Ask The Wizard #33
I looked over your expected payouts for the various deuces wild pay schedules, but I did not find the particular schedule I was looking for. Could you tell me the expected payout for a deuces wild with the following schedule:
Royal flush - 840
Four deuces - 200
Wild Royal - 20
Five of a kind - 12
Straight flush - 9
Four of a kind - 5
Full house - 3
Flush - 2
Straight - 2
Three of a kind - 1
I would do this myself, but I am unable to use the necessary software, as I am not a windows user.
The return is 99.0251%.
I thought the deal in video poker was a stack of 10 cards (actually 52 cards) which is dealt from the "top". The first five cards are my original hand are taken "off" the stack and if, e.g., I draw two cards irrespective of where these two cards "appear" in my hand -- they are replaced with the next two cards from the stack. My brother-in-law says no -- he says that the five exposed cards are dealt with the next five cards "behind" them, and will replace the discarding of the "exposed" card. Thus making the card being drawn a function of the "position" of the discard. Obviously, in the second case the original order of the stack is "violated", but will this "violate" the overall random process of the deal? I don't like the sound of the second case, but can't figure out why. Is there a (bad) conditional probability being added to the mix?
It is my understanding that the remaining 47 cards are continuously being shuffled until the player makes a decision what cards to draw. So, the draw cards are not predestined at all. Mathematically speaking, it doesn't make any difference.
Some internet casinos offer multi-player Caribbean stud poker. Do you think a team of determined players with good computers could beat the game? If a team were to occupy all five places at a table, they could see into half the deck. A computer could call the optimal play based on seeing 26 cards (5 per player plus the dealer's up card). Thanks again for the gambling advice -- I'm a long-standing fan.
Somebody else asked this is a past column. The book Finding the Edge presents a paper titled 'An Analysis of Caribbean Stud Poker' by Peter Griffin and John Gwynn Jr. There they state that if seven players colluded perfectly they would enjoy a 2.3% player advantage. However, they don't state what the edge would be in a five-player game. I suspect that the odds would swing back to the house.
Can you please explain what the term "Law of mathematical averages" mean? Thank you, and keep up the good work.
I think what you are referring to is actually called the "Law of Large Numbers." This states that for a random sample of n random variables with mean x, that the sample mean xn converges to x as the size of the sample approaches infinity. We can think of the outcome of a bet as a random variable. This law tells us that as the number of bets becomes very large the average result will get closer to the house edge.
Still love your site! I always turn to your site when I'm having questions, most of the time I will find the answer but not always. When playing basic strategy blackjack I understand that I will have ups and downs and over the long run I will roughly break even, my question is what is really "over the long run"? A month, a year, five years? Any ideas?
Thanks for the kind words. You ask a good question for which there is no firm answer. It is more a matter of degree, the more you play the more your results will approach the house edge. I recently replaced my blackjack appendix 4 with some information about the standard deviation which may help. For example this table shows that if you play 10,000 hands of blackjack the probability is 90% of finishing within 192 units where you started after subtracting the expected loss due to the house edge. So in 10,000 hands you are likely to win or lose less than 2% of total money bet due to random variation. However if we go up to one million hands the probability is 90% of an 0.2% variation due to luck. In general the variation in the mean is inversely proportional to the square root of the number of hands you play. All of this assumes flat betting, otherwise the math really gets messy.
I have read everything you have to say about the Fortune side bet but I can't decide whether it's totally stupid to play it or not. I know the odds are long, but it's still a fun "what if" bet. If you have any thoughts about that I would love to hear them.
Gambling is supposed to be fun, so if you think you get 39.5 cents worth of fun every $5 you bet, then you should play it. That is how much you will give up in the house edge, assuming no other players.
InterCasino Double Bonus has the following pay table for 5 coins bet. What is the return of this game? Royal-4200
St Flush-250
4 Aces-750
4/2,3,4-450
FourKind-250
Fullhouse-40
Flush-25
Straight-20
3Kind-10
2Pair-10
JacksBetter-5
The return is 99.9367%.
Does a video poker game, whether it be jacks or better or any wild version, play like an actual deck of real cards? In other words, the payoff schedule on the front of the machine determines what exact payback this machine has or can this be tinkered with inside with computer chips making the payoff schedule meaningless? I always thought this would be dishonest until I read an article in Strictly Slots magazine that it has and can be done. If this is true you could have two identical video poker machines side by side that have different house edges like. I know casinos can and in regular slot machines. If this is so then all this VP payback percentage on payoff schedules that I have been reading for years in magazines, software and books is useless.
I am very confident that any respectable maker of video poker machines makes them fair and accurate. Is it possible there are dishonest machines or chips out there? Sure. I would be interested to read the article you refer to.
What are the odds of having two four of a kinds and a straight flush dealt to the same player in a Texas hold 'em game with ten players in 50 hands?
The probability of a four of a kind in seven cards is 0.00168067, the probability of a straight flush is 0.00027851. If x is the probability of a four of a kind and y is the probability of a straight flush then the probability you ask for is combin(50,2)*48*x2*y*(1-x-y)47. The answer comes out to .0000421845, or 1 in 23,705.