What is the probability of two overs vs a pocket pair

[Mikey]

No no no....... not 50/50.

Hand 1: AK vs 88

I know that usually when we look at pocket pairs vs overs we think of the odds of either one winning is 50/50, but I want to know what the odds are that the AK will outdraw the 88 on the flop.

Instead of playing a 5 card game, play it only with 3 cards, only a flop can be dealt.

Then play it as a 4 card game.

Can you guys show the calculations too with how you came up with the answer.

Thanks in advance.

[Tommy LaKey]

Hi Mikey,
The first step I use in determining odds is comparing known cards to unknown cards. In this case, you have AK and 88 - a total of 4 "known" cards and 48 "unknown" cards. Let's start with the AK combination.....
We know there are 3 A's and 3 K's not shown, so your chance of hitting an A or K is 6 out of the 48 "unknown" cards or 1 chance in 8 of getting one more A or K (48/8). The 8's are somewhat different.....
There are only two 8's left in the deck. Your chance of hitting an 8 is only 2 chances in 48 or 1 chance in 24 (48/2). In this example, the AK has a 3-1 advantage of winning with a high pair in a showdown.
Hope this helps - good luck!

[dabluebery]

Unless I'm way off here, since you're playing just a 4 card game, the only outs AK has are to hit an Ace or a King. Is that what you'd want calculated? The chance of hitting an Ace or a King on the flop? If so, it's;

1-((42/48)*(41/47)*(40/46)) which is about .336 or 33.6%.

To be more accurate, you'd then have to calculate the chances that an (A or K) drop AND an 8 drops, giving pocket 8's a set. Is that what you're looking for?

[BruceZ]

[ QUOTE ]
Hi Mikey,
The first step I use in determining odds is comparing known cards to unknown cards. In this case, you have AK and 88 - a total of 4 "known" cards and 48 "unknown" cards. Let's start with the AK combination.....
We know there are 3 A's and 3 K's not shown, so your chance of hitting an A or K is 6 out of the 48 "unknown" cards or 1 chance in 8 of getting one more A or K (48/8). The 8's are somewhat different.....
There are only two 8's left in the deck. Your chance of hitting an 8 is only 2 chances in 48 or 1 chance in 24 (48/2). In this example, the AK has a 3-1 advantage of winning with a high pair in a showdown.

[/ QUOTE ]

88 does not need to draw an 8 to win. AK is not a 3-to-1 favorite, but a 2.25-to-1 underdog on the flop, and a 1.69-to-1 underdog on the turn.

3 card solution:

P(AK wins) =

[ 6*C(40,2) + C(6,2)*40 + C(6,3) + 2*(3+3) ] / C(48,3)

= 30.7% = 2.25-to-1

The first 3 terms are for 1, 2 or 3 A/K, and no 8. The final term is for a single 8 (2 ways) and either AA (3 ways) or KK (3 ways). This is all divided by C(48,3) total flops. This ignores QJT which adds another 0.37%.


4 card solution (ignoring straights and flushes):

P(AK wins) =

[ 6*C(40,3) + C(6,2)*C(40,2) + C(6,3)*40 + C(6,4) + 2*(3+3)*40 + 2*C(6,3) ] /C(48,4)

= 37.2% = 1.69-to-1.

The first 4 terms are for 1,2,3 or 4 A/K with no 8. The fourth term is for a single 8 (2 ways) and AAx or KKx (3+3)*40, and the final term is for a single 8 with 3 A/K, making either a full house or quads. This is all divided by C(48,4) total 4-card boards.

[BruceZ]

Obviously, if an 8 falls, AA or KK doesn't do AK any good. Here are the corrected calcs which removes 1 term from each case, but the final answer doesn't change much.

3 card solution:

P(AK wins) =

[ 6*C(40,2) + C(6,2)*40 + C(6,3) ] / C(48,3)

= 30.6% = 2.26-to-1

The first 3 terms are for 1, 2 or 3 A/K, and no 8. This is all divided by C(48,3) total flops. This ignores QJT which adds another 0.37%.


4 card solution (ignoring straights and flushes):

P(AK wins) =

[ 6*C(40,3) + C(6,2)*C(40,2) + C(6,3)*40 + C(6,4) + 2*C(6,3) ] /C(48,4)

= 36.9% = 1.71-to-1.

The first 4 terms are for 1,2,3 or 4 A/K with no 8. The final term is for a single 8 with 3 A/K, making either a full house or quads. This is all divided by C(48,4) total 4-card boards. Also, QJTx will win an additional 1.1% of the time, including x=8.

[BruceZ]

For the 4-card case, AK can also win when the board has 2 pairs both higher than 88. This adds another 1.2% to the cases we have already considered.