Rare hand

[rxphil]

So, playing super triple play jackpots, I held three 6s and drew quad 6s on ALL 3 HANDS! I was trying to find the odds on this happening... Anyone know?

[TripleTriple]

I believe it's 1 in 12,564. Congrats!

[olds442jetaway]

Ditto. I am probably wrong, but I got about one in about 106,600

[Come Back Kid]

Wow, nice one!

[dinghy]

I am probably wrong, but I got about one in about 106,600

I would need to see your work on that one. I'm just a little off from TripleTriple's number. For any one line, it's 23.5 because you have 2 chances drawing from 47 cards.

For all three lines, 23.5^3 = 12,978.

[New2vp]

I'll give dinghy my vote for accuracy: 1 in 12,977.875 exactly or 12,976.875 to 1; but since TripleTriple was close and quite quick, he gets the speedy award.

It looks to me like olds was sort of working with a 1 in 47 chance to get the quads, which makes sense since only 1 card in the remaining deck completes the quad, but he forgot to take into account that you get to draw 2 cards from the deck looking for the remaining 6-spot. 47^3 is 103,823 which is just a few percent away from his suggestion. Olds, were you using your slide rule on that one? I remember when it was a luxury to get the 3rd decimal place correct.

Of course, rxphil, all of these odds and probabilities are associated only with the draw ... after the deal. So, if you want to jazz these numbers up a bit, you will get trip sixes dealt about once in 615 hands (not including those times when you have a full house or quads). So you can multiply (or divide) all those numbers by 615. Or if you counted other quads as well as quad sixes, you would get such a deal more frequently, about once every 47.33 hands.

In any event, rxphil, we all congratulate you and hope you enjoyed the rush. You might not see something like this again on your own play for quite a while.

[onemoretry]

I'll give dinghy my vote for accuracy: 1 in 12,977.875 exactly or 12,976.875 to 1; but since TripleTriple was close and quite quick, he gets the speedy award.

Speedy, but not quite right. I am curious as to how that 12564 number was calculated.

[New2vp]

I am curious as to how that 12564 number was calculated.

My guess is that he set the probability of drawing quads as (1/47 + 1/46) instead of 2/47. If you take the reciprocal of the cube of that number, you get about 12,563.72.

[Eduardo]

I ran a more detailed analysis than these other guys and the answer is nobody on this forum will ever see that happen, just you.

[TripleTriple]

I am curious as to how that 12564 number was calculated.

My guess is that he set the probability of drawing quads as (1/47 + 1/46) instead of 2/47. If you take the reciprocal of the cube of that number, you get about 12,563.72.

Yes, because there is only a single card needed to get the desired hand, the second card is irrelevant. In that case you have to look at it as two separate events, which have different probabilities. I'm not saying this is right, I'm saying that is my rationale. I'll dig out my stats book sometime...