possible to predict ace pairs?

[BigT]

The casino i go to has a side bet where for £1 you get £100 for 2 aces. Its a 6 deck hand shuffled game with what i think is 70% penetration. Is it possible to keep some sort of ace count to know when there is an increased chance of getting ace pairs? I have tried ace tracking using 2 key cards but with 6 decks there are so many key cards that could just be from another deck.

Cheers

[FLASH1296]

If the chance of being dealt paired aces is roughly 143 to 1; a patron who is willing to accept 100 to 1 is a dolt.

You are searching for a way to reduce the odds. IF you tracked aces well enough to know that you had a 1 in 9
chance of getting dealt one, then your odds of getting the second ace would make this proposition a 8 x 12 = 96 =
95 to 1 odds. A tidy profit of 5% IF you can get to that 1 in 9 (12.5%) chance on a singleton ACE.

[Automatic Monkey]

175.7826087 to 1.

An ace sidecount will do the job, but it's probably not worth it if you can't put down any more than a GBP.

[London Colin]

Flash, I think your arithmetic has gone a little awry in various places.

It might be easier, and less confusing, to do all the work in terms of probabilities, and convert to odds at the end if desired.

By my calculation the odds of being dealt a pair are 1/13 * 1/13 = 1/169 (or 168 to 1) for infinite deck; or 24/312 * 23/311 = 0.0057 (or approx. 175:1) for 6 decks.

And 1 in 9 is not 12.5%! (That would be 1 in 8, or 7 to 1 odds.)


I think I know the promo in question, and there are other, higher variance payouts (including a progressive jackpot) for greater numbers of aces. (I forget the details.)

But, just focusing on the single pair, I believe it actually pays 99:1, in that you don't get your stake back. So we need a 1/100 chance of getting the pair in order to break even.

So if the chance of a random ace is 1/13, the chance of the first, tracked ace would have to be 1/(100/13) = 1/(7.7), in order to break even.

[London Colin]

Originally Posted by Automatic Monkey

175.7826087 to 1.

An ace sidecount will do the job, but it's probably not worth it if you can't put down any more than a GBP.

Like Flash, my assumption is that tracking/steering the aces would be the only way to potentially profit from this. (For those that can do such things.)

If you go purely by the density of the remaining aces, it doesn't seem like there would be very many opportunites to make a +EV bet.

[Automatic Monkey]

Originally Posted by London Colin

Like Flash, my assumption is that tracking/steering the aces would be the only way to potentially profit from this. (For those that can do such things.)

If you go purely by the density of the remaining aces, it doesn't seem like there would be very many opportunites to make a +EV bet.

Right. But if you normally sidecount aces and you're playing the game anyway, it's surely worthwhile to know the point where you would make the bet.

[London Colin]

Originally Posted by London Colin

So if the chance of a random ace is 1/13, the chance of the first, tracked ace would have to be 1/(100/13) = 1/(7.7), in order to break even.

Or would it? This kind of thing always confuses me.

The 'tracked' ace could be the first or second card, as could the random ace. Not sure if that invalidates my above calculation.

[London Colin]

Originally Posted by Automatic Monkey

Right. But if you normally sidecount aces and you're playing the game anyway, it's surely worthwhile to know the point where you would make the bet.

That's true.

Equally, I suppose, if you are an ace tracker then you get an added benefit from the sidebet, on top of the usual benefits of your art.

[tezzadiver]

Originally Posted by London Colin

Flash, I think your arithmetic has gone a little awry in various places.

It might be easier, and less confusing, to do all the work in terms of probabilities, and convert to odds at the end if desired.

By my calculation the odds of being dealt a pair are 1/13 * 1/13 = 1/169 (or 168 to 1) for infinite deck; or 24/312 * 23/311 = 0.0057 (or approx. 175:1) for 6 decks.

And 1 in 9 is not 12.5%! (That would be 1 in 8, or 7 to 1 odds.)


I think I know the promo in question, and there are other, higher variance payouts (including a progressive jackpot) for greater numbers of aces. (I forget the details.)

But, just focusing on the single pair, I believe it actually pays 99:1, in that you don't get your stake back. So we need a 1/100 chance of getting the pair in order to break even.

So if the chance of a random ace is 1/13, the chance of the first, tracked ace would have to be 1/(100/13) = 1/(7.7), in order to break even.

Any idea of the payout on that progressive jackpot Colin? I have just come to hear about this promo myself.

Am I correct to say 4 suited aces sweeps it?

[London Colin]

Originally Posted by tezzadiver

Any idea of the payout on that progressive jackpot Colin? I have just come to hear about this promo myself.

Am I correct to say 4 suited aces sweeps it?

Sorry, I don't know. I hope to pay a visit in the next couple of days; I'll try and remember to note down all the details. (Something I've forgotten to do on two previous visits.)

From memory, there are three tiers of fixed payouts (I think for any 2, 3, or 4 aces), plus the jackpot. I don't think the jackpot was for 4 suited aces; it may have been something like 3 ace of hearts, but all the above may be misremembered.

You can also get 10% of the jackpot if you are playing the sidebet when someone else at the table wins it. (Hopefully this is in addition to the winner's 100% payout, rather than deducted from it!)