Another senseless question

I just didn't want to say "stupid question," because every teacher I ever had taught me there's no such thing.

Anwyay, last night (or rather very early this morning), I dreamt I was playing 10 hand VP. at one point I got 10c, Jc, Qc, Ac and 4h. I held the suited four to a royal, and wound up drawing royals on two hands.

So the question is: what are the odds of this actually happening?

Quote: Nareed

I just didn't want to say "stupid question," because every teacher I ever had taught me there's no such thing.

Anwyay, last night (or rather very early this morning), I dreamt I was playing 10 hand VP. at one point I got 10c, Jc, Qc, Ac and 4h. I held the suited four to a royal, and wound up drawing royals on two hands.

So the question is: what are the odds of this actually happening?

Hitting two 47 to 1 shots in ten attempts wouldn't seem to be all that difficult.

I've done it on ten-play VP twice, while wide awake.

(46/47)⁸ × (1/47)² × combin(10, 2) = 1.7151%, or about once every 58 times you find yourself in that situation.

(Hopefully I did that right.)

I assume you're talking about the odds of making the royal on the draw, not the odds of being dealt four to the royal.

To expand on MKL's response, the odds of making at least two royals on those 10 draws is ( 1 / 47 ) ^ 2 = 0.000452694

The odds of making exactly two royals out of those 10 draws is ( ( 1 / 47 ) ^ 2 ) * ( 46 / 47 ) ^ 8 ) = 0.00038114

Oops. Yeah, JB's right.

mutiply by combin ( 10 , 2 ) = 45

I.E. There are 45 combinations of having 2 winners and 10 losers.

Quote: DJTeddyBear

Oops. Yeah, JB's right.

Oh! I must say I expected a lot less.

Now I've something to shoot for. I have been dealt 4 to a royal three or four times, and I got a royal with deuces once plus nothing the other times.

To your other post, the question does assume you've already been dealt 4 to the royal.