Questions re: Probability of Higher Flush by WoO

Quote: podski

So...Wizard 'n' miplet - any thoughts on the other bit...

If we incorrectly assume that each player is dealt cards from a separate deck, then if the probability that one player beats you is p, then the probability that at least one player out of n will is 1-(1-p)^n. My table is based on that assumption.

For example, the probability of rolling at least one six in 10 rolls of a die is 1-(1-(1/6))^10 = 1-(5/6)^10=0.838494417.

Quote: Wizard

...the probability that at least one player out of n will [beat you] is 1-(1-p)^n. My table is based on that assumption.

So, I believe, is mine.

Do I take it then that factoring in the effect of card removal on the chances each opponent after the heads-up calculation is too onerous?

Or is it just that it makes bugger-all difference?

Sorry for being a pest; my thirst for knowledge gets the better of me sometimes.

Cheers!
p

Quote: podski

Do I take it then that factoring in the effect of card removal on the chances each opponent after the heads-up calculation is too onerous?

Yes. If you wish to factor that in, then a simulation is the easiest way to go.

Coolio.

My thanks for your input on this, Sir. 'tis most appreciated!

Cheers!
p