Probability by # of players at the table

[Onaflag]

I don't know how to calculate this one. Say there's a 5% chance of you being dealt a certain hand. If playing HU, then I suppose the chance of your opponent being dealt that hand would be 5%. I'm friggin' smart, eh?

How about when you have two opponents, or three, or four, etc? How is that calculation done?

Bonus question - In 3 Card Poker there is a 75% chance (74.38 actually) you'll be dealt nothing which means no pair or higher, so you could still have high card which is not worthless, but let's not consider that. When it is three handed, what are the chances of your two opponents having nothing? Four handed, etc?

Thanks a bunch. I suck at this stuff.

Onaflag.......

[Albert Moulton]

[ QUOTE ]
Say there's a 5% chance of you being dealt a certain hand... How about when you have two opponents, or three, or four, etc? How is that calculation done?


[/ QUOTE ]
Each player, regardless of how many players, would have an equal 5% chance of getting that hand unless you added some additional information, like some dead cards or knowledge of your own hole cards that might increase or decrease that 5% for the others.

[Gullanian]

75% 1 person has nothing
0.75^2 chance that 2 people have nothing
0.75^3 chance that 3 people have nothing

[Onaflag]

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^

[/ QUOTE ]

That's not on my calculator, I don't think.

I'm a nitwit with this. English, please.

Onaflag........

[Gene Paulson]

just as in a problem of (5%chance of winning)you multiply each so .05x.05x.05 for three or thus with 3 5%x33.3%. chances go down steeply with more in so the 55 HU would be a lot less with you and two others.