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Best of Donald Catlin
I recently received an email from my old friend and colleague The Midnight Skulker. Every now and then the Skulker sends me an interesting fact, a puzzle, or an interesting question. This time it was a question and I gave the Skulker the wrong answer. He caught my error and proceeded to solve the problem correctly. I’ll give his solution below but first I have to develop some simple mathematics that will be used in the solution.
The sum of the geometric series having the common ratio r, with 0 < r <1, is well known and is given by
1/(1 – r) = 1 + r + r2 + r3 + … (1)
We will also need the sum S of the series given by
S = 1 + 2r + 3r2 + 4r3 + … (2)
Multiplying the expression in (2) through by r we obtain
rS = r + 2r2 + 3r3 + 4r4 + … (3)
Subtracting the expression in (3) from that in (2) yields
S – rS = 1 + r + r2 + r3 + … (4)
Factoring S from the left side of (4) and noting that the righthand side is the series in (1) gives us
(1 – r)S = 1/(1 – r) (5)
Solving (5) for S and then replacing S by the expression in (2) we have
1/(1 – r)2 = 1 + 2r + 3r2 + 4r3 + … (6)
(6) is the relation that we will need.
Now to the Skulker’s question. Suppose that we have two players laying bets at Craps as follows. Player A lays 40 dollars on the No 4. Player B lays 20 dollars each on the No 4 and the No 10. A cycle for player A is simply a resolution of that bet. When that bet is resolved it is replaced if it lost or left up if it won. A cycle for player B is the resolution of both bets. If they both win, which they do simultaneously, the cycle is complete and the bets are left up. If one of them loses, it is not replaced until the other bet is resolved, which then completes the cycle. This second bet is then replaced as well if it also lost or left up if it won (and only the first bet is replaced). It seems clear that A will have more cycles than B. The question posed by the Skulker is: How much more frequently does A cycle than B?
Let us define the random variable X to be the number of times A cycles for each cycle of B. We want to calculate the expected value of X, E(X). The relevant numbers are 4, T (for ten), and 7. The tables below list the relevant sequences of rolls for each value of X.
For X = 1:
1.1 - 7 (both players win)
1.2 - 4,T (A completes a losing cycle and starts a new one while B loses on 4 and then B loses on T to complete B’s cycle; A is unaffected)
1.3 - T (B loses one bet and both players will complete their cycles on a 4 or 7)
For X = 2:
2.1 - 4,7 (A loses then wins completing 2 cycles; B loses one bet and then wins completing one cycle)
2.2 – 4,4,T (A completes two losing cycles and starts a new one; B loses one bet and wins one bet completing one cycle)
For X = 3:
3.1 - 4,4,7 (A loses twice then wins completing three cycles; B loses one bet and wins one bet completing one cycle)
3.2 – 4,4,4,T (A completes 3 losing cycles then starts a new one; B completes a cycle losing both bets while A is unaffected)
For X = n:
n.1 - n – 1 4s followed by a 7
n.2 - n 4s followed by a T
The probabilities involved are ¼ for the 4 and T and ½ for the 7. Since the rolls are independent the probability that X = n, denoted p(n), is given by
p(n) = (1/4)n-11/2 + (1/4)n1/4
p(n) = (1/4)n-11/2 + (1/4)n+1 for n > 1 (7)
p(1) = 1/2 + 1/4 + (1/4)2 = 13/16 (8)
We can rewrite (7) as
p(n) = (1/4)n-1(9/16) (9)
The expected value of X is thus given by the infinite series
E(X) = 1p(1) + 2p(2) + 3p(3) + 4p(4) + … (10)
So by (8) and (9) expression (10) can be evaluated as
E(x) = 13/16 + (9/16)[ 2(1/4) + 3(1/4)2 + 4(1/4)3 … ] (11)
In order to use expression (6) with r = ¼ we add and subtract a 1 in front of the series:
E(X) = 13/16 + (9/16)[-1 + 1 + 2(1/4) + 3(1/4)2 + 4(1/4)3 … ] (12)
Hence, using (6) with r = 1/4 we have
E(X) = 13/16 + (9/16)[ -1 + 1/(1 – 1/4)2]
E(X) = 20/16 = 1.25 (13)
So A will have, on average, 25% more cycles than B. If you want to see what motivated the Skulker’s question and his solution see http://www.crapsforum.com/viewthread/2770/P30/#9729. See you next month.
Don Catlin can be reached at email@example.com
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