anonymous" .

Thanks for the kind words. To be honest nobody cares much about click-through rates any longer. So don't feel obligated to click through the banners if it is just for show. To answer your question, in the United States the probabilities are very close to 50.5% boy and 49.5% girl. Assuming no other information is known by the betting community the player advantage on the boy bet would be .505*1.05 - .495 = 3.53%. It could be that somebody with inside knowledge is betting on a girl. Another theory is that some people incorrectly believe you can tell the gender by the shape of the mother’s belly, and these people are betting on a girl. Personally I’m going to leave this one alone.

Richard M. from Silver Spring, MD

I hope you’re happy, I spent all day on this. After writing and running a simulation I find that the probability that any 3 lines will contain 11 or more marks is 86.96%! That isn’t even giving the other side a fighting chance. You can go up to 12 marks and still have a probability of 53.68% of winning, or an advantage of 7.36%. However, I think you have the wrong side of the empty row bet. The probability of at least one empty row is only 33.39%, better to take the other side of no empty rows. While I was at this I did lots of other probabilities and put them in a new page of keno props. Here is a list from that page of these and other good even-money bets. The good side is listed.

### Even Money Keno Props

Prop | Probability of a Win |
House Edge |
---|---|---|

No row will have 5 or more hits | 53.47% | 6.94% |

Greatest number of hits in a column will be exactly 4 | 55.2% | 10.4% |

Every row to have at least one mark | 66.61% | 33.23% |

Number of empty columns will not be 1 | 54.08% | 8.15% |

Top/bottom to have 9 to 11 marks | 56.09% | 12.17% |

3 lines (rows and/or columns) will contain 12 or more marks | 53.68% | 7.36% |