A Quantitative Study on Average Number of Spins of Two-Player Dreidel
AA Quantitative Study on Average Number of Spinsof Two-Player Dreidel
Thotsaporn “Aek” ThanatipanondaScience Division, Mahidol University International CollegeNakornpathom, Thailand [email protected]
June 28, 2019
Abstract
We give a high precision approximation of the average number of spins ina simplified version of a two-player version of the game Dreidel. We also makea conjecture on the average number of spins of the full version of the game.
In 2006, Robinson and Vijay [1] showed that the average number of spins of theDreidel with k ≥ n nuts is O ( n ). In thispaper, we give a quantitative analysis of the average number of spins of a two-playersimplified version of this game, where players start with a and b nuts, respectively. Motivation: Gambler’s Ruin:
We will first motivate our work with a well known statistical concept, gambler’sruin, which is interpreted in terms of a game that starts with a gambler who has zerochips. For each play, s/he could gain one chip or lose one chip with equal probability p = q = 1 /
2. The game continues until the gambler gains M chips or loses N chips.We are interested in the expected number of plays until the game ends.Let G ( a ) be the expected number of plays until the game ends when the gamblerinitially has a chips. The system of recurrence relations that G satisfies is as follows: G ( M ) = G ( − N ) = 0 , a r X i v : . [ m a t h . HO ] O c t ( a ) = 1 + 12 G ( a + 1) + 12 G ( a − , − N < a < M.
The solution to this system is well known, G ( a ) = ( N + a ) · ( M − a ) , − N ≤ a ≤ M, (1)i.e. [2]. This beautiful result is a cornerstone of probability models and is the sourceof inspiration for our research problem. A Dreidel is a four-sided spinning top, and the game associated with it is playedduring the Jewish holiday of Hanukkah. Each side of the Dreidel bears a letter ofthe Hebrew alphabet: gimel, hay, nun and shin. This is an n -player pot game whereplayer i begins with a i nuts, contributing or taking away nuts from the pot as thegame progresses. At the start of the game, everyone donates one nut to the pot andtakes their turn to spin the Dreidel. The player who spins the Dreidel takes a certainaction depending on how the Dreidel lands: • Gimel: Player takes the whole pot, after which everyone donates one nut tothe pot and then the next person spins. • Hay: Player takes the smaller half of the pot and the next person spins. • Nun: Player takes and gives nothing and the next person spins. • Shin: Player gives one nut to the pot and the next person spins.The game ends when one person has all of the nuts in their possession.We consider the simplest non-trivial version where two players spin with only twooutcomes: Gimel (player takes the whole pot) and Shin (player gives one to the pot),2nd we answer the question of how long the game will last if the two players startwith a and b nuts, respectively.Let D ( a, p, b ) denote the average number of spins until the game ends where the firstplayer has a nuts, the second player has b nuts, and p nuts are in the pot. We canderive the following recurrences for the simplified game: D ( a, p, b ) = 1 + 12 [ D ( b − , , a + p −
1) + D ( b, p + 1 , a − , (2) D (0 , p, b ) = D ( a, p,
0) = 0 , for any a ≥ , p ≥ , b ≥ . Example.
Some values of D ( a, p, b ) when a + b + p = 6 are D (1 , ,
4) = 3316 , D (1 , ,
3) = 52 , D (1 , ,
2) = 94 , D (1 , ,
1) = 1 ,D (2 , ,
3) = 5716 , D (2 , ,
2) = 3 , D (2 , ,
1) = 32 ,D (3 , ,
2) = 154 , D (3 , ,
1) = 178 , D (4 , ,
1) = 94 . Experimental Math in Action:
We are interested in the average number of spins until the game ends, T ( a, b ) := D ( a, , b ) . It would be too much to ask for an exact formula for T ( a, b ) because we can seethat the recurrence (2) is quite complicated. The best we could hope for is a goodapproximation of it. The first thing that comes to mind is to compare T ( a, b ) with theresult of gambler’s ruin (i.e., see (1)). It is reasonably safe to guess that T ( a, b ) ≤ ab for a, b ≥ D ( a, p, b ), we see that the sequence∆ a := D ( a + 1 , p, b ) − D ( a, p, b )quickly converges to a constant for each p, b. A similar thing happens for∆ b := D ( a, p, b + 1) − D ( a, p, b ) . This implies that ∆ a ≈ , ∆ b ≈ . In other words, D ( a + 2 , p, b ) − D ( a + 1 , p, b ) + D ( a, p, b ) ≈ T ( a + 2 , b ) − T ( a + 1 , b ) + T ( a, b ) ≈ . As a consequence, we learn that T ( a, b ) is almost linear in both a and b . T ( a, b ) mustbe of the form T ( a, b ) = c · ab + c · a + c · b + c + (cid:15) a,b , (3)where (cid:15) a,b is some small error in which we shall see that it is exponentially small in a and b , i.e. the leading terms are 1 / a and 1 / b . This is a fairly nice observation,but how could we prove it rigorously?
Symbolic Computation in Action:
Since we are only interested in T ( a, b ) (in the situation where p = 2), we will rewritethe system of equations (2) recursively in terms of T ( a, b ) only. Example: T (1 ,
3) = 1 + 12 [ T (2 ,
2) + 0] ,T (2 ,
2) = 1 + 12 [ T (1 ,
3) + D (2 , , (cid:20) T (1 ,
3) + 1 + 12 [0 + 1] (cid:21) = 1 + 12 + 14 + 12 T (1 , ,T (3 ,
1) = 1 + 12 [ T (0 ,
4) + D (1 , , (cid:20) T (0 ,
4) + 1 + 12 [ T (1 ,
3) + 0] (cid:21) = 1 + 12 + 12 T (0 ,
4) + 14 T (1 ,
3) = 1 + 12 + 14 T (1 , . The process appears to get out of hand pretty quickly. But we can sort and checkthese relations with the actual values using the computer. As a result, we obtain T ( a, b ) = min (2 a − , b − (cid:88) i =0 (cid:18) (cid:19) i + min ( a,b ) (cid:88) i =1 T ( b − i, a + i )2 i − + min ( a,b +1) (cid:88) i =2 T ( a − i, b + i )2 i − . (Key)The first sum comes from both players landing on shin (pay one) until one of themruns out of nuts. The second/third sum comes from the first/second player landingon gimel (takes the whole pot) respectively for the first time. The Guess and Check Method:
So far so good! Equation (Key) is the key! We simply plug in the conjecturedequation (3) into (Key). We consider two different cases according to the upperlimits of the sums: case 1: a ≥ b + 1, case 2: a ≤ b. T ( a, b ) fits (Key),where the terms with 1 / a and 1 / b have been ignored. All of these cases producethe same solution where c = 12 /
19 and c = c + 2 /
19 and there are no restrictionson c and c . The Approximation of T ( a, b ) : The two free variables c and c could be approximated using the least squaresmethod. In principle, we should work on the two cases separately. However, doing itall at once gives a very nice approximation, so we choose to do it this way. We usethe least squares method over the values of T ( a, b ) where 30 ≤ a, b ≤
60 to obtain T ( a, b ) ≈ (cid:101) T ( a, b ) = 1219 ab + c a + (cid:18) c + 219 (cid:19) b + c , in which c = − . . . . and c = 2 . . . . . This approximation is astonishing, for example, | T (100 , − (cid:101) T (100 , | < − . The approximation of D ( a, p, b ) !: We don’t only have a good approximation for every starting position but also a goodapproximation for every position!! For any numbers of nuts a and b , we notice thatthe sequence h a,b ( p ) := D ( a, p + 1 , b ) − D ( a, p, b ) (4)converges to a constant. Moreover these constants are linear in a and b i.e. h a,b ( p ) = s · a + s · b + s . (5)By combining these two ideas i.e. (4) and (5), it is reasonable to set up the solutionform of D ( a, p, b ) as D ( a, p, b ) = T ( a, b ) + ( p − s · a + s · b + s ) + (cid:15) a,p,b . (GOOD GUESS)This is a nice idea! But how can we prove it? We make use of the original recurrencerelation D ( a, p, b ) = 1 + 12 D ( b − , , a + p −
1) + 12 D ( b, p + 1 , a − . Plug in (GOOD GUESS) on both sides and equate the coefficients of ab, a, b and theconstant term. The conjectured equation fits. We obtain that s = 419 , s = 819 and s = c − . The final solution is D ( a, p, b ) ≈ ab + (cid:18) c + 4 p − (cid:19) a + (cid:18) c + 8 p − (cid:19) b + c + ( p − (cid:18) c − (cid:19) , c and c were defined earlier. We stress that the approximation works betterfor large a and b . The Main Conjecture:
After successfully finding a good solution to the simplified Dreidel game, it is naturalto try to see whether the same method works for the full version (that is, where allfour outcomes can occur). Let us denote the average number of spins for the fullversion by Dr ( a, p, b ) . We notice that each of the sequences∆ a := Dr ( a + 1 , p, b ) − Dr ( a, p, b )and ∆ b := Dr ( a, p, b + 1) − Dr ( a, p, b ) . converge to constants as before. Therefore, the same technique should work oncewe figure out relations similar to (Key). For now, we only conjecture how long theDreidel game lasts and denote Q ( a, b ) := Dr ( a, , b ) . Conjecture 1 (Dreidel conjecture) . Q ( a, b ) ≈ (cid:101) Q ( a, b ) = 2 . · ab − . · a − . · b + 2 . . We obtain (cid:101) Q ( a, b ) by the least-squares method on 15 ≤ a, b ≤ . This functionagrees with Zeilberger [3], if we let a = b = nuts − . The approximation also comeswith incredible accuracy, for example | Q (35 , − (cid:101) Q (35 , | < − . For a practical perspective, we may assume that it takes 10 seconds per play, if bothplayers start with 10 nuts, an average game will last 28.10 minutes. And if bothplayers start with 15 nuts, an average game will last 69.33 minutes.
References [1] T. Robinson, S. Vijay,
Dreidel Lasts O ( N ) Spins , Advances in Applied Math-ematics 36, 85–94, 2006.[2] Sheldon M. Ross,
Introduction to Probability Models , Academic Press, sixthedition, 1997.[3] D. Zeilberger,