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Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Need Help Calculating Odds - Same Pair Six Times Making Huge Hands

08-29-2022 , 06:38 AM
Trying to help a friend who is looking for the odds of this happening:

Quote:
Here's what I'm trying to calculate: Over the span of 300 hands, I was dealt the same pocket pair six times, and here's what happened each time:

1) Triple-barrel bluff (that was probably value on the river)
2) Turned a set
3) Flopped a set
4) Flopped a set
5) Flopped a set, turned quads
6) Flopped quads
Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Quote
08-29-2022 , 10:20 AM
Over the next 300 hands played? Or just a cherry-picked string of 300 hands in the future?
Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Quote
08-29-2022 , 10:44 AM
Quote:
Originally Posted by Didace
Over the next 300 hands played? Or just a cherry-picked string of 300 hands in the future?
Just the chance of it happening at all over the course of 300 hands
Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Quote
08-31-2022 , 08:57 PM
Didace's point is that your friend has probably played thousands or more hands, and thus had multiple opportunities for this or something similar to happen in one of his 300-hand subsets. However, I'll go ahead and answer it as though we're talking about the next 300 hands played.

I interpret it as, "What's the chance of at least one pocket pair occurring at least 6 times, accompanied by at least 5 sets/boats/quads using the pair?" I'll just give an approximate answer.

For the pocket pairs, start with 1-BinomialCDF(300, 1/221, 5) and call that p. The chance is between [1 - (1-p)^13] and 13p, and I'm guessing it's closer to the former, but I'll go with the overestimate here (13p) because I'll be underestimating in the next part.

The chance of hitting a set or better is 1-(45C2)/(50C2) = q
The chance of it happening at least 5 out of 6 times is 6(1-q)q^5 + q^6
(But the pocket pair could have been dealt more than 6 times, or two different pairs could have occurred 6+ times, which is why that's an underestimate.)

Multiply that by 13p for an answer of 1 in 21955
Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Quote
09-01-2022 , 06:18 AM
Quote:
Originally Posted by heehaww
Didace's point is that your friend has probably played thousands or more hands, and thus had multiple opportunities for this or something similar to happen in one of his 300-hand subsets. However, I'll go ahead and answer it as though we're talking about the next 300 hands played.

I interpret it as, "What's the chance of at least one pocket pair occurring at least 6 times, accompanied by at least 5 sets/boats/quads using the pair?" I'll just give an approximate answer.

For the pocket pairs, start with 1-BinomialCDF(300, 1/221, 5) and call that p. The chance is between [1 - (1-p)^13] and 13p, and I'm guessing it's closer to the former, but I'll go with the overestimate here (13p) because I'll be underestimating in the next part.

The chance of hitting a set or better is 1-(45C2)/(50C2) = q
The chance of it happening at least 5 out of 6 times is 6(1-q)q^5 + q^6
(But the pocket pair could have been dealt more than 6 times, or two different pairs could have occurred 6+ times, which is why that's an underestimate.)

Multiply that by 13p for an answer of 1 in 21955
Thanks, I'll pass it along to my friend.
Need Help Calculating Odds - Same Pair Six Times Making Huge Hands Quote

      
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