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Author Topic: Olympic Games chance of victory  (Read 2691 times)

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Iron_curtain

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Olympic Games chance of victory
« on: November 24, 2013, 10:18:22 pm »
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I've tried to calculate the chance of winning the Olympics as sponsor and arrived at 81.25%. Is this right? The calculation is below:

X = 1/36 + 2/36 + 3/36 + (4/36)X

Solving for X, it gives X = 0.1875, which is the chance of losing the olympics, thus 1 - X = 0.8125

If the above is right, then the average VP gain off of playing Olympic Games is 1.625.
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Eruantalon

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Re: Olympic Games chance of victory
« Reply #1 on: November 25, 2013, 06:26:57 am »
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All cases of losing the Olympics by phasing player are following rolls (phasing player first):
(1,4) (1,5) (1,6) (2,5) (2,6) (3,6), thus 6/36=1/6.

So there is 5/6 of +2 VP's and 1/6 of -2 VP's.

5/6*2=10/6
1/6*2=2/6
10/6-26=8/6=1,33 on average.

(remember, ties are re-rolled)
« Last Edit: November 25, 2013, 06:29:44 am by Eruantalon »
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Jack Rudd

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Re: Olympic Games chance of victory
« Reply #2 on: November 25, 2013, 12:52:06 pm »
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All cases of losing the Olympics by phasing player are following rolls (phasing player first):
(1,4) (1,5) (1,6) (2,5) (2,6) (3,6), thus 6/36=1/6.

So there is 5/6 of +2 VP's and 1/6 of -2 VP's.

5/6*2=10/6
1/6*2=2/6
10/6-26=8/6=1,33 on average.

(remember, ties are re-rolled)

Not so! You forgot to factor in (1,3), (2,4), (3,5) and (4,6) followed by one of the six rolls above.

If X is the event "phasing player wins", then the actual calculation is P(X) = 26/36 + (4/36)P(X), which cancels out to P(X) = 13/16, and an actual expected payoff of 1.25 VP.
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Iron_curtain

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Re: Olympic Games chance of victory
« Reply #3 on: November 25, 2013, 02:58:08 pm »
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All cases of losing the Olympics by phasing player are following rolls (phasing player first):
(1,4) (1,5) (1,6) (2,5) (2,6) (3,6), thus 6/36=1/6.

So there is 5/6 of +2 VP's and 1/6 of -2 VP's.

5/6*2=10/6
1/6*2=2/6
10/6-26=8/6=1,33 on average.

(remember, ties are re-rolled)

Not so! You forgot to factor in (1,3), (2,4), (3,5) and (4,6) followed by one of the six rolls above.

If X is the event "phasing player wins", then the actual calculation is P(X) = 26/36 + (4/36)P(X), which cancels out to P(X) = 13/16, and an actual expected payoff of 1.25 VP.

Right! I miscalculated the average payoff. It should be (0.8125 * 2) - (0.1875 * 2) = 1.25 VP
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Eruantalon

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Re: Olympic Games chance of victory
« Reply #4 on: November 25, 2013, 04:04:13 pm »
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Yeah, I forgot, that rolls that tie aren't victories of phasing player..
« Last Edit: November 26, 2013, 02:29:23 am by Eruantalon »
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pietshaq

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Re: Olympic Games chance of victory
« Reply #5 on: November 26, 2013, 02:58:41 am »
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1.25 VPs is the answer. My off-topic question is: does it really matter?

I mean: imagine a card changes so that it gives the sponsor +1 or +3 to the roll. Or that ties are not re-rolled. Or both. Do you see a scenario that really changes your decision about playing this card for the event or not in the game? Or even changes a decision whether to give Olympic Games or something else in exchange for Missile Envy?
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discomute

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Re: Olympic Games chance of victory
« Reply #6 on: December 20, 2013, 04:32:04 am »
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1.25 VPs is the answer. My off-topic question is: does it really matter?

I mean: imagine a card changes so that it gives the sponsor +1 or +3 to the roll. Or that ties are not re-rolled. Or both. Do you see a scenario that really changes your decision about playing this card for the event or not in the game? Or even changes a decision whether to give Olympic Games or something else in exchange for Missile Envy?

well... on some occasions when you have ordinary headline events
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