Ask the Wizard #167
There is a drawing for a $27,000 car, with tickets being sold at six for $500.00, or one for $100.00. 68 tickets have been sold, and tomorrow is the deadline for purchase. I know that for a 50% probability of winning, I must spend $5666.44, and for a 66.66% of winning, I must spend $11,332.88 (Right?). How much should I spend (or how many tickets must I buy) to virtually ensure that I "win" the car? (90%? 95%?) Is this raffle worth playing, or must I spend the cost of the car?
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
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...
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
[((t2 +69t+68)/((t+69)*(t+68))] − [(t2+69t)/((t+68)*(t+69))] = 0.003086
68/((t+68)*(t+69)) = 0.003086
((t+68)*(t+69)) = 220.32
t2+137t+4692 = 22032
t2 +137t - 17340=0
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.
Great site! Your blackjack strategy card is the best one I’ve seen. Is it available any where? Theone I’m currently using does’t have surrender on it, what am I loosing by using this?
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.
Fantastic site — A gambling bible no less. Which would be the better rule for the player in an otherwise normal blackjack game with six decks. The first rule is the insertion of two jokers in the shoe. If the player gets a joker in his first two cards it is an automatic winner, paying even money. If a joker comes out at any other time, including to the dealer, it is burned. The second rule is a Five-Card Charlie.
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.
I’ve been playing 100-hand bonus deuces wild video poker at Bodog. When I have no 2s in my hand but an inside straight possibility you say to hold the inside straight chance. Typically I’ll end up with maybe 25 coins back. When playing hands with garbage in them I’ve thrown them away and often I’ve ended up with at least that much if not better. So I was wondering, should play be modified a bit when playing 100 hands at once? Or is it actually something with the pay table that causes that one play not to be quite right?
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.
Just thought you might find this interesting. At Bodog they offer the following bet, “Will Britney Spears and Kevin Federline's second baby be a Boy or Girl?” The odds on a boy are +105, and for a girl −145. Last I checked this has been 1:1 since the beginning of mankind. I'd like to know who is taking the -145 side of this one. Love the site, visit often, and click your sponsors in appreciation.
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.
There is an interesting way that one can play keno, though not in the way the state intended. Bet that at least 11 of the 20 numbers will appear in 3 rows; horizontal, vertical or a combination. Stress that there are 18 rows. Many times the sucker will play. A variant of this bet is that one row will be blank. I hope that you can use this. You have an excellent, informative site. Note that one needs a bankroll though not a large one. 10 to 15x the largest bet that you are going to cover is enough.
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
of a Win
|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%|