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SideCash
12-21-2004, 05:58 PM
If you have two people who both have a 1 to 10 chance of hitting something, how do you calculate the odds of either person hitting?

Cobra
12-21-2004, 06:44 PM
Assuming the first person winning doesn't effect the second person winning just add them. That gives you 20% chance either one will win. I am assuming your 1 to 10 is a probability not the odds of winning.

elitegimp
12-21-2004, 06:52 PM
you figure out the chance that neither hits (9/10 * 9/10 = 0.81 or 81%), and subtract from 1 (or 100%) to get .19 or 19%.

That includes the 1/100 (1%) chance they both hit, subtract it out if you want exactly one person to hit.

gaming_mouse
12-21-2004, 07:42 PM
[ QUOTE ]
Assuming the first person winning doesn't effect the second person winning just add them. That gives you 20% chance either one will win. I am assuming your 1 to 10 is a probability not the odds of winning.

[/ QUOTE ]

Cobra, this is incorrect. Elitegimp's answer is right.

I see that you are new to the forum, and I do not mean this to be rude, but you really should not be answering questions like this yet. This is a very basic calculation, and answering it incorrectly just propogates a common error.

gm

BTW: You could only add them if they were mutually exclusive. However, by definition, mutually exclusive events affect each other. So your answer is wrong even given -- or I should say especially given -- the explicit assumption you make.

mannika
12-21-2004, 07:46 PM
[ QUOTE ]
you figure out the chance that neither hits (9/10 * 9/10 = 0.81 or 81%), and subtract from 1 (or 100%) to get .19 or 19%.

That includes the 1/100 (1%) chance they both hit, subtract it out if you want exactly one person to hit.

[/ QUOTE ]

This is true if the two people hitting is completely independant, however, in a game like poker, this assumption is false. Either the players share outs, in which case the chance of one hitting is dependant on the other (one hitting is more likely if the other one hits), or the players share no outs, in which case the chance of one hitting is STILL dependant on the other (one hitting is less likely if the other one hits).

Therefore, your calculation is correct in terms of independant trials, but in a poker situation, this is likely not the case.

gaming_mouse
12-21-2004, 07:49 PM
mannika,

this is a good point. however, to the theoretical question that the OP posted (which implied independence, at least to me), it's worth noting that elitegimp's answer is correct.

gm

elitegimp
12-21-2004, 07:50 PM
[ QUOTE ]
[ QUOTE ]
you figure out the chance that neither hits (9/10 * 9/10 = 0.81 or 81%), and subtract from 1 (or 100%) to get .19 or 19%.

That includes the 1/100 (1%) chance they both hit, subtract it out if you want exactly one person to hit.

[/ QUOTE ]

This is true if the two people hitting is completely independant, however, in a game like poker, this assumption is false. Either the players share outs, in which case the chance of one hitting is dependant on the other (one hitting is more likely if the other one hits), or the players share no outs, in which case the chance of one hitting is STILL dependant on the other (one hitting is less likely if the other one hits).

Therefore, your calculation is correct in terms of independant trials, but in a poker situation, this is likely not the case.

[/ QUOTE ]

that's a good point, and I should have mentioned the assumption of independence when I made it.

BruceZ
12-22-2004, 02:43 AM
[ QUOTE ]
mannika,

this is a good point. however, to the theoretical question that the OP posted (which implied independence, at least to me), it's worth noting that elitegimp's answer is correct.

gm

[/ QUOTE ]

The original question in no way implied independence. If we are talking about hitting a hand in Hold'em, the hands are virtually never independent, though they may sometimes be approximately so. The hands are often mutually exclusive when only one can hit his hand. In that case we add the probabilities (to get 20% in this case). We cannot completely answer the original question since we are not given the probability of both hitting something. The answer is 20% minus the probability of both of them hitting.

gaming_mouse
12-22-2004, 03:45 AM
[ QUOTE ]
The original question in no way implied independence. If we are talking about hitting a hand in Hold'em, the hands are virtually never independent, though they may sometimes be approximately so. The hands are often mutually exclusive when only one can hit his hand. In that case we add the probabilities (to get 20% in this case). We cannot completely answer the original question since we are not given the probability of both hitting something. The answer is 20% minus the probability of both of them hitting.

[/ QUOTE ]

Eh... maybe you're right. It seemed to imply independence to me for some reason, but re-reading the post I'll admit that it's pretty unfounded.

In any case, my post was in response to Cobra's assumption of mutual exclusivity, which is very misleading to someone learning probability. Perhaps my assumption of independence was equally misleading.

The best response would have been, as you say, "20% minus the chance of them both hitting."

gm

HesseJam
12-23-2004, 06:33 AM
[ QUOTE ]
I see that you are new to the forum, and I do not mean this to be rude, but you really should not be answering questions like this yet. This is a very basic calculation, and answering it incorrectly just propogates a common error.

[/ QUOTE ]

I do not agree.

i) everybody has a right to take a shot at an answer, even a wrong one ii) this certainly increases his learning experience and iii) it very likely increases the learning experience for others who would have made the same mistake provided they check back to see what the original poster answers.