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Two games, both use un-biased 6 sided dice.

game A, Sam throws one die 4 times. He wins if he rolls at least a 6 game B, he has 24 turns, and each time he rolls two dice simultaneously. He wins if he rolls at least one "double six"

Which game is Sam most likely to win?

I have a good idea of how to work this out and i have done majority of this i just wanted to double check that i am correct.

What i have done:

game A sample space = $24$ $(1,2,3,4,5,6 * 4)$

E = all outcomes of rolling a 5

|E| = 4 $((6),(6),(6),(6))$

probability (6) = $1/6$

Game B Sample space = |S|=$36*4 = 864$

E = all outcomes with (6,6) |E| = $26$

probability (6,6) = $24/864$ = $1/36$

Is my working out correct?

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The sample space is to some degree at our disposal. However, it is convenient, if we can manage it, to have a sample space in which all outcomes are equally likely.

For Game A, a convenient space is the set of all ordered quadruples $(a,b,c,d)$, where each of the symbols ranges over the numbers $1$ to $6$. The numbers $a,b,c,d$ record, in order, the result of the first toss, the second, and so on.

The sample space has $6^4$ elements. We count the favourables, the sequences which have at least one $6$. It is easier to count the sequences that have no $6$. There are $5^4$ of them.

So the probability of at least one $6$ is $\frac{6^4-5^4}{6^4}$. This is approximately $0.517746913$. That is a good approximation of the probability of winning Game A.

Now we find the probability of at least one double $6$ in $24$ tosses. Here the sample space is the set of all sequences of length $24$, where the elements of the sequence are all ordered pairs $(x,y)$, where $x$ records the number on the gree die, and $y$ records the number on the blue die. There are $36^{24}$ such sequences.

There are $35$ equally likely ways to throw two dice and not get a double $6$. So there are $35^{24}$ "bad" sequences of length $24$ consisting of bad double-throws only. Thus the probability that in $24$ double throws there will be at least one double $6$ is $\frac{36^{24}-35^{24}}{26^{24}}$. Thus the probability of winning Game B is about $0.491403876$.

Game A is somewhat more favourable for Sam than Game B.

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  • $\begingroup$ I have a question for game B as to why the total number of possible outcomes is $26^24$ ? Shouldnt it be $36^24$ $\endgroup$ – I hate math Sep 24 '14 at 23:26
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    $\begingroup$ It depends how you count. I chose equally likely outcomes. When you toss a red die and a green die, there are $36$ equally likely outcomes. For $24$ tosses, that gives $36^{24}$. There are other ways of solving. Like probability of a double-six is $1/36$. The probability of not double-six in one toss is $35/36$. The probability this happens $24$ times in a row is $(35/36)^{24}$. The probability it doesn't happen (so we get at least one double-six) is $1-(35/36)^{24}$. Same as what I wrote in the answer. $\endgroup$ – André Nicolas Sep 25 '14 at 0:09
  • $\begingroup$ Another question i have is how would i prove the following argument? A double six in a single turn in game B is 1/6 as likely as rolling a six in one turn in game A. But there are 6 times as many turns in game B as game A. Thus the two games are equally good bets $\endgroup$ – I hate math Sep 25 '14 at 5:19
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    $\begingroup$ Well, the detailed calculations show that this intuitively appealing argument is not correct. What is correct is that the expected (mean) number of $6$'s in the first game is equal to the expected number of double-$6$ in the second game. The question you ask is very interesting. It is closely related to a question the Chevalier de Méré asked Pascal, in the early days of probability theory. Back then, there was a degree of confusion between expectation and probability. $\endgroup$ – André Nicolas Sep 25 '14 at 5:25
  • $\begingroup$ what ive done is $(36^1-35^1)/(35^1)$ = 0.028% and $(6^1-5^1)/6 = 1/6 = 0.16$ not quite sure what to do with this information $\endgroup$ – I hate math Sep 25 '14 at 6:04
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Your sample spaces are not correct. For the first one, you have six possible rolls each time, so the size of the sample space is $6^4$, but that is not needed. The chance of getting at least one six in four rolls is one minus the chance of getting no sixes. What is the chance of not getting a six on one roll? Now you need to succeed at that four times in a row. Similarly for the second problem, what is the chance of not getting double sixes on one roll (of two dice)? To not get a double six in 24 rolls, you need to succeed at this 24 times in a row.

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