Gambling Guide
Ask the Wizard #148
Paul from Portland
The following table shows the probability in a single-player game. Remember, the dealer will not play out his hand if the player busts first. If you add players the dealer’s probability of busting will go up because there will be a greater probability of at least one player not busting.
Bust Probabilities
| Decks | Soft 17 | Player | Dealer |
|---|---|---|---|
| 6 | Stand | 15.72% | 24.07% |
| 6 | Hit | 15.68% | 24.40% |
Ashlee from Bemidji
The joke is that you can tell a man is cheating if he is happy. Seriously, look for any changes in his behavior. If he spends less time with you, looks better, goes to different places, or does different things then he probably cheating.
Scott from Toronto
My blackjack appendix 7 can be used to approximate the answer to this question. Adding the ace would favor the player by 0.005816/24 = 0.00024. Removing a two would favor the player by 0.003875/24 = 0.000161. So the total effect would be 0.000404, or a decrease in the house edge of 0.04%.
Andreas from Edmonton
The probability that intelligent life exists anywhere in the galaxy I believe is very high. The Drake Equation seeks to estimate the number of incidents of intelligent life in the galaxy, which depending on the numbers you put into comes up with a figure of about a million. However there is also no good evidence that these civilizations have ever visited us or made contact. So the famous Fermi Question is "Where is everybody?" I do think the lack of evidence of other intelligent life casts some doubt on the Drake Equation but I would still put the number of intelligent civilizations in our galaxy in the ballpark of 1000. That is just our galaxy, there are billions of galaxies out there. However the distance between galaxies is so vast there is really not much point in discussing travel or communication between them. So to answer your question I would say roughly 99.9%.
Ken from Tenian Island
No. If splitting is the right play you should do it every time, and if not never. So this rule is moot if you play properly.
Dan from San Diego
If we assume all hands are played to the end the probability of any given player having a royal flush is 4*combin(47,2)/combin(52,7) = 1 in 30,940. To make things simple lets assume each hand is independent. The probability of at least one player in 10 having a royal flush would be 1-(1-(1/30940))10 = 0.00032, or 1 in 3094.
Jennifer from Hillsboro, Oregon
Tell that bum to hit the road.
