Ask the Wizard #135
If an online casino offered this deal would it still be profitable? Over a one-month period, you take the entire profit/loss of each player. If the player is in profit you, you give them an extra 10% of their profit and deposit it to their accounts at the end of the month. If the player has lost money, you give them 25% of their loss back and credit this back to their account at the end of the month. Would the online casino still be profitable? What is the risk? Can anyone do the math behind this?
Max from London
If the players were smart, no! Assume the player made just one bet on an even-money bet in single-zero roulette. The players expected profit would be pr(win)*(1.1) - pr(lose)*0.8 = (18/37)*1.1 - (19/37)*0.8 = 12.43%. In an online environment a high percentage of players grind through their entire deposit until they lose everything, so you might be okay. However I think the pros will smell blood and attack it like and the casino would get killed.
I have never gambled online before but wanted to give it a try-specifically at Pharoah's because of their generous video poker pay tables. I was not able to make a deposit because the only bank I have cards with forbids online gambling and every transaction I've attempted has been blocked. I really wanted to give this a try (and support some of your sponsors) but have had no luck. Before I give up completely on the idea, can you tell me if there are any ways around this problem? Also, have any recent determinations been made recently on the legality and regulation of online gambling? Thanks for your time.
Steve from Gresham, Wisconsin
Thanks for attempting to patronize my advertisers. Not many casinos accept credit card transactions, at least from U.S. players. In my opinion the most convenient way for U.S. players to get funds into an Internet casino is via Neteller. Similar to Paypal, Neteller is an online bank, but unlike Paypal, Neteller honors transactions with Internet casinos. I'm not the best person to speak on the legality topic but as far as I can tell nothing has changed but there is still no federal law that specifically says gambling on the Internet is illegal. Efforts have been made to pass such a law but the same bill has yet to pass both houses of congress.
I have tried, quite unsuccessfully, to locate on your excellent site and elsewhere how to properly calculate a confidence interval based on 1, 2, or 3 standard deviations and number of hands played with Expected Value. For example, I’d like to learn how to calculate the +/- standard deviations for playing $10/hand at blackjack for 300 hands using the .50% house edge for basic strategy. Thanks for your GREAT support and advice.
Peter from Orlando
The expected loss would be 300*$10*0.005 = $15. As I state in my blackjack appendix 4 the standard deviation is 1.17 (based on Atlantic City rules). The standard deviation on 300 hands at $10 each would be 3001/2 * $10 * 1.17 = $202.65. So here are the confidence intervals on the expected win for 1, 2, and 3 standard deviations:
1 standard deviation (68.27% probability): -$15.00 +/- $202.65 = -$217.65 to $187.65
2 standard deviations (95.45% probability): -$15.00 +/- 2*$202.65 = -$420.30 to $390.30
3 standard deviations (99.73% probability): -$15.00 +/- 3*$202.65 = -$622.95 to $592.95
I would deeply appreciate it if you could answer my question for me. I have e-mailed several poker pro’s including the Canadian one (Blount). Not one has answered my question. Most of them never even wrote back, including Blount. My question is - Could you please tell me the formula for figuring out the odds and percentages of getting the first two cards in hold’em and the percentage of that particular hand beating the other hands-assuming you know what they are-like you see on TV. I already know the formula and the easy way of figuring out you making your hand after the flop. I’m aware of the poker calculator but I would like to know the formula for my own knowledge. You hear about the pro’s knowing all the odds. I’m beginning to think it’s a bunch of bunk because not one person has answered my question. Thank you very much for taking all the time to read this.
Don from Niagara Falls, Ontario
There is no easy formula. Personally my program cycles through all the remaining cards and records how the number of hands that win for each player and takes a percentage based on those totals. I imagine everyone else either does that or is random simulation based.
(Bluejay adds: As for your doubting that pros really know the poker odds because they didn’t write back to you -- didn’t it occur to you that another likely explanation is that they didn’t care to serve as a free helpdesk to the whole world? Britney Spears must be a fraud because she never wrote back to me, either.)
Just for simplicity, let’s say there are 322 cups on a table and one has a ball under it. What are the chances that I will pick the ball if i pick a cup 75 times (and the cups don’t go away after I pick it up, it’s always a random pick with 322 cups). At first I just thought to say 75/322, but I realized that was incorrect, as 322 picks does not result in a 100% chance of getting the ball because I could pick a million times and not get the ball.
John from Miami
Your answer would be correct if you removed cups after an incorrect pick. Since you leave the cups on the table each pick as a 1/322 chance of being right, or 321/322 of being wrong. The probability of 75 picks being wrong is (321/322)75 = 79.193%. So the probability of getting at least one correct in 75 picks is 100% - 79.193% = 20.807%.
In Texas hold em, if two players are dealt a pocket pair pre-flop, what are the odds of each of these players flopping a set(three of a kind)?
Bob from Cincinnati
Let’s assume you have a pair of aces. Before considering that the other player has another pair the probability of flopping a three of a kind is the [nc(one ace)*nc(two ranks out of 12)*nc(one suit out of 4)2 + nc(any other three of a kind)]/nc(any three cards), where nc(x) = number of combinations of x. This equals [2*combin(12,2)*42+12*combin(4,3)]/combin(50,3) = (2112+48)/19600 = 11.020%. Now lets assume the other player has any other pair, but not the same as yours. Then probability becomes [2*(combin(11,2)*42 + 11*2*4 + 11*combin(4,3)]/combin(48,3) = 11.4477%.
Tequila Poker looks like fun. I know you get a lot of questions but I just had to ask: what would you do if you got dealt four aces?
Bill from Columbia, Maryland
The expected value of playing High Tequila is 115.904, while Tequila Poker is only 16. So you definitely play High Tequila.
Do my two best friends really love each other and if they do, what will happen to one of their partners?
This one is so lacking in details I can’t do much with it. What I will advise is to worry less about your friends and more about yourself. Let them work this out and if you’re not asked to participate then don’t.