Of Hand Cricket and Game Theory

The true genius of a man can be recognized by the dedication with which he analyses the trivial.

If you are an Indian student with an active interest in Cricket and a passive desire to actually play the sport, it’s highly likely that you, at least at one point in time, have entertained yourself with a short game of hand-cricket.

Hand cricket is a two-player, simultaneous action hand game, wherein the two players, one the batsman and other the bowler simultaneously make hand gestures that represent a score from 1-6. If the action played by both the players is the same, the batsman is deemed out, otherwise the score of the batsman is added to his total score.

Now, the beauty of the game lies in the non symmetricity  of its pay-offs.
It’s this fact that makes it different from other similar games, and makes it of particular interest to this post. 

This post would limit itself with the strategies that the batsman can apply, assuming the practice of a random-uniform strategy by the bowler.
Let me define a random-uniform strategy as one wherein the bowler adopts each play(hand gesture) randomly, and with equal probability.

Also, for the purposes of this post, a play is defined as the strategy adopted ( score played ) by the batsman for a single event.

The payoff of the event for the batsman are as follows:

Bowler’s Play –>
Batsman’s Play	Payoff	1	2	3	4	5	6
	1	Out	1	1	1	1	1
	2	2	Out	2	2	2	2
	3	3	3	Out	3	3	3
	4	4	4	4	Out	4	4
	5	5	5	5	5	Out	5
	6	6	6	6	6	6	Out

Thus, unlike the beloved rock-paper-scissor, the game does not exhibit symmetry in its payoffs. The batsman has more of a incentive to play a higher score than a lower one.

Now, given this framework, how would the batsman adopt his strategies to ensure an optimum score, and what score is a good score?

If both the batsman and the bowler were to base their hand score selection purely on randomness, i.e to say both choose their play uniformly (random-uniform) , we get the mean (expected) score of the batsman to be as 17.5

To consider further, how the scores would be distributed, I simulated a considerable large number of games, and here’s what came up:

The summary statistics are as follows:

Mean	Median	Mode	Variance
17.5	11	0	383.4

The mean is 17.5, which is the same as the theoretical mean for such distribution. The mode is 0, with the theoretical probability of 16.67%. Also, the positively skewed nature of the score is readily apparent.

Here’s a density plot of the simulated scores:

and the score quantiles are as follows:

Quantile	Score
Minimum	0
10%	0
20%	2
30%	5
40%	8
50%	11
60%	16
70%	21
80%	29
90%	43
Maximum	263

Loosely interpreted, if you manage a score of 43 batting first, the chances of you losing the match are only 10%.

A Note:

It is worth noting that the expected score of the batsman can be increased by only selection of higher score plays. For example, if a batsman only mixes his play between the scores 5 and 6, he can obtain an expected score of 27.4.
However, it is futile to discuss such strategies, since their adoption would trigger the adoption of a similar imitation strategy by the bowler, thus leading to a drop in the expected score below the uniform-random expected score of 17.5. (in the case above, imitation by the bowler lowers the expected score to a mere 5.5)

Still, something’s amiss.

If something can be stated about humans, it’s the fact that we are risk averse. We strive more to minimise our expected losses than to maximise our gains.

Let’s consider game theory now. There is no pure strategy equilibrium for the game.

Nor is there a strongly or weakly dominant strategy for the batsman.

But the batsman can follow a mixed strategy such that each play would give him the same expected score given the random-uniform strategy of the bowler, and make him indifferent to the play selected by the bowler.

This mixed strategy choice selection can be computed ass follows:

Play	Probability
1	40.82%
2	20.41%
3	13.61%
4	10.20%
5	8.16%
6	6.80%

The summary statistics for the simulated scores using the equilibrium strategy are:

Mean	Median	Mode	Variance
12.5	8	0	192.8

Thus equilibrium, and a lower variance can be obtained, though at a cost of lower expected scores.

Here’s a density plot of such simulated scores to sign off with: