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Ask the Wizard #279
JyBrd0403
Let's solve for the two loss case first.
Let x be the expected number of future flips starting from the beginning or after any win.
Let y be the expected number of future flips after one loss.
We can set up the following two equations:
(1) x = 1 + .5x + .5y
The one represents that the player must flip the coin to change states. There is a 50% chance of a win, remaining in state x. There is a 50% chance of a loss, advancing to state y.
(2) y = 1 + .5x
From state y again, the 1 represents the flip at that point. There is a 50% chance of a win, reverting to state x. There is a 50% chance of a loss, ending the game, necessitating no additional flips, so it is an implied 0.5*0.
Multiply both equations by 2 and reorder to get:
(3) x  y =2
(4) x + 2y = 2
Add the two equations to get:
(5) y = 4
Plug that into any equation from (1) to (4) and get x=6.
For the three loss case, define the three possible states as:
Let x be the expected number of future flips starting from the beginning or after any win.
Let y be the expected number of future flips after one loss.
Let z be the expected number of future flips after two losses.
The initial equations are:
x = 1 + .5x + .5y
y = 1 + .5x + .5z
z = 1 + .5x
We can set up the initial states in matrix form as:
0.5  0.5  0  1 
0.5  1  0.5  1 
0.5  0  1  1 
If you remember your matrix algebra, we can solve for x as determinant(A)/determinant(B) where
A =
1  0.5  0 
1  1  0.5 
1  0  1 
B =
0.5  0.5  0 
0.5  1  0.5 
0.5  0  1 
0.5  0.5  0 
0.5  1  0.5 
0.5  0  1 
Excel has a handy determinant function: =mdeterm(range). In this case x = mdeterm(matrix A)/mdeterm(matrix B) = 1.75/0.125 = 14.
We can use recursion for additional consecutive losses. Let's consider 4. We know from above it will take 14 flips on average to get 3 losses in a row. At that point the coin will be flipped again, with a 50% chance of starting over. So:
x = 14 + 1 + x/2
x/2 = 15
x = 30
In other words, add one to the previous answer and then double.
It is not difficult to see the pattern. The expected number of flips to get n losses in a row is 2^{n+1}2.
This question was raised and discussed on my forum at Wizard of Vegas.
Jon M. from Philadelphia, PA
Thanks. I can't tell what that is a patent for, but interesting to know that the business of inventing casino games goes that far back.
Jonathan F.
Given two cards of different ranks, the probability of making a full house are 1 in 121.6. The odds of making it on the river are 1 in 207.
The odds of making such a hand on the river two out of two times is 1 in 43,006.
The odds of this happening with the same two starting cards, in rank only, are 1 in 3,564,161.
The odds of this happening with exactly 102 both times is 1 in 295,379,826.
P90
I asked a former Nevada gaming regulator and casino president about this. He said it would be treated as an "all in" kind of situation, much the same way an accidental disconnection is handled in Internet poker.
In other words, a side pot would be made of chips in the middle at the time of death. Then, any additional bets would be set aside in a separate pot. If the dead player had the highest hand, then he would win the side pot. Win or lose, any chips he had on the table after the hand would be set aside for the deceased's estate.
This question was raised and discussed on my forum at Wizard of Vegas.
soulhunt79
As long as you are not slowing down the game, particularly when there are big bettors at the table, you can usually show how you would set your tiles and ask the dealer, "Is this how you would do it?" It will also depend on how patient the dealer is and/or whether the other players seem to object. One dealer I know didn't like to be asked because she said it confused her when she had to set her own hand. With any difficult game, if you're a beginner I would recommend trying to get a table to yourself the first time, so you don't inconvenience other players with a lot of questions.
Regarding the second question, if the player is going against the traditional house way, the house way will be correct 80.2% of the time. That other 19.8% is another reason why pai gow is such a difficult game to master.
This question was raised and discussed on my forum at Wizard of Vegas.
"Anonymous" .
For those unfamiliar with the game, both the attacker and defender will roll 1 to 8 dice, according to how many armies they each have at that point in a battle. The higher total shall win. A tie goes to the defender. If the attacker loses, he will still retain one army in the territory where he initiated the attack. For this reason, he must have at least two armies to attack, so if he wins one can inhabit the conquered territory and one can stay behind.
The following table shows the probability of an attacker victory according to all 64 combinations of total dice.
Probability of Attacker Win
Attacker  Defender  

1 Army  2 Armies  3 Armies  4 Armies  5 Armies  6 Armies  7 Armies  8 Armies  
2  0.837963  0.443673  0.152006  0.035880  0.006105  0.000766  0.000071  0.000005 
3  0.972994  0.778549  0.453575  0.191701  0.060713  0.014879  0.002890  0.000452 
4  0.997299  0.939236  0.742831  0.459528  0.220442  0.083423  0.025450  0.006379 
5  0.999850  0.987940  0.909347  0.718078  0.463654  0.242449  0.103626  0.036742 
6  0.999996  0.998217  0.975300  0.883953  0.699616  0.466731  0.259984  0.121507 
7  1.000000  0.999801  0.994663  0.961536  0.862377  0.685165  0.469139  0.274376 
8  1.000000  0.999983  0.999069  0.989534  0.947731  0.843874  0.673456  0.471091 
The next table shows the expected gain by the attacker, defined as pr(attacker wins)*(defender dice)+pr(defender wins)*(attacker dice 1). It shows the greatest expected gain is to attack with 8 against an opponent with 5.
Net Gain of Attacker Win
Attacker  Defender  

1 Army  2 Armies  3 Armies  4 Armies  5 Armies  6 Armies  7 Armies  8 Armies  
2  0.675926  0.331019  0.391976  0.820600  0.963370  0.994638  0.999432  0.999955 
3  0.918982  1.114196  0.267875  0.849794  1.575009  1.880968  1.973990  1.995480 
4  0.989196  1.696180  1.456986  0.216696  1.236464  2.249193  2.745500  2.929831 
5  0.999250  1.927640  2.365429  1.744624  0.172886  1.575510  2.860114  3.559096 
6  0.999976  1.987519  2.802400  2.955577  1.996160  0.134041  1.880192  3.420409 
7  1.000000  1.998408  2.951967  3.615360  3.486147  2.221980  0.098807  2.158736 
8  1.000000  1.999847  2.990690  3.884874  4.372772  3.970362  2.428384  0.066365 
Michael M. from Las Vegas
Yes! I loved it. I think Wally was playing tiles. Here are my reasons:
 Wally looks like the kind of nonAsian player one would find at a tile table.
 Dilbert is the scientific type, who normally are sticklers for using correct terminology. Calling pai gow poker "pai gow" is incorrect and lazy. I know most people do so anyway, but I have higher expectations for Dilbert.
 In the second frame, Dilbert says that pai gow is "a difficult game to learn after a few adult beverages." Note that he said "learn," not "play." Pai gow poker is not that hard to learn. If you understand how to play poker, then pai gow poker can be easily explained in under a minute. Meanwhile, tiles is difficult to learn as well as play.
 The cartoon came out on Chinese New Year. This may have been an inside joke.
In the unlikely event Scott Adams should read this, I would welcome a definitive answer.
This question was discussed on my forum at Wizard of Vegas.