[Math] Should I throw the dice again if I have rolled 4

probability

My math skills are very basic so it might be a stupid question, I had a discussion with my brother in law and now we have a 'math problem'.
We were playing a game with dices and he threw 4. The challenge was to throw the highest number, you can stop or throw again once, you cant see what the opponent has thrown, you both reveal after finishing. He said if you have 4 you have a 50% change the next time to throw the same or higher.

(4,5,6) vs (1,2,3)

and should throw again.

But I said that I won't throw again on 4 because you already threw 4 and you are not likely to throw 4, 2 times in a row and therefore I would stop at 4. Am I right or is he and do you have a 50 % chance on throwing 4 or higher again?

Game rules

You throw a normal dice 1/6
You can choose to throw again or keep the current value
Your opponent cant see your value, you cant see his.
The one with the highest value wins.

Short version:

If I threw 4, how big is the chance I throw 4 or more the next time I throw the dice, and should i take that chance?.

Thoughts

Average of 1 dice is 3.5, if i throw 4 im above Average and am more likely to throw below average next time.

Best Answer

The chance of rolling a four or higher on your next roll is independent of your original roll. The fact you just rolled a four doesn't make it any less likely to roll one again. So your chance of rolling a four or higher is indeed 1/2, since you have three ways of rolling a four or higher and six total outcomes. 3/6=1/2.

So to answer your full question, if you roll again, you have a 1/2 chance of doing the same or better. However your chance of doing the same or worse is the number of ways to roll a four or lower (4), divided by the total outcomes (6). 4/6=2/3 is about 67%, so it would not be better to roll again. You are right to not roll again because you odds of rolling higher or the same are worse than your odds of rolling lower or the same.

Related Solutions

Probability – Most Unfair Set of Three Nontransitive Dice

It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.

From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.

You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:

0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4

0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4

0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6

0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8

0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6

0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8

0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6

0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7

There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.

P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.

[Math] Guessing the other dice number, given two dice were rolled, and one of them rolled a 3

This kind of question is very sensitive to how the roller decides what to say. Maybe the dice are different colors and he will tell you the number on the green die. You know nothing about the other die, so all numbers are equally probable. Maybe he will tell you the higher number. Now the only rolls possible are $13,23,33,31,32$ so you should bet on $1$ or $2$. Maybe he will pick a random die and tell you the number on it. Now he could have rolled $33$ and been forced to say $3$ or rolled $3x$ or $x3$ and chosen to say $3$. You have the same chance of $x$ as $3$ for the other die so bet on anything you like. If you don't know how the roller chooses what to say it is not a mathematical question. There is not enough information to answer.

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