a *process* question about odds

[cnfuzzd]

At my home game tonight we started discussing pocket pairs and flopping sets. I spouted off my rote knowledge that the odds are 7.5:1 against, or around one every ten times, or so. Someone else questioned that, and said they thought it was "supposed" to happen one out of every six times. I got down to the math and got as far as below. I'm not really sure where to go from there, and think this is probably a converting odds to percentage question.

2 cards to improve to set on flop.

(1/25)*(2/49)*(1/24)=0.000068 (this is the part of math that made me stop with trig. i hate having answers that make little sense to me because i have partial recall of something i learned years ago)

So what is this number, and what is the best thing to do about it.

Thanks for any help in answering this question.

peace

John nickle

[BruceZ]

7.5-to-1 are the odds of flopping a set or better, including a full house or quads. This is 1 minus the probability of NOT flopping one of the 2 cards needed for the set.

1 - (48/50 * 47/49 * 46/48) = 11.76% = 1 in 8.5 = 7.5-to-1.

[Mike Haven]

let's say you have a pair of sevens

there are two more sevens left in the other fifty cards

if you draw a card (say the first card of the flop) there are 48 chances out of 50 that it isn't a seven

if you then draw another card there are 47 chances out of 49 that it isn't a seven either

and if you draw a third card there are 46 chances out of the 48 cards left that it isn't a seven

multiply these chances together to get (48/50) x (47/49) x (46/48) or (48x47x46)/(50x49x48) = 103776/117600 = 0.882

multiply this 0.882 by 100 to get the percentage chance of you not flopping a third seven = 88.2%

if you don't flop a seven 88.2% of the time obviously you do flop a seven 100-88.2 = 11.8% of the time

so you flop a seven 11.8 times in 100 tries or (11.8/11.8) in (100/11.8) tries or 1 in 8.47 tries, say 1 in 8.5 tries

the odds of flopping a seven are 88.2/11.8 = 7.47/1, say 7.5 to 1

[Im Just A Bill]

So, this covers not getting another 7 (or getting another depending on how you look at it), but what about flopping trips to give you the full house. Can you help me understand the calculation for that piece?

Bill

[BruceZ]

I assume you mean the case where all 3 flop cards are the same rank, and we hold a pair, so we flopped a full house. There are 48 ways to pick the first flop card out of 50, then the next card must be the same rank, so there are 3 remaining choices for that card out of 49, and then 2 choices left for the last card out of 48.

48/50 * 3/49 * 2/48 = 0.24% = 1 in 408 = 407-to-1.

So it is pretty negligible.

Here's another way to look at it. There are 12 ranks that can make trips, and there are 4 ways to make trips with each rank (since there are 4 ways to pick the card that is left out). So there are 12*4 = 48 flops out of a total of 50*49/2 = 19,600 possible flops. 48/19,600 = 0.24% = 1 in 408 = 407-to-1 exactly as before.

This kind of full house was not included in the odds of 7.5-to-1 computed previously. That only included cases where we flop a set. If we include these full houses, then the odds become 7.3-to-1. Note that this type of full house is usually a much weaker hand since it is vulnerable to any pair higher than your pockets.

[Im Just A Bill]

Yep, that's what I was looking for. Thanks for the info.