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Ask the Wizard #117
"Anonymous" .
The best piece of advice anywhere on this site may be this: The first round, ALWAYS PICK PAPER. That is because amateur players tend to pick rock the first time. Just hold out your hand in each position, one at a time, and you’ll see that rock is the most comfortable and natural choice. If you play repeated rounds you should pick whatever would beat your opponent the last round with probability less than onethird. This is because I believe amateurs repeat less than onethird of the time. If playing a pro who you fear can get into your head then randomize by looking at the second hand of your watch, divide the number of seconds by three and take the remainder, then map the remaider as follows 0=rock, 1=scissors, 2=paper (or any other mapping as long as determined in advance). So the next time you go to a restaurant Dutch style I suggest playing a single round for the check and then pick paper. You can thank me later.
Jay F.
The rule of thumb when it comes to comps is that the casinos give back some percentage, usually onethird. So if your goal is to get the room with as little expected loss as possible then whatever game offers the lowest house edge is what you should play. You will probably earn that room faster and with less bankroll volatility playing pai gow or pai gow poker. However the house edge is higher so your expected loss will be greater than in blackjack. In my opinion you should play whatever you would play if there were no comps at all. Then consider comps as icing on the cake.
"Anonymous" .
I hope not. If anyone brings such a lawsuit I hope the casino wins. As long as the casinos are operating honestly and fairly, which in general I believe they are, then if the player loses more than he can afford it is his own fault. I’m not a lawyer but nobody here in Vegas seems very worried about this.
Kevin H.
The casino had the right to do this. However in my opinion it was a bad business decision. Not only did the casino waste time resolving this mess but as you point out it resulted in bad feelings on the part of all players. This just goes to show the folly of following rules religiously. Personally I think rules should be weighed against common sense.
"Anonymous" .
The easiest counting method is what I call the "eyeball" method, in which if you see lots of small cards come out then increase your bet, and vise versa. However this is better suited to one and two deck games. For 6 or 8 deck games I would recommend the ace/five count. This requires only counting aces and fives. According to Ken Uston in Million Dollar Blackjack this give the player an additional 0.5% with just a 1 to 3 unit bet range. That is enough to overcome the house edge in most games.
"Anonymous" .
The probability of throwing seven sixes with seven dice is (1/6)^{7} = 1 in 279,936. So the car would have to have a value of £279,936 or more for this to be a good bet. Even your average Rolls Royce is not worth this much, so I would say that was a terrible bet.
[Bluejay adds: Uh, yeah, but I think the point was that it was for charity. What’s more fun: Donating £1.00 to charity and getting nothing back but the good feeling of helping out, or donating £1.00 and getting the good feeling plus the longshot chance of winning a car?]
Chris
No! Not only do betting systems not overcome the house edge but they can’t even put a dent in it. Nor can they increase the house edge. All they can do is affect volatility. Since it sounds like you like a volatile, exciting game than your system is fulfilling its purpose. Just don’t expect to win.
"Anonymous" .
It isn't often I say this but I tried for hours but the math on this one was simply over my head. So I turned to my friend and math professor Gabor Megyesi. Here is his formula for any "drought" problem.
 Let p be the probability of winning any given hand.
 Let d be the length of the drought.
 Let n be the number of hands played.
 Set k=dp and x=np.
 If k=1 then let a=1, otherwise find a such that k=ln(a)/(1+a). (a is a negative number, if k>1 then 1 < a < 0, if k < 1 then a < 1, and a needs to be calculated to high accuracy.) [Wizard’s note: This kind of solution can be easily found in Excel using the Goal Seek feature under the tools menu.]
 if k=1 then let A=2, otherwise let A=(1+a)/(1+ak).
 The probability of no drought of length d in n hands is approximately Ae^{a}^{x}.
In this particular problem p=1/40391, d=200000, n=1000000, k=4.9516, x=24.758, a=0.0073337, A=1.03007. So the probability of no drought is 1.03007*e^{0.0073337*24.758} = 0.859042. Thus the probability of at least one drought is 10.859042 = 0.140958.
Here is Gabor Megyesi's full 5page solution (PDF). Thanks Gábor for your help.
I did a random simulation of 32,095 sets of one million hands. The number with at least one drought was 4558, for a probability of 14.20%.
"Anonymous" .
It would depend on the specific skill factor of the players. Without knowing that, but assuming the skill level is equal among players, I would have the bank option rotate from player to player.
"Anonymous" .
It is true the casino busts more often if the dealer hits a soft 17. However the dealer also gets fewer seventeens, which is not a very good hand. It is to the dealer’s advantage to hit a soft 17 for the same reason the player should always hit or double on a soft 17. A 17 is a lousy hand, and whether the player or the dealer hitting a soft 17 offers two chances to improve upon it.
"Anonymous" .
 Five of a kind: 6/6^{5} = 0.08% (obvious)
 Four of a kind: 5*6*5 = 1.93% (five possible positions for the singleton * 6 ranks for the four of a kind * 5 ranks for the singleton).
 Full house: combin(5,3)*6*5/6^{5} = 3.86% (combin(5,3) positions for the three of a kind * 6 ranks for the three of a kind * 2 ranks for the pair).
 Three of a kind: COMBIN(5,3)*COMBIN(2,1)*6*COMBIN(5,2) / 6^{5} = 15.43%. (combin(5,3) positions for the three of a kind * combin(2,1) positions for the larger of the singletons * 6 ranks of the three of a kind * combin(5,2) ranks for the two singletons.
 Two pair: COMBIN(5,2)*COMBIN(3,2)*COMBIN(6,2)*4 / 6^{5} = 23.15% (combin(5,2) positions for the higher pair * combin(3,2) positions for the lower pair * combin(6,4) ranks for the two pair * 4 ranks for the singleton.
 Pair: COMBIN(5,2)*fact(3)*6*combin(5,3) / 6^{5} = 46.30% (combin(5,2) positions for the pair * fact(3) positions for the three singletons * 6 ranks for the pair * combin(5,3) ranks for the singletons.
 Straight: 2*fact(5) / 6^{5} = 3.09% (2 spans for the straight {15 or 26} * fact(5) ways to arrange the order).
 Nothing: ((COMBIN(6,5)2)*FACT(5)) / 6^{5} = 6.17% (combin(6,5) ways to choose 5 ranks out of six, less 2 for the straights, * fact(5) ways to arrange the order.
"Anonymous" .
Thanks for the compliment. I’m afraid I know of no source, including myself, that shows code for game analysis. It took me years to get my blackjack engine to work perfectly (splits when the dealer had a 10 or ace showing was very tricky). An easier way to get the house edge for blackjack is to write a random simulation. One of these days I would like to write a book on how I analyzed the games, but I’m afraid only you would buy it.