[Closed] small math problem

Ok firstly it is out of 7 spades, not 8. 13-6 = 7.
Think about it...
There are 6 chances that one of them has a higher spade, because there are 6 hole cards held by opponents that you have to worry about.

I'd work it out like this.
First card of first hand = 4/45
if no high spade then 2nd card of first hand = 4/44
if no high spade then 1st card of 2nd hand = 4/43
etc. 4/42
etc. 4/41
if no high spade then 2nd card of 3rd hand = 4/40
ADD (not multiply!!!) those all together and you get probability of at least 1 spade.

You could do that if you want a precise calculation, or just go 4/45 x 6.

you said it;s 7 , 13-6 =7 . what 6? there are 5 spades . 4 on the board and 1 in my hand so 13-5=8. and i guess the rest of the math is okey, ill think about it. anyone else?

the probability of some1 having higher spade then you isnt that important.. Ask yourself, what kind of hand would bet out after 4th spade comes up ?
do you expect some with 5 of spades to fire and expect 4s to call ?

common man, i know that. but really 10 dollar sng;s , you wont believe that some players are that bad.

Originally posted by GooRukYONG
the probability of some1 having higher spade then you isnt that important.. Ask yourself, what kind of hand would bet out after 4th spade comes up ?
do you expect some with 5 of spades to fire and expect 4s to call ?

LOL I agree to some extent but you would need to see the bet sizing and patterns. I've bet in situations similar when I don't have a spade in my hand, sometimes you win just by betting. Keep in mind he said decent bet. I would never call a huge bet or allin or anything though.

Originally posted by tevere
common man, i know that. but really 10 dollar sng;s , you wont believe that some players are that bad.

+1 It's more likely that he has 2 under cards and is value betting. These limits are so horribly bad (and unpredictable).

If you assume the other players have any 2 cards with equal probability given the betting patters (probably not a great assumption!) then the easiest way of finding the probability that someone has a higher spade is by finding the probability that nobody has a higher spade first.

This is (41/45)*(40/44)*(39/43)*(38/42)*(37/41)*(36/40) which is roughly 0.55,
so the probability someone DOES have a higher spade is 1 - 0.55 = 0.45.

There are 6 cards in your opponents' hands, each of which is one of the 45 remaining cards in the deck. The probability the first is not one of the 4 higher spades is (45-4)/45 = 41/45. The probability that neither the first or the second are higher spades is P(first is not a high spade)*P(second is not a high spade GIVEN that the first is not a high spade). If we know that the first is not a high spade, then the chance that the second is not a high spade is 40/44 (since one card is now accounted for by the first card). If you continue this process you get the formula above.

You should never be adding probabilities in situations like this...if you have a 1% chance of winning something and you try it twice, the chance that you win at least once is NOT 2%!!! It is 1-(1-.01)*(1-.01) = .0199. The difference here is negligible but if you have two coinflips, the probability you win one is not .5+.5 = 1 but .75 (even though it may seem like .35 sometimes :P)

thanks, that's what i needed, just the math not any other strategy discussion.