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I'm trying to assess whether the number of doubles (dice) being rolled by my opponents on a backgammon website are fair; to the naked eye they most definitely aren't but I need some science to back up my claims.

Using the CHISQ.TEST function in Excel 2010, I'm getting a value returned of 0.038. What does this actually mean?

For background info, the source data is

Total dice rolls = 3962 1-1 = 114 2-2 = 116 3-3 = 107 4-4 = 125 5-5 = 134 6-6 = 131

Many thanks for any pointers!

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  • $\begingroup$ Your six dice rolls does not sum to 3962 (I get 727). Can you clarify? $\endgroup$ Mar 21, 2021 at 10:55
  • $\begingroup$ those dicerolls are his non-doubles. only the doubles are specified. $\endgroup$
    – Sirius
    Mar 21, 2021 at 10:57
  • $\begingroup$ As per Sirius, total dice rolls = 3962, the doubles only are specified $\endgroup$ Mar 21, 2021 at 11:08

2 Answers 2

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This seems like a good case for the binomial test

You'd expect $1/6$ of cases to be a double when you roll two dice.

I don't have excel in front of me, but judging from google searches it seems to have the binom.dist function available for this.

You are observing 727 doubles out of the 3962 roll total.

The probability of this many should then be 1 minus this:

=BINOM.DIST( 727, 3962, 1/6, TRUE ) 

If excel is worth its salt, this should give: 0.9976782. Your p-value then is 1 - this (for a one-sided test), which is less than 0.05 and you shold reconsider ever playing with your friend again!

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  • $\begingroup$ What's the significance of the 0.05 figure? $\endgroup$ Mar 21, 2021 at 11:42
  • $\begingroup$ A p-value of 0.05 is deemed suffiient to reject your null hypothesis (which is that 1/6th of throws should be doubles in this case). It's an established norm, the history of which goes beyond the scope of this question. $\endgroup$
    – Sirius
    Mar 21, 2021 at 11:53
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I don't know about excel, and what that returned value of 0.038 is. At least, excel documentation ought to tell you if it is meant as the observed chi-squared value, or a p-value.

But using R I get:

N <- 3962
> doubles <- c(114, 116, 107, 125, 134, 131)
> str(chisq.test)
function (x, y = NULL, correct = TRUE, p = rep(1/length(x), length(x)), 
    rescale.p = FALSE, simulate.p.value = FALSE, B = 2000)  
> chisq.test(c(N-sum(doubles), doubles), p=c(5/6, rep(1/36, 6)))

    Chi-squared test for given probabilities

data:  c(N - sum(doubles), doubles)
X-squared = 13.118, df = 6, p-value = 0.0412

so some evidence of non-uniformity. It is usefull to look at the individual cell values (contributions to the chi-squared statistic):

 obs <- c(N-sum(doubles), doubles)
 names(obs) <- c("offdiag", "1-1", "2-2", "3-3", "4-4", "5-5", "6-6")
 p <- c(5/6, rep(1/36, 6))
 E <- N*p
 (obs-E)^2 / E
  offdiag       1-1       2-2       3-3       4-4       5-5       6-6 
1.3461215 0.1413708 0.3210780 0.0848337 2.0293062 5.2095182 3.9858938 

so the largest contributions to chi-squared comes from 5-5 and 6-6. The test in the answer by @Sirius only considers doubles as a group, not individually, in difference from this chi-squared test that considers the doubles individually.

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