Gambling Guide
Ask the Wizard #135
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.
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.
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
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.)
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%.
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%.
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.
"Anonymous" .
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.
