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Mordredd |
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Post subject:
 Posted: May 10, 2004 - 05:10 AM
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Joined: Mar 03, 2003
England
Posts: 728
Location: England
Status: Offline
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Yep, I made a bit of a fool of myself on that one didn't I.  
Thanks for the reference Bevan. I'll look it up, although I think I've pretty much worked it all out now anyway.  |
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Post subject:
Posted: May 14, 2004 - 11:55 PM
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Joined: Oct 24, 2003
Posts: 1671
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Here's another example for you that i've just thought about to show that predictions made with fewer initial data can still be accurate.
You're playing poker with 3 of your friends. You each receive 5 cards from a 52-card deck deck and lo and behold, you get 3 aces! The other 2 cards are mismatched so you discard them and ask for 2 new ones. What's your chance of getting a 4th ace?
Since we don't know what the other players have (they may or may not be holding the last ace), we'll assume that they don't have a single card and that everything not in your hand is still in the deck. You receive 2 cards from this 47-card deck, so your chance of getting an ace is :
(1/47) + (46/47)*(1/46) = 4.255%.
You could go about this another way and take into account the fact that there are 15 cards currently in the other players' hands and not available to you. However, since you don't know what these cards are, this changes nothing in the problem and the result will be the same. Here it is in case you don't believe me. 15 of the 47 cards that are not in your hand are in theirs, and the remaining 32 are still in the deck. So there's a 15 in 47 chance of the last ace being in their hands, in which case you have no chance to get it. Then there's a 32 in 47 chance that the ace is still in the deck, in which case you can still get it will the following probability :
(32/47) * [ (1/32) + (31/32)*(1/31) ] = 4.255%.
You see, the result is the same. This example shows you that things we don't know can be taken out of the equation without affecting the result. If you don't know, at the time you create your team, what the other team's FF will be when you meet them, then you can take it out of the equation and still get a perfectly valid result.
A similar problem would be to ignore the other players and say you were playing poker solitaire, and ask what the chance is of getting a pair to complete your 3 aces, assuming you're not getting another ace. This is what i started typing first but i realized that it got complicated near the end so i decided to present an easier problem first which is what you got above. With that out of the way, i think you can now understand the second example better so i'll give it to you, if only so that i didn't type it in vain.
Let's say that you're playing poker. The deck has 52 cards. You receive 5 cards, and 3 of them are aces (the other 2 cards are mismatched). You keep the 3 aces and exchange the other two cards. Assuming you don't get another ace, what's your chance of completing this with a pair to make a full house?
The chance is 3 in 47. Since we don't know whether you'll receive a card that matches one of the two that you discarded, we can ignore those and pretend that you never had them in the first place and that the pack contained 49 cards after you discarded. Actually, since i said "assuming that you don't get another ace", one of those cards (the last ace) is not available to you so that the deck can be thought to have 48 cards. Whatever card you receive first, there will be 3 matching cards in the remaining 47 so that your chance of getting a pair will be 3 in 47, or 6.383%.
Now, you could go about this differently and take into account the cards that you discarded. Again assuming that you don't receive an ace, the deck now has 46 cards available to you, and 45 after you receive the first one. The first card you receive has a 6 in 46 chance of matching one of these. If this happens, there will be only 2 matching cards remaining for that one, giving a 2 in 45 chance of a pair. In the other 40 of 46 cases, you'll get a card different from those you discarded and there will be 3 matching cards for this one, giving a 3 in 45 chance of a pair. Put all this together and you get a total chance of
(6/46)*(2/45) + (40/46)*(3/45) = 6.377%.
The result is slightly different. Why? Because we assumed that we knew nothing of the 2 cards you discarded which was wrong : you knew that they were mismatched. If you truly didn't know anything about them that is relevant to our problem, then they could be safely taken out of the equation and you'd get the proper result. The chance of them forming a pair is 3 in 47 and the chance of them being mismatched is 44 in 47. Here's the equation that you would obtain :
(44/47) * [ (6/46)*(2/45) + (40/46)*(3/45) ]
+ (3/47) * [ (2/46)*(1/45) + (44/46)*(3/45) ]
= 6.383%.
I hope this was enough to convince you now. |
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Post subject:
Posted: May 16, 2004 - 03:18 PM
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Joined: Feb 10, 2003
Undisclosed
Posts: 2696
Location: Undisclosed
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Help! Nurse! I think my head exploded!  |
_________________
_____ and rankings - that is all
#27 of the "24 club" (due to some dodgy accounting)
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Doubleskulls |
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Post subject:
Posted: May 16, 2004 - 03:30 PM
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Ex-Rulz Committee
Joined: Mar 05, 2003
Undisclosed
Posts: 2627
Location: Kent, UK
Status: Offline
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Actually I can't see how that proves anything. With a deck of cards you know the probabilities of a given card being in the deck. Whether cards may be held in your opponent's hand is irrelevant - you know they exist and are in play.
With FF is that generally you don't know FF of your opponents - so the only way to produce reasonable stats is to ignore it all together or make assumptions.
Now if you know your likely opponent's FF then you can factor that in, if you don't then you are left with estimates and assumptions. |
_________________
Ian 'Double Skulls' Williams
SLOBB
NAF Racial Results
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Post subject:
Posted: May 16, 2004 - 03:50 PM
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Joined: Oct 24, 2003
Posts: 1671
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Doubleskulls wrote: With FF is that generally you don't know FF of your opponents - so the only way to produce reasonable stats is to ignore it all together or make assumptions.
Exactly! This is what i've been saying all along!
And if you can't even make a reasonable guess, then the only possible alternative is to ignore it. Fortunately, you can still make pretty good calculations even if this is the case. |
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