Hold'em probabilities

[Whatsmytotal]

I know this is probably a simple probability question but here goes...

What are the chances of at least one of 3 specific cards appearing on the flop?

(e.g., probability of at least one Ace, King or Queen in 3 draws from a standard deck.)

Someone was offering even money that any one would come up. I'm wondering who holds the advantage. Thanks

[moo321]

13 ranks. 3/13 chance of one of those ranks hitting * 3 cards on the flop = 9/13 chance. Odds are one of those ranks will hit.

[shadroch]

Originally Posted by moo321

13 ranks. 3/13 chance of one of those ranks hitting * 3 cards on the flop = 9/13 chance. Odds are one of those ranks will hit.

Don't think you are using the correct formula here. Suppose the flop was 5 cards instead of three. You take 3/13 and multiply by 5 and you have 15/13,or better than a 100% chance. You know that can't be right.

[miplet]

Originally Posted by Whatsmytotal

I know this is probably a simple probability question but here goes...

What are the chances of at least one of 3 specific cards appearing on the flop?

(e.g., probability of at least one Ace, King or Queen in 3 draws from a standard deck.)

Someone was offering even money that any one would come up. I'm wondering who holds the advantage. Thanks

http://wizardofodds.com/askthewizard...wizard196.html second question.

Before any cards are seen, the probability of any three ranks not appearing on the flop are combin(40,3)/combin(52,3) = 9880/22100 = 44.71%. So this guy had a 10.59% advantage.

[moo321]

Originally Posted by shadroch

Don't think you are using the correct formula here. Suppose the flop was 5 cards instead of three. You take 3/13 and multiply by 5 and you have 15/13,or better than a 100% chance. You know that can't be right.

Yeah, I need to change the denominators to 52-51-50. But it's still obvious the bet's wrong.

[callipygian]

Originally Posted by moo321

Yeah, I need to change the denominators to 52-51-50. But it's still obvious the bet's wrong.

(12/52)+(12/51)+(12/50) = 0.706. That's not right either.

The fault in this reasoning is that the addition only works for small probabilities. If you had a 1/1000 chance of something, and two opportunities, the probability is about 2*1/1000. But if you have 1000 chances, the probability is not 1000*1/1000.

The core of the fallacy is that you've double-counted the probability of something happening twice. When the probabilities are small, the probability of something happening twice is small, and can be ignored. When the probabilities are big, the probability of something happening twice is large, and cannot be ignored.

The best way to think about the problem is to take the inverse case - the probability that NONE of the three cards are ace, king, or queen. In this scenario, the probability is exactly (1-12/52)*(1-12/51)*(1-12/50) = 0.447, as the wizardofodds points out. Then you subtract that from 1 to find the probability that ONE OR MORE of the three cards is an ace, king, or queen. 1-0.447 = 0.553.

If you remember the distributive property from algebra, you can see where this number starts to differ from your equation above.

1- (1-12/52)*(1-12/51)*(1-12/50)
= 1 - 1*1*1 + 12/52*1*1 + 1*12/51*1 + 1*1*12/50 - 12/52*12/51*1 - 12/52*1*12/50 - 1*12/51*12/50 + 12/52*12/51*12/50
= (12/52 + 12/51 + 12/50) - (12/52*12/51 + 12/52*12/50 + 12/51*12/50) + (12/52*12/51*12/50)
= 0.706 - 0.166 + 0.013

Note that the first term is what you got above. It's an overestimation of how often AKQ comes up. The second term is the correction for two high cards coming up - you subtract it out. But then the third term is the correction to the correction - you need to add back in the probability of all three cards coming up.

When you add up all the contributions, you get the correct answer: 0.706 - 0.166 + 0.013 = 0.553.

If instead of 12/52, 12/51, 12/50, you had smaller numbers, you can see now that the second and third terms start disappearing very quickly. For instance, instead of saying any ace, king, or queen, let's just calculate the probability that the AKQ of spades will come up.

Exact method:
1-(1-3/52)*(1-3/51)*(1-3/50) = 0.166

Approximation:
(3/52)+(3/51)+(3/50) = 0.176

The smaller the individual probabilities, the more additive they are.

[InPlay]

Originally Posted by Whatsmytotal

I know this is probably a simple probability question but here goes...

What are the chances of at least one of 3 specific cards appearing on the flop?

(e.g., probability of at least one Ace, King or Queen in 3 draws from a standard deck.)

Someone was offering even money that any one would come up. I'm wondering who holds the advantage. Thanks

There are at least 14 outs after the flop has a better than 50% chance to catch one of its card.

[InPlay]

This article should help you out figuiring out the odds.

http://en.wikipedia.org/wiki/Poker_p...s_hold_'em)

[callipygian]

Originally Posted by InPlay

This article should help you out figuiring out the odds.

http://en.wikipedia.org/wiki/Poker_p...s_hold_'em)

Your URL tags are messed up.

http://en.wikipedia.org/wiki/Poker_p..._hold_%27em%29

[InPlay]

Originally Posted by callipygian

Your URL tags are messed up.

http://en.wikipedia.org/wiki/Poker_p..._hold_%27em%29

Thanks for correcting thats the URL I was pointing to.