# A Short Dice Puzzle II

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?

• Hey athin. I'm just going to be incredibly pedantic here, but your description doesn't include ever looking at the result. It says the die is hidden throughout. – AJFaraday Jan 28 '19 at 15:00
• Unless I'm missing something, the answer seems incredibly obvious... – Chris Sunami supports Monica Jan 28 '19 at 17:33
• This could be a lot more interesting if you had two identical unfair dice. – MooseBoys Jan 28 '19 at 17:43
• @Chris Sunami, I don't know.. but I and my friends were trying to solve them in minds, and without proper calculation we thought the answer was quite opposite than the correct one. – athin Jan 29 '19 at 0:09
• @athin: A good way to mentally solve this is to exaggerate the probabilities. Let's say we're looking at an empty food bowl and there are two possibilities: (1) the dog ate it (almost definitely true) (2) aliens stole the dog food (astronomically unlikely). Every day, you look at an empty food bowl and make an assumption. Does it intuitively seem likely to you that intentionally guessing "aliens" once in a while is going to increase your chances to be correct? If there was a 50/50 split between the options, it wouldn't matter. But it matters when probability vastly favors one over the other. – Flater Jan 29 '19 at 12:29

Strategy 1 is better than strategy 2.

Explanation:

Using strategy 1, you will on average get $$\frac{600\times5}6 = 500$$ 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 $$\frac{500\times5}6 + \frac{100\times1}6 = 433.\bar3$$.
This is obviously less than the 500 from strategy 1 so strategy 1 would be the better pick.

• @Gregor please don't post spoilers in comments. I feel like my method is exactly the same as yours with a slight different order of operations. – AHKieran Jan 28 '19 at 16:23
• Sorry for spoilers, deleted. I suppose I just miss seeing the probabilities multiplied/added to get the overall percentages each way, which seems like the general answer. – Gregor Thomas Jan 28 '19 at 16:26

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 $$\frac{100k}{6}+\frac{500(6-k)}{6}=\frac{3000-400k}{6}=500-66\frac23 k$$ times in 600.

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

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 chance\n");
printf("1. %d   %f\n", wins1, wins1 / 36.0);
printf("2. %d   %f\n", 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.