let it ride math question

When dealt a non-paying pair what percent chance do you have of turning that into a winning hand? Please provide the math that supports the answer. I don't want to know the house edge, just the percent chance of this becoming a winning hand. Thanks.

I am assuming that you do not have the information on other people's hands, and therefore the only information to be coming out is two cards from the dealer.

If you have a non-winning pair, you can get a winning pair by the dealer pulling a pair of its own, a card matching the 3rd non-pair card in your hand, or one or two cards matching your pair.

The number of combinations available to the dealer is 49 x 48 / 2 = 1,176

Let's call your hand xxy where xx is a pair and y is the 3rd card.

The odds of 4 of a kind is 1 /1,176
The odds of a dealer pair other than x or y is 3 x 2 x 11 = 66 / 1,176
The odds of full house xxyyy is 3 / 1,176
The odds of full house xxxyy is 6 / 1,176
The odds of a three of kind xxx is 4 x 11 x 2 / 1,176 = 88 / 1,176
The odds of two pair xxyy is 6 x 11 / 1,176 x 2 = 132 / 1,176
The odds of a non winning hand is 8 x 55 x 2 = 880 / 1,176

Therefore the odds of a winning hand is 25.17%, which is why you don't let a non-paying pair ride.

Thank you for the information. This settles a bet with my son provind Dad was right again.

I understand that for that specific hand it is a bad bet, but statistically over the course of those 1176 hands if you let the first bet ride, I believe you would come out ahead vs not letting a non-paying pair ride. (You would pull back the second bet if you had a still had a non-paying hand, therefore for each of those 1176 scenarios you would have 1 more bet riding by playing a non paying pair)

Quote: qwerty

Thank you for the information. This settles a bet with my son provind Dad was right again.

I understand that for that specific hand it is a bad bet, but statistically over the course of those 1176 hands if you let the first bet ride, I believe you would come out ahead vs not letting a non-paying pair ride. (You would pull back the second bet if you had a still had a non-paying hand, therefore for each of those 1176 scenarios you would have 1 more bet riding by playing a non paying pair)

Just because a non-paying pair will only give you a winner 25% of the time (and I understand boymimbo you were just answering the original question), doesn't mean in and of itself that you shouldn't let it ride. Only if the EV of the payouts of the 25% of hands that give you a win (some are not 1:1, you'll get more 3OAKs) doesn't overcome the added bets should you not let it ride. I didn't do that math though.

The Wizard did the math on WOO:

I trust your Let it Ride advice is correct but I still like raising on a low pair with three cards. Often I have seen these turn into paying hands. So how much is it costing me to raise on low pairs?

With a low pair your expected value on the initial bet is -7.40%. So if your original bet was $10 then letting it ride with a low pair will cost you an extra 74 cents. May 13, 2004

The let it ride is the oldest bet in gambling. For centuries players with little money would 'let it ride' because after the first win, you're now betting the casinos money and if you lose, all you lost was your initial bet.