Easy as pi...

DeMango,

I believe the sidebet you meant is the one called "Pair Square", for which details can be found at the Wizard of Odds site at this URL:

http://wizardofodds.com/blackjack/appendix8.html#pairsquare

Since you asked specifically about the 25-10 paytable, I'm assuming the SB is on a DD table. If by chance you've found a casino that offers that paytable on a 6D (or even better, an 8D) game, email me the location and I'll immediately bust this post ;-)

First of all, forget about keeping track for the 25:1 payout.

Instead, just concentrate on the 10:1 payout, and realize that one-seventh of the wins will receive a 15-unit bonus: thus, effectively you're playing a SB that pays 10+15*(1/7) = 12.143... to 1. I say 1/7th since, if your first card is a 6C, then of the remaining 7 6's in the deck, one will be a 6C, and the other six will be other 6's.

The SB is even-money if you win it once every 12.134 + 1 = 13.143 times: you'll lose 12.134 on the losses, and win 12.134 on the one win, so you'll break even. Thus, you need the odds of winning the SB to be 1/13.143 = 7.609%.

Off-the-top, though, your odds of winning are only 7/103 = 6.796%. Thus, our off-the-top edge is (6.796%-7.609%)/7.609% = -10.68%, or a HOUSE EDGE of 10.68%. That's pretty high.

Note, though, that if we remove ALL of the 5's and 6's (and NONE of the other ranks) from the DD, then the odds of winning the SB rise to 7/87 = 8.0460%: thus, it becomes +EV with our edge being (8.0460%-7.609%)/7.609% = +5.74%.

So, how do you beat this SB? Just keep track of how many cards of each of the thirteen ranks are still in the deck. Let's let Ni be the number of cards of rank "i" remaining, with i=1 for Ace, i=2-10 for 2's through 10's, i=11 for J, i=12 for Q, and i=13 for K. Then the total number of cards is Ntot = sum(Ni), where the sum goes from i=1 to 13. At the start of each round, calculate the EV of the SB as follows:

EV = sum(Ni/Ntot)*(((Ni-1)*(12.143...)-(Ntot-Ni))/(Ntot-1))

Where the 12.143... is 10+15*(1/7), and again the sum goes from i=1 to 13. If EV>0, bet the "Pair Square"!

Example: let's say we have 4, 6, 7, 6, 3, 5, 8, 2, 4, 7, 8, 1, 6 cards left in each of the 13 ranks at the start of the round, for a total of 67 cards left. Then, applying the above formula gives a SB EV of 0.46%: woo hoo... bet the farm!

Of course, it helps to have JSTAT's calculating abilities to do the required math in the time allotted ;-)

Hope this helps!

Dog Hand

P.S. You DO recall that "pi" is irrational... right?