Patrick Jamet
France
Paris
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Steve,
I understand that you demonstrated that it can’t happen more than twice in game. I agree but I don’t understand where you go. I say that it may happen once in a game, at any of the five turns. The chance that it happens in a game is 3 out of 5, or 3 times every 5 games, not 3 times in a game. English is not my native language. Excuse me if I wasn't clear.
(72-13)/72 is the probability that it does not happen in a turn. So, this number elevated at the power 5 is the probability that it does not happen in a game. And one minus this number is the probability that it happens at least once. The fact that it can’t happen twice can’t modify this number.
That said, a better approximation should be : 1 - C(72-13,5) / C(72,5) = 64% (a little more than the previous 63%) where C(x,y) is the number of combinations of y elements among x.
As you noticed, this approximation does not take into account the linkage factor. In fact, I should think in terms of permutations of 6 elements among 9. There are P(9,6) = 60480 possibilities for 6 ordered tiles. To count the number of favorable cases is a bit complicated (a computer program would do it easily) but I will try.
If I decide that a series includes one favorable couple, there are 5 positions for it (turn 1, 2, etc) and there are P(7,4) possibilities for the remainder of the series. So, 5 x P(7,4) = 5 x 840 = 4200. This is the number of series of 6 tiles including a given couple like (9,4).
There are :
• 4200 series of 6 tiles including (9,4)
• 4200 series of 6 tiles including (10,4)
• Series with (10,5) should not include series with (9,4) or they would be counted twice.
How many permutations with (9,4) and (10, 5) and 2 other tiles among 5 ?
There are P(4,4) x C(5,2) = 24 x 10 = 240
• Series with (10,6) should not include series with (9,4) = 1 x 240
• Series with (11,4) should not include series with (10,5) and (10,6) = 2 x 240
• Series with (11,5) should not include series with (9,4), (10,4) and (10,6) = 3 x 240
and so on… (details on request)
I found that 46 x 240 series must be subtracted or they would be counted more than once. So the number of favorable series are 13 x 4200 - 46 x 240 = 43560, or 72% of 60480.
In conclusion, I believe that this unfavorable situation happens 72% of the time, almost 3 times out of 4.
You believe it’s way too much and impossible? I drew 6 tiles manually 20 times, and got a total of 12 or higher 16 times (80% because I’m unlucky). Would you try and come back with your results?
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Steve Duff
Canada
Ottawa
Ontario
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Pyjam wrote: Steve,
I understand that you demonstrated that it can’t happen more than twice in game. I agree but I don’t understand where you go. I say that it may happen once in a game, at any of the five turns. The chance that it happens in a game is 3 out of 5, or 3 times every 5 games, not 3 times in a game. English is not my native language. Excuse me if I wasn't clear.
Yes, I think I was misunderstanding you. I agree that a turn with a 12+ base value or higher will happen most games. I think that's clearly what the designers were aiming for, with the tile system and values they chose.
Because of that, I find most players keep some dollars handy, so they will be able to deal with high values. I just haven't seen it as a problem in the game. It's just a gaming element you have to plan for and deal with.
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Patrick Jamet
France
Paris
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Interesting and quite funny. First reaction was : it never happens. Second reaction was : it rarely happens. Third reaction is : it’s a feature.
It’s probably a feature indeed, one I don’t like a lot because if it doesn’t happen, it was better to spend my money, if it happens, it was better to keep it. As it happen thrice more often than not, I know what do to now. Note that the 3-player game plays differently.
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Dan Schaeffer
United States
Unspecified
Illinois
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I think there was some initial misunderstanding of what you were talking about, hence the changing reactions. Your original claim seemed to be that in a 4P game, the fourth player will always get shafted. I think the reality is more that, while there is a pretty good chance that on A turn of any particular game, whoever happens to be the fourth player for THAT turn may get stuck with high action costs, that is both part of the design and a possibility that the players can plan for.
What you didn't provide math for is how often it occurs that the 4th player at the beginning of the game is stuck with a 12+ action cost. If that happened 60-75% of the time, that might be a problem, but I suspect it is far rarer.
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Steve Duff
Canada
Ottawa
Ontario
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Pyjam wrote: Interesting and quite funny. First reaction was : it never happens. Second reaction was : it rarely happens. Third reaction is : it’s a feature.
I thought you were saying it happens again and again and again in the *same game* So yeah, I was arguing hard against that, it's only 5 rounds.
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Jamie Pollock
Scotland
Edinburgh
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May as well add my voice to those who say I don't find this to be a major factor. Plan accordingly, have money available, and be cognisant of player order. Simples.
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Daniel Corban
Canada
Newmarket
Ontario
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It's immediately clear to anyone playing this game that you need to have some money on hand for security. Not to mention for buying crew/captains and possibly a plan. I find that this is where the tension in the game lies. The actual actions you perform provide a little variety, but its the decisions as to when to take actions and before whom that is the meat of the game.
Sometimes you just have to take start player. If you know you are going to be low on cash in the following round, and the cash reserves on the board are going to be low, take start player. I find that start player is actually relatively weak compared to the others, which is undoubtedly why it includes a 2VP bonus. Even then, I tend to avoid it, but sometimes you should probably take it.
Also, I find that the game is better with four players.
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