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This question/claim was from a manga but I have trouble understanding it. The situation is this: two people are gambling by throwing dice.

They define the numbers 4,5,6 as U (short for UP) and 1,2,3 as D.

The two people write three letters (such as UDD, DUD, etc. Thus, there are 8 methods in total) each. Then, a third party throws the dice until one of the two predictions come true.

For example if I wrote (UDD) and the opponent wrote (DDD), and the dice throws were 5,4,2,1 that leads to me winning.

At first, I thought that the probabilities of winning would be pure luck, but the manga claimed that certain predictions have an advantage. Can anyone explain why probabilities(?) for some predictions are higher?

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Probably this is a version of Penney's game. The second player has advantage over the first player.

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  • $\begingroup$ Wow, that is counterintuitive. I mean, even if both players submit their sequence at the same time it would mean one sequence could "beat" another. The chart says that $HHH$ and $TTT$ are strictly dominated, right? Does this make my answer wrong? $\endgroup$ – SK19 Feb 20 '18 at 6:29
  • $\begingroup$ @SK19 $HHT$ does not win over $HHH$, so $HHH$ is not 'strictly dominated'; if you want to analyse the game deeper, you may be interested in the full game table (text is in Russian, but the table is just numbers). $\endgroup$ – kludg Feb 20 '18 at 6:50
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If we assume that each die roll is done independent of the others and further assume that each side has the same probability, then this experiment is just the same as throwing a perfect coin until one party wins. There is no way to predict which of the eight predictions would be more likely because no one is.

BUT we are dealing with a "real" die here (real in the Manga universe), and real dice don't really have each side with same probability (because they are not perfect, may have little cuts or a site weights more or something similar) AND our inability to predict how the die is thrown is bad due to too many variables (how often is the die shaken in the hand etc.). So if the manga characters would have some strong, supernatural sense of all the surroundings and how the one who throws the die does it, they could make predictions. But this is really, really far fetched.

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  • $\begingroup$ Well personally I found that this manga (at least in this place) didn't take into those numerous other factors such as "the weight of certain sides of the die" or "the way the die is thrown" but still thanks for a possible explanation $\endgroup$ – Taegyu Ahn Feb 20 '18 at 6:20
  • $\begingroup$ It's true that each dice role have the same probability, but that doesn't mean that when you role the dice until one of you gets the roles they chose you have equal chances of winning. Numberphile made a great video about this exact question. I'd also recommend looking up 'non-transitive games'. Here is a link to the video: youtu.be/SDw2Pu0-H4g $\endgroup$ – Ido Fraenkel Feb 20 '18 at 7:48
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First of all, you procedure is exactly like flipping a coin and getting heads or tails, so you should phrase your question that way. If you were only to throw the coin three times, everyone would have an equal chance of winning, but because you throw continuesly until someone gets their letters, certain letters have an advantage over other letters. For example, DHH has an advantage over HHH (because of you get two heads, there is a chance a tail was thrown before). In general, for any letters ABC, you can choose letters C'AB to win the other player (c' is the opposite of c). Here is a link to numberphile's video explaining this in further detail: https://youtu.be/SDw2Pu0-H4g

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  • $\begingroup$ I get that DHH has an advantage over HHH but you explained as if this is because two heads have already been thrown, and yet you also claimed that C'AB can always won ABC. But what is the AB part is not the same two characters? does that still work? $\endgroup$ – Taegyu Ahn Feb 20 '18 at 6:12
  • $\begingroup$ The difficulty in your approach is that the predictions are made before the first throw, so conditional probabilities are not to be applied here, I guess. $\endgroup$ – SK19 Feb 20 '18 at 6:14
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It is a binomial experiment. Let $X$ is a number of UPs. When die is rolled three times, i.e. $n=3$, then: $$P(X=0)=\frac18; \ P(X=1)=\frac38; \ P(X=2)=\frac38; \ P(X=3)=\frac18.$$ So selecting one or two UPs have higher chances.

Similarly if $n=4$, the combinations are: $$C(4,0)=1; \ C(4,1)=4; \ C(4,2)=6; \ C(4,3)=4; \ C(4,4)=1.$$ So the winning choice is to have two UPs and two DOWNs.

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