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A standard $6$-sided die is put inside a cup. The cup is shaken and put upside-down, hiding the die.

You are going to guess what is the number on the top of the die.

But, instead of given $6$ options, there are only $2$ options for you:

• It is $6$, or


• It is not $6$ (i.e. it is between $1$ to $5$).

After you choose your guess, the cup is opened, and you will win if the die matches with your guess.

You are going to play this game $600$ times, and you want to win as many as possible.

There are these two possible strategies:

1. Always go for the second option (not $6$), or


2. You "imitate" the roll: you bring a new die and roll it.
   
   If it rolls $6$ then guess $6$, if not $6$ then guess not $6$.

Which strategy will you choose? Do you have any better strategy?

Answer:

Strategy 1 is better than strategy 2.

Explanation:

Using strategy 1, you will on average get $frac{600times5}6 = 500$ correct guesses.

Using strategy 2, this is in effect, the same as guessing 'not 6' 500 times and '6' 100 times. But they wouldn't always line up, so your overall average number of successes would be $frac{500times5}6 + frac{100times1}6 = 433.bar3$.

This is obviously less than the 500 from strategy 1 so strategy 1 would be the better pick.

If you choose the '6' option you are right 1 in 6 times. If you choose the 'not 6' option, you are right 5 in 6 times. If in every 6 throws you choose the '6' option k times, then you will have chosen the '6' option $100k$ times and the 'not 6' option $100(6-k)$ times over all 600 throws. This equates to being right $$frac{100k}{6}+frac{500(6-k)}{6}=frac{3000-400k}{6}=500-66frac23 k$$ times in 600.


Your option 1 sets $k=0$ and option 2 sets $k=1$, so option 1 gives the better results.

Answer

It is a fallacy that you can use one random event to predict another
random event.


A common example is when throwing a coin that keeps coming up heads,
assuming that the chance of the next flip being tails increases. It
remains 50/50 and has nothing to do with the previous throw (unless it
is a trick).


The odds are calculated in another answer, here is a little program
that examines the situation empirically. It doesn't matter how many
times the dice are thrown so to avoid the notorious
PRNG
the 36 possible combinations are examined.

 

 #include <stdio.h>

 

 int main(void)

 {

     int throw1, throw2;

     int wins1 = 0, wins2 = 0;

 

     for(throw1 = 1; throw1 <= 6; throw1++) {

         for(throw2 = 1; throw2 <= 6; throw2++) {

 

             // always choose < 6

             if(throw1 < 6)

                 wins1 += 1;

 

             // choose the same as the other throw

             if((throw2 < 6 && throw1 < 6) || (throw2 == 6 && throw1 == 6))

                 wins2 += 1;

         }

     }

     printf("   wins chancen");

     printf("1. %d   %fn", wins1, wins1 / 36.0);

     printf("2. %d   %fn", wins2, wins2 / 36.0);

     return 0;

 }

 

Program output:

 

    wins chance

 1. 30   0.833333

 2. 26   0.722222

 

So strategy (1) is better.

Is there a better strategy?

No, you cannot beat the odds.

Answer

It doesn't matter how many times you will play - 600, 7000, or 800000.


The first strategy gives you a 5/6 probability to win in each attempt.


If you choose the second one you win only if numbers on both dices match.

And it gives you a (1/6)(1/6)+(5/6)(5/6) = 26/36 probability to win which
is obviously less than 5/6 (30/36).



So the first strategy is definitely better than the second.