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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.

Sven van den Boogaart
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  • When does the game stop? Is the setup that you throw once, and if you are not satisfied you can have another go and that's it? – String Aug 21 '14 at 20:12
  • Yes thats the idea. 'The game is different but i thought that would be a clear example' – Sven van den Boogaart Aug 21 '14 at 20:13
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    Yes, of the times when it matters which course you take, 3 of them you do worse, and 2 of them you do better. So holding is better. – Thomas Andrews Aug 21 '14 at 20:14
  • @ThomasAndrews: That is what I thought! – String Aug 21 '14 at 20:14
  • @ThomasAndrews so you are not less likely to throw a 4, 2 times in a row? – Sven van den Boogaart Aug 21 '14 at 20:16
  • The probability question is clear. The events are independent and indeed a 50% of getting four or higher. But the odds will be $3:2:1$ of declining, improving, and maintaining status quo respectively ... – String Aug 21 '14 at 20:16
  • No, definitely not. Die rolls are "independent." This is something you can actually try with dice, if you are willing to write down a lot of results. What you roll before it doesn't matter. @SvenB – Thomas Andrews Aug 21 '14 at 20:18
  • @ThomasAndrews: I take it, you practiced that a lot and can tell from personal experience ;) – String Aug 21 '14 at 20:23
  • When I was a young kid, I did a lot of experiments with a toy roulette wheel, trying to figure out a way to beat it. Lots to learn doing experiments like that :) @String – Thomas Andrews Aug 21 '14 at 20:24
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    Does anything change if we consider your opponent gets two rolls to tie/beat a four? I.E. What is the probability your opponent gets 4 or more from 2 rolls? – Vincent Aug 21 '14 at 20:36
  • @ThomasAndrews: Yes! I just imagined how one could sit down for hours in a subconscious state of mind, like a trance, only counting occurrences of 4 without thinking too much. And approximately one person out of $82817975$ trying this experiment with $100$ throws will experience NO occurrence of $4$, thus turn very skeptic towards the idea that $4$ is just as likely as other outcomes ... – String Aug 21 '14 at 20:38
  • @Vincent only if they know you score i guess else they will stop at 4 to ;) – Sven van den Boogaart Aug 21 '14 at 20:39
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    The objective of the game is not well-defined. What is it exactly that we want to maximize? So we make the problem more explicit. If we quit after the first throw, we get in dollars the number showing. If we choose to toss again, we get in dollars what showed the second time. We want to maximize the expectation. Then we can find the optimal strategy. – André Nicolas Aug 21 '14 at 20:53
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    That's called the gambler's fallacy. – UserX Aug 21 '14 at 21:13
  • Almost all of the expert responses here assume a fair die and independence between tosses. In the real world, I don't think either of these would be the case. The degree to which the die is not balanced, and the influence of one roll result on the next may be negligibly small, but it's worth noting that they are there to promote a better understanding of probability modeling versus reality. – 2'5 9'2 Aug 22 '14 at 04:20
  • @alex.jordan In very few plausible situations involving dice rolling would rolling a 4 give a lower probablity for subsequent 4's to be rolled. – Taemyr Aug 22 '14 at 07:14
  • maybe a stupid thought, but if you throw 2 dices what is the average (i would think 12/2 =6) so if i throw 4 im likelier to throw 2 ? – Sven van den Boogaart Aug 22 '14 at 10:56
  • It seems like there are now several (valid) different conclusions drawn about the game. Could you specify the rules in more detail? Especially the order in which events happen ( I assumed Player 1 Roll 1, Player 1 Roll 2, Player 2 Roll 1, Player 2 Roll 2) – Dennis Jaheruddin Aug 22 '14 at 12:45
  • Who decides first whether to re-roll? If your opponent decides not to re-roll, he probably has 4 or more. If you only have a 4, then you are not going to beat him without re-rolling. Are you allowed to re-roll until the very moment you reveal? What if your opponent reveals quickly, such that you didn't have time to decide whether to re-roll? I think the rules need to be a bit more clear in that respect, before the question can be answered. – kasperd Aug 22 '14 at 12:52
  • @DennisJaheruddin i added the rules hope its more clear now. – Sven van den Boogaart Aug 22 '14 at 12:53

10 Answers10

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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.

mphy
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    Yes, the chance of rolling four or higher is $\frac12 = 50%$. But that should be compared to the probability of rolling four or lower which is $\frac46 = 67%$. Whether to roll again or not will depend on the exact rules, though. The expectation value of the next roll is $3.5$ which is less than $4$. – Jeppe Stig Nielsen Aug 21 '14 at 22:09
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    @JeppeStigNielsen The expected value does not matter at all in my view. If you had a dice with faces $1,2,3,4,5,10^6$, then the expected value would be quite different, but the probabilities unchanged. (of course, this also depends on the unstated rules of the game; I have no idea if the actual result has an in-game meaning or if it's used just for comparisons). – Federico Poloni Aug 22 '14 at 10:04
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    @FedericoPoloni With faces {1,2,3,4,5,6}, the value of a face is the number of face values that it beats or equals. The expected value for the number of rolls that a throw beats is meaningful. – Gilles 'SO- stop being evil' Aug 22 '14 at 11:44
  • @Gilles Try {3,3,3,4,5,6} then, which satisfies your property. If you insist in saying that this is "meaningful", I'd like to hear an argument in its favor. – Federico Poloni Aug 22 '14 at 11:53
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    Your answer is irrelevant to the problem, which is "Should I roll again on a four?" And Federico is right: the expected value of the roll is also irrelevant, since the only thing that matters is whether it is bigger, smaller, or the same as your opponent's roll. The margin of difference is not important. – TonyK Aug 22 '14 at 12:31
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    Actually, what's important to the scope of the question is whether on not rolling again will improve winning odds, not expected score. – Cthulhu Aug 22 '14 at 13:04
  • Considered downvoting for incompleteness, but then realized it reasonably answers the short version of the third question edit. Would still recomend updating the answer to cover the full question. – Dennis Jaheruddin Aug 22 '14 at 13:32
  • @FedericoPoloni When I made that comment, the rules of the game were entirely unclear. You could not see if the "size" of your victory mattered. You could not tell for sure if the mean value, or the median, or neither, was relevant (of course for a usual "uniform" die the mean equals the median). So I was commenting on that particular version of the answer which I felt was uncomplete (even if the question was uncomplete too then). After that question has been clarified, TonyK has submitted an excellent answer. – Jeppe Stig Nielsen Aug 22 '14 at 18:13
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If you and your brother-in-law both have the same strategy, then the game is even. So the only question is, what happens if you re-throw on a 4 and your brother-in-law doesn't? Here is a table of the probability of each outcome:

                   1      2      3      4      5      6
You               1/9    1/9    1/9    1/9    5/18   5/18
Brother-in-law    1/12   1/12   1/12   1/4    1/4    1/4

The probability of a tie is $3 \cdot \dfrac{1}{9} \cdot \dfrac{1}{12} + \dfrac{1}{9} \cdot \dfrac{1}{4} + 2 \cdot \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{7}{36}$

The probability that you win is $\dfrac{1}{12} \cdot \dfrac{8}{9} + \dfrac{1}{12} \cdot \dfrac{7}{9} + \dfrac{1}{12} \cdot \dfrac{6}{9} + \dfrac{1}{4} \cdot \dfrac{5}{9} + \dfrac{1}{4} \cdot \dfrac{5}{18} = \dfrac{29}{72}$

The probability that you lose is $\dfrac{1}{9} \cdot \dfrac{11}{12} + \dfrac{1}{9} \cdot \dfrac{10}{12} + \dfrac{1}{9} \cdot \dfrac{9}{12} + \dfrac{1}{9} \cdot \dfrac{1}{2} + \dfrac{5}{18} \cdot \dfrac{1}{4} = \dfrac{29}{72}$

So it makes no difference!

TonyK
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  • I am not sure which situation this describes, perhaps the decision to reroll the four made in advance (regardless of the previous rolls). – Dennis Jaheruddin Aug 22 '14 at 12:43
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    The OP has edited the question to clarify the situation: you can't see what your opponent has rolled. – TonyK Aug 22 '14 at 12:53
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    @DennisJaheruddin: The other player can use a strategy (here always re-rolling 1, 2 and 3) to improve their odds of having a high roll. So your aim is not to maximise your average score, but to maximise the probability of drawing or beating the opponent. – Fillet Aug 22 '14 at 13:37
  • Indeed, another example of the phenomenon that the average roll cannot be used to guide the comparison between dice rolls is the (in)famous set of non-transitive dice. – Willie Wong Aug 22 '14 at 13:56
  • @TonyK I understand the answer now, if you could edit it I can correct my unjust vote. – Dennis Jaheruddin Aug 22 '14 at 13:58
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Going back the the question of should you throw again I would say the answer is no you should not. If you roll a 4 the chance that you win is 50%, the chance that you lose is 33.33% and the chance that you tie is about 16.66%. So if you roll a 4 you don't lose 66.66% of the time.

KBusc
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    I don't think rolling a 4 gives you a 50% chance to win. Are you taking into consideration the opponent's ability to re-roll? The opponent will certainly re-roll a 1, and likely a 2. Assuming your opponent's strategy is to re-roll the 1 and nothing else, if you rolled a 4 your chance of a win would be 5/12, and your chance of a win or a draw would be 7/12. If your opponent re-rolls on 2 as well, your chances get even worse. That being said, I agree that you shouldn't re-roll a 4 although my reasoning is that you are more likely to decrease your roll (3/6) than increase your roll (2/6) – Matt Aug 22 '14 at 03:13
  • I considered this right after I posted this answer but I had to leave. I do agree with you though and I'm glad someone took the time to add that. – KBusc Aug 22 '14 at 12:14
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Answer updated after edit: It does not matter

An analytical solution has already been given by @TonyK. As such I have just added the simulation code to represent this situation:

% Generate first rolls and potential second rolls for each strategy
strategy1= randi(6,2,1000000);
strategy2= randi(6,2,1000000);
% Determine whether to reroll or to keep
idx_keep1 = strategy1(1,:)>=4; %Keep the 4
idx_keep2 = strategy2(1,:)> 4; %Don't keep the 4
% Use the first value if you keep, the second if you reroll
strategy1(2,idx_keep1)= strategy1(1,idx_keep1);
strategy2(2,idx_keep2)= strategy2(1,idx_keep2);
% Check the statistics
h=hist(sign(strategy1(2,:)-strategy2(2,:)))

Results of strategy1 vs strategy 2

If you reroll the 4 and the other player does not, these are the resulting probabilities:

  • Chance that you win: 40.3%
  • Chance that you draw: 19.4%
  • Chance that you lose: 40.3%

Answer before edit:

How to win the game

For completeness I will repeat that if you roll again, you have indeed got a 50% chance of throwing at least four. But here is why you should not do it!

Assuming the game is you may roll once or twice, then your opponent rolls once looks at your score and may decide to roll again, and that the one with the highest score wins and it is otherwise a tie:

Opponent roll 1:

  • There is a 2/6 chance that he will roll above 4
  • There is a 1/6 chance that he will roll 4 (and thus should stop as the second roll is expected to be below 4)
  • There is a 3/6 chance that he will roll below 4 (and thus should roll again)

Opponent roll 2:

  • There is a 2/6 chance that he will roll above 4
  • There is a 1/6 chance that he will roll 4
  • There is a 3/6 chance that he will roll below 4

Resulting probabilities

These are the probabilities given that you don't reroll after getting a 4

  • Chance that your opponent wins: 2/6+3/6*2/6 = 50%
  • Chance that your opponent draws: 1/6+3/6*1/6 = 25%
  • Chance that your opponent loses: 3/6*3/6 = 25%

Conclusion

If you roll again you are expected to roll lower than what you have, I didn't do the math here but intuition tells me that in the resulting situation (you roll once, your opponent rolls twice) you will face worse odds than you have now (with your above average roll of 4). The simulation below seems to confirm this:

Simulation to calculate odds if you do throw again

r1= randi(6,1,1000000);
r2= randi(6,2,1000000);
r2(2,r2(1,:)>=4)=0;
r2 = max(r2);
h = hist(sign(r1-r2))

Note that it is Matlab. If you want to simulate the odds of beating 4, just replace r1 by 4 in the last line.

The results for if you do throw again are approximately:

  • Chance that your opponent wins: 2/6+3/6*2/6 = 56%
  • Chance that your opponent draws: 1/6+3/6*1/6 = 17%
  • Chance that your opponent loses: 3/6*3/6 = 27%
Community
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Dennis Jaheruddin
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  • Conclusion If you roll again you are expected to roll lower than what you have, . why the chance is 50% right? – Sven van den Boogaart Aug 22 '14 at 10:54
  • @SvenB Perhaps I should have formulated it like so: If you roll again, your expected outcome is lower than what you currently have. (Because 4 is an above average roll). – Dennis Jaheruddin Aug 22 '14 at 11:49
  • I did do the math (see my answer), and it turns out that whether you re-throw or not is irrelevant. – TonyK Aug 22 '14 at 12:32
  • @TonyK It seems like you consider a different game than me. I consider the situation where the second player already knows the outcome of the first players throw (and uses that information in a smart way). – Dennis Jaheruddin Aug 22 '14 at 12:41
  • I see now what happened: the OP edited the question to make this clear, but it was after you posted your answer. – TonyK Aug 22 '14 at 12:51
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    There seems to be a typo in your code? idx_keep2 should be defined relative to strategy2, and not strategy1, right? – Willie Wong Aug 22 '14 at 13:35
  • That is to say, your code seems to be requiring player two to reroll whenever player one rolls 4 or lower, while player one only rerolls whenever player one (i.e. himself) rolls 3 or lower. A little bit of computations shows that in this scenario the probability of a tie is $1/6$, the probability of a player one win is 13/24 and the probability of a player two win is 7/24. These numbers agree with your numerics, but the scenario is definitely not what the OP had in mind! It is easy to see that this game is rigged in favor of player one. – Willie Wong Aug 22 '14 at 13:53
  • @WillieWong Thanks, it was a simple typo, though it goes against my intuition my results now agree with those by TonyK. – Dennis Jaheruddin Aug 22 '14 at 13:55
  • @DennisJaheruddin: to further mess up your intuition :-) you can take a look at http://en.wikipedia.org/wiki/Nontransitive_dice The average rolls of the three dice are all the same (5), but die A beats die B 5/9 of the time, B beats C 5/9 of the time, and C beats A 5/9 of the time (no ties as there are no repeated numbers between dice). – Willie Wong Aug 22 '14 at 13:59
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You have exactly the same probability of getting a 4 on every single throw. The next dice doesn't care what you threw last. It doesn't look at the other dice, and think "well, 4 is already taken sooo...".

We call this "independent events".

Alec
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2

The question is not if I get a higher die roll or not but the question is "Is the probability that I beat the opponent higher".

Case 1: opponent rolls again on a 1,2,3 and keeps on a 4,5,6

He has a 1 if he rolls a 1, 2 or 3 in the first throw and a 1 in the second. That has a probability of $\frac{3}{6}*\frac{1}{6} = \frac{3}{36}$.

With the same probability he has a 2 or 3.

He has a 4 when he rolls a 4 with the first throw or a 1, 2 or 3 in the first throw and a 4 in the second throw. That has a probability of $\frac{1}{6}+\frac{3}{6}*\frac{1}{6} = \frac{9}{36}$

With the same probability he has a 5 or 6.

That means without throwing new you win when your opponent has a 1, 2 or 3. That has a propability of $\frac{3}{36}+\frac{3}{36}+\frac{3}{36}=\frac{9}{36}=\frac{1}{4}=0.25$

When you throw again you win on a 2-6 if your opponent has a 1 on 3-6 if your opponent has a 2 ...

That has a propability of $\frac{3}{36}*\frac{5}{6}+\frac{3}{36}*\frac{4}{6}+\frac{3}{36}*\frac{3}{6}+\frac{9}{36}*\frac{2}{6}+\frac{9}{36}*\frac{1}{6}+\frac{9}{36}*\frac{0}{6}=\frac{63}{216} = \frac{7}{24}\approx 0.29$

That means you should throw again.

Case 2: opponent rolls again on a 1,2,3,4 and keeps on a 5,6

That means he has a 1/2/3/4/5/6 with a probability of $\frac{4}{36}/\frac{4}{36}/\frac{4}{36}/\frac{4}{36}/\frac{10}{36}/\frac{10}{36}$

Without rethrowing you win with a probability of $\frac{4}{36}+\frac{4}{36}+\frac{4}{36}+\frac{4}{36}=\frac{16}{36}=\frac{4}{9}\approx0.44$.

With rethrowing you win with a propability of $\frac{4}{36}*\frac{5}{6}+\frac{4}{36}*\frac{4}{6}+\frac{4}{36}*\frac{3}{6}+\frac{4}{36}*\frac{2}{6}+\frac{10}{36}*\frac{1}{6}+\frac{10}{36}*\frac{0}{6}=\frac{66}{216} = \frac{11}{36}\approx 0.31$

That means you should not throw again.

Conclusion

The conclusion is that you should do the opposite of what your opponent does to maximize your winning probability.

But you can do the same calculations to show that doing the opposite of your opponent maximize the losing probability also. Only the tying probability decreases.

In fact the losing probability increases the same amount as the winning probability.

If a loss gives the same amount of minus points as the win gives plus points it doesn't matter if you reroll a 4 or not.

If only a win gives points then you should reroll a 4 when your opponents doesn't reroll a 4.

Timbo
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    Nice note on the variability. This especially matters if you play something like 'first to five points wins and a tie counts for half'. Then you could choose to play it safe if you are close to winning the match. -- Sidenote, if only a loss gives your opponent a point as well the last line is not relevant. – Dennis Jaheruddin Aug 22 '14 at 15:04
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He said if you have 4 you have a 50% change the next time to throw the same or higher.

This is true, but is misleading. It's also true that you have a 4/6 chance to throw the same or lower.

If you wish to maximize the expected value, this is the strategy to follow:

The expected value each time you roll is $3.5$ (since this is the average of the values, and they have equal probabilities). Since your roll of $4$ is above this, rerolling is expected to lower your score, not raise it. You should reroll $(1,2,3)$ and hold $(4,5,6)$. Note that the rolls are completely independent events, so rolling a 4 the first time does not decrease the probability that you'll roll a 4 the second time.

Whether this actually results in you beating your opponent more than 50% of the time depends on his strategy, which likely depends on what he knows of your strategy.

Tim S.
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As André Nicolas commented, whether on not to throw again depends on the precise objective of the game. Vincent also asked whether or not the opponent has the choice of rolling twice makes any difference.

If we assume that all wins give the same payoff, and a loss and a draw are both worthless, then it turns out that it doesn't matter whether the opponent gets to roll twice, you are still better off keeping a four (and re-rolling a three).

If we assume the opponent rolls once, then, if I roll $i$, the probability of him winning is $\frac{6-i}{6}$. If he is allowed to roll twice, then there is a $\frac{6-i}{6}$ chance that he wins on his first roll and likewise on his second roll, thus the probability of the opponent winning is $\frac{6-i}{6}+\frac{i}{6}\cdot\frac{6-i}{6}=\frac{6-i+i(6-1)}{36}=\frac{(6-i)(6+i)}{36}$.

Plotting these curves gives:

probability of opponent winning

The horizontal lines show the expected value, assuming your roll is uniformly distributed, which it would be if you roll again. You can see that if you roll a four, your opponent will have a lower probability of winning than the expectation if you rolled again. So it is better to not roll if you have a four (under the assumptions that loss and draw are worthless and all wins are equal).

Joel Bosveld
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To answer the short version: "If I threw 4, how big is the chance I throw 4 or more the next time I throw the dice"

The answer is 50%.

The unspoken questions seems to be "if I don't know what my opponent threw and I threw a 4, should I roll again or 'stick'?"

This boils down to 'what are the chances of improving on a 4?' and the answer to THAT is ⅓ (33.3333%). Since this is less than 50%, you might as well just stick on a 4 (because you have a 33⅓% chance of improving but a 50% chance of getting worse).

Nagora
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Interestingly, it seems that, in the limit of n, the Nash optimal strategy for a die with n-faces is to rethrow whenever your first trie is less than $\frac{n}{\phi}$ where $\phi$ is the golden ratio.

You should not rethrow a 4, but if these were 20 sided die, you would want to rethrow anything below 12.

Arthur B.
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