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Multiline Mathematics6 April 2003
One phenomenon to hit casino games during the past ten years or so is that of the multiline. One of the first was Three Hand Blackjack and then came Multiline Video Poker. For all I know, there are some multiline games out there that I have yet to see. It is interesting to see what effect these types of games have on both the hit frequency and the house advantage and that is what I will address in this article.
First of all suppose that we start with a simple game in which there are two player payoffs: x1 with probability p 1 and x2 with probability p2. It follows that the hit frequency p for this game is given by
p = p1 + p2 (1)
and the expected return to the player is
exp1 = p1x1+ p2 x2 (2)
The house edge here is, of course, 1 - exp1.
Now suppose that we offer this game to the player as two independent games with the same probabilities and payoffs and costing two units to play. What is the hit frequency in this case? Would it be twice p? Well, that sounds plausible but in fact is wrong. Let's look at the situation and see if we can determine the right answer. To help us out let's define q = 1 - p; q is the probability of not wining anything in the first game.
The player of this "two line" game can either win line 1 and lose line 2 with probability pq or lose line 1 and win line 2 with probability qp or win both lines with probability p2 . In other words the hit frequency is
Hit Frequency = 2pq + p2 (3)
Replacing q with 1 - p and cleaning up the algebra gives us
Hit Frequency = 2p - p2 (4)
So the hit frequency has increased but clearly this number is a bit less than 2p.
Had we created a 3 line game the hit frequency would be 3 pq2 + 3qp2 + p3 which reduces to 3p - 2p2 - p2q , a number a bit smaller than 3p.
In light of these results could it be that expected return for multiline games are not multiples of the single line? Well, I could quote a theorem about expected value to settle this one quickly (and I'll do this below) but I think it is kind of fun to see the algebra work out. For two lines the player could win x1 on line 1 and lose on line 2 with probability p1q, win x2 on line 1 and lose on line 2 with probability p2 q, win x1 on line 2 and lose on line 1 with probability qp 1, win x2 on line 2 and lose on line 1 with probability qp 2, win x1 on line 1 and x2 on line 2 with probability p1p2, win x 1 on line 2 and x2 on line 1 with probability p 1p2, win x1 on both lines with probability p12, or win x2 on both lines with probability p22. Whew! The expected return is therefore
Expected Return = 2p1qx1 + 2p2 qx2 + 2p1p2 (x1 + x2) + 2p12x1 + 2p22x2 (5)
Collecting the x1 and x2 terms together we obtain
Expected Return = (2p 1q + 2p1p2 + 2p 12)x1 + (2p2q + 2 p1p2 + 2p22)x 2 (6)
Factoring 2p1 out of the first term and 2p2 out of the second term we have
Expected Return = 2p 1(q + p2 + p1)x 1 + 2p2(q + p1+ p2 )x2 (7)
Since p = p1+ p2 and q = 1 - p it follows that q + p1+ p2 = 1 and so (7) is just
Expected Return = 2p1x1 + 2p 2x2 = 2Exp1 (8)
Since the 2 line game costs 2 units to play, the expected return per unit is exactly the same in the two line game as in the one line game.
This last result is really no surprise. In fact the calculation isn't really necessary. If you think of two players, one playing line 1 and the other playing line 2, each for one unit, either player is completely unconcerned with the results of the other player so the expected return for each player is just Exp1. The total expected return for the two of them is just twice the expected return for one of them. This argument relies on the independence of the two lines but, as we'll observe below, the adding of expected values does not require independence. In fact, if f1 and f2 represent payoff functions for two different games it is a general theorem that the expected return for f1 + f2 is exactly the same as the sum of the expected returns for each of them, whether or not they are independent games .
What if we have an n-line game? Well, from our last observation it is clear that without doing any computations as we did above, the expected return of the game is just n times the expected return for one line. The hit frequency? The best way to calculate this is not as we did earlier but rather to do the following. If q is the probability of losing any one line, then qn is the probability of losing all n lines providing, as above, that the lines are independent from each other. In this case the hit frequency is just 1 - qn.
Now consider a simple game played with a two-by-two matrix as
Each cell contains the integers 1 through 4, chosen randomly, with equal probability of 1/4. There are six lines here: two horizontal, two vertical, and two diagonal. A win occurs on any line if the same integer occurs twice. Notice that the lines are not all independent. A player wagering on the line consisting of Cell 2 and Cell 3 would certainly be pleased if the top line and the left vertical line contained winners because that would mean that his line was also a winner.
On any particular line, there are 4 ways to win out of 4 x 4 or 16 possibilities, so the probability of a win on any one line is 1/4. If we paid out 3 units on such a win (a win of 2 units since it costs 1 unit to play), the house edge would be 1 - 3/4 or 25%. From our discussion above we would expect the 6-line game to have an expected return of -6/4. On a per unit basis this would represent a 25% edge. We'll check this below.
The hit frequency is quite another matter. Since the lines are not
independent we cannot use the trick discussed above. Rather, we have to
look at each separate configuration. There are 44 or
256 possible ways to fill in the 4 cells, each equally likely. If only
one line is a winner then the other two cells can be filled in in 6
ways. Since there are four ways to win on a give line this means that
each line is a single winner 24 ways. Since there are 6 lines there are
6 x 24 or 144 single line winners. Double lines can occur as two
horizontals, two verticals, or two diagonals, each occurring 12 ways.
Three-line winners can occur, for example, as having the same integer in
cells 1, 2, and 3 and a different integer in cell 4. There are 4
choices for the integer in cells 1, 2, and 3 and then 3 choices for the
integer in cell 4, 12 ways altogether. There are four such
configurations for 3-line winners so altogether there are 48 of these.
Clearly there are four 6-line winners. In order to lose all lines you
have to have a different integer in each cell so there are 24 losers.
This means that the hit frequency is 2332/256 or 90.625%. Note that
this is not 1 - (3/4)6, which is approximately 82.2%.
In the table below I tabulate the numbers computed above and compute
the game's return
The number 1152 represents the total units paid back to the player in 256 plays if everything happens as probability theory dictates. The cost to the player will be 256 x 6 units or 1536. The amount lost is 1536 - 1152 or 384. The loss per unit is 384/1536, which is exactly 25%, the same as the expected return on one line.
I know of no way to calculate hit frequency in these dependent situations using the single-line hit frequency. In fact, suppose we play the above game using only three integers 1, 2, and 3. The hit frequency on one line is 1/3 yet the hit frequency for the 2 x 2 matrix game is 100%. Weird! See you next month.
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