2

I've recently came across very interesting question and I'm looking for the simplest solution. It is also very interesting to see calculations for the expected amounts for both cases.

Alice wants to join her school's Probability Student Club. Membership dues are computed via one of two simple probabilistic games.

The first game: roll a die repeatedly. Stop rolling once you get a five followed by a six. Your number of rolls is the amount you pay, in dollars.

The second game: same, except that the stopping condition is a five followed by a five.

Which of the two games should Alice elect to play? Does it even matter?

Eugene D. Gubenkov
  • 123
  • 4
  • 1
    See https://stats.stackexchange.com/questions/12174 for what is essentially the same question but in a slightly simpler context. – whuber Sep 23 '19 at 18:31

3 Answers3

1

Alice should play the first game

Both games are identical until Alice rolls a 5. At that point, on the next roll, she has a 1/6 chance of rolling a 5, and a 1/6 chance of rolling a 6. So her chances of winning at this point are identical between both games, but what makes this interesting is what happens if she does not win immediately after rolling her first 5.

In the second game, suppose Alice rolls a 5 and then a non-5. She's now back to square one, having to roll a 5 again before looking for her second 5. She will require a minimum of two more rolls to win. In contrast, in the first game, it's possible for Alice to roll a 5 followed by a non-6, which could be a 5. In the second game, failure to win doesn't necessarily revert the player to square one - there's a possibility she could still be just one roll away from winning.

After rolling the first 5, a player of the first game has a 1/6 chance of winning, a 1/6 chance of trying again, and a 4/6 chance of starting over. A player of the second game has a 1/6 chance of winning and a 5/6 chance of having to start over. The first game will take fewer rolls to win, on average.

Nuclear Hoagie
  • 5,553
  • 16
  • 24
  • Did you mix up the two games in your description at the end? – whuber Sep 23 '19 at 18:32
  • 1
    @whuber I don't think so - I'm somewhat confusingly using the names "first game" and "second game" based on the order in the question. The first game wins on 5+6, the second wins on 5+5 - the first game (rolling a 5 and then a 6) will take fewer rolls on average. – Nuclear Hoagie Sep 23 '19 at 18:36
  • Thank you, can you also elaborate on exact tosses expectation calculation for both cases? – Eugene D. Gubenkov Sep 24 '19 at 04:31
1

Here is a result using simulations in R, summary shows some statistics for number of throws.

First case:

res1=replicate(1e4,{
  tmp=sample(1:6,500,replace=T)
  for(i in 2:500){
    if(tmp[i-1]==5 & tmp[i]==6){
      return(i)
    }
  }
})

and the results:

summary(res1)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   12.00   25.00   36.01   49.00  476.00

Second case:

res2=replicate(1e4,{
  tmp=sample(1:6,500,replace=T)
  for(i in 2:500){
    if(tmp[i-1]==5 & tmp[i]==5){
      return(i)
    }
  }
})

and the results:

summary(res2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   2.00   13.00   30.00   41.97   57.00  350.00

And so, the expected outcome for the first case is 36, for the second it is 42.

user2974951
  • 5,700
  • 2
  • 14
  • 27
  • Can you elaborate? Does it mean that games are equivalent in terms of Expectation? It should not be the case, as far as I'm aware first game has expectation of 36 rolls, and the second has 42 rolls. – Eugene D. Gubenkov Sep 24 '19 at 08:07
  • @EugeneD.Gubenkov My mistake, I misunderstood the question. I didn't see that the 6 and 5 have to be consecutive. I will fix shortly. – user2974951 Sep 24 '19 at 08:12
-1

Each roll is an independent event, it will not matter Pr(x=6|x-1=5) = Pr(x=6) and Pr(X=5|x-1=5) = Pr(X=5) , now assuming its a fair dice, Pr(X=5) = Pr(X=6) = 1/6 the same. So in terms of expected tries ( which follows negative binomial, number of tries to get prob success two independent event 1/6*1/6 = 1/36) will be the same.

Yash
  • 89
  • 1
  • 7