Expected cost of a Pink Chip

[jar]

When playing at the Trop, a friend payed someone at the table $3 for a $2.50 pink chip as a souvenir of the legendary pink chip game, which is currently on hiatus.

This got me to wondering, what is the expected value of playing BS $5 blackjack until one wins a pink chip? I'm not sure the BJ rules at the trop, but I assume they're pretty bad, since it's AC. Anyone know where to find blackjack rules for variuos casinos online?

[eobmtns]

Lots of clubs offer pink $2.50 chips, and in those that do you can get as many as you want by asking for "check change." If you can find somebody willing to buy them for $3, you have achieved a +EV of 20%. If you feel that you must earn one by playing $5 BJ, regardless of your strategy the odds of getting dealt a BJ (thus receiving $7.50) are on the order of 20-1. Probably the net expected loss by then, assuming perfect basic strategy (and not counting cards), would be on the order of $5-10, less the cost of your free drink.

Stanford Wong (bj21.com) publishes a newsletter of BJ rules in effect at various casinos, and Michael Shackelford (wizardofodds.com) has a table of the effect of the various rules on the house PC. There is probably other information on BJ rules out there that you could obtain if you were to STFW (Search the Friendly Web).

[IsaacW]

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If you feel that you must earn one by playing $5 BJ, regardless of your strategy the odds of getting dealt a BJ (thus receiving $7.50) are on the order of 20-1. Probably the net expected loss by then, assuming perfect basic strategy (and not counting cards), would be on the order of $5-10, less the cost of your free drink.

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No way is the expected loss this high if you play perfect basic strategy. The question that jar is asking, I believe, is if it would be "cheaper" to buy a pink chip by playing $5 blackjack than by paying someone $3 for one.

According to the wizard most AC games carry a house edge of 0.43% for playing perfect BS.

In an 8-deck game (guessing here), the probability of getting blackjack (assuming no knowledge of the state of the shoe) is (32*128/2)/(416c2) = 2.37%, or about 41.1:1. So expect to give on average 42.1*$5 = $210.50 in action before receiving a blackjack. The expected loss on this action is $0.905. Of course, the variance in blackjack is pretty severe, and you may lose much more than this (or win a couple hundred /images/graemlins/grin.gif) when winning any particular pink chip.

Over a large number of trials, it is far cheaper to earn the chips playing $5 blackjack than to buy the $2.50 tokens for $3 each /images/graemlins/tongue.gif Who knew?

[pzhon]

You get blackjack about 1/21 of the time. You win with blackjack about 1/22 of the time. On average, you will play 22 hands before your first win with blackjack. I'm not sure what the house advantage is in live games, but if it is 0.5%, you will pay an average of 0.11 bets before you win with blackjack, $0.55.

That must be added to the nominal value of $2.50, for a total cost of $3.05. It looks like it is cheaper to pay $3 unless the house advantage is lower than about 0.45%.

It may be cheaper to play for it if you can get the chip in other ways, such as by taking insurance or surrendering. I'm not sure what the optimal strategy is if you can get a pink chip in these ways.

[IsaacW]

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You get blackjack about 1/21 of the time. You win with blackjack about 1/22 of the time.

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What am I missing in my calculation then? Let's assume a single deck. To get blackjack you must get exactly an ace and a "ten" card. There are 4 aces and 4 each of tens, jacks, queens, and kings. There are, therefore, 4*(4 + 4 + 4 + 4)/2 = 32 combinations that give you blackjack. There are 52 choose 2 = 1326 combinations of starting cards that you can get. The probability of getting blackjack is then 32/1326 = 2.41%. The odds against this happening are (1/0.0241 - 1):1 = 40.44:1. Thus you will get blackjack once every 41.5 hands, on average.

[fnord_too]

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You get blackjack about 1/21 of the time. You win with blackjack about 1/22 of the time.

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What am I missing in my calculation then? Let's assume a single deck. To get blackjack you must get exactly an ace and a "ten" card. There are 4 aces and 4 each of tens, jacks, queens, and kings. There are, therefore, 4*(4 + 4 + 4 + 4)/2 = 32 combinations that give you blackjack. There are 52 choose 2 = 1326 combinations of starting cards that you can get. The probability of getting blackjack is then 32/1326 = 2.41%. The odds against this happening are (1/0.0241 - 1):1 = 40.44:1. Thus you will get blackjack once every 41.5 hands, on average.

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Why are you dividing by 2? There are 64 black jack hands, not 32.

[IsaacW]

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Why are you dividing by 2? There are 64 black jack hands, not 32.

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There are 64 permutations of blackjack hands, but only 32 combinations of blackjack hands, because order is not important.

The math comes out the same way when you use permutations. There are 4*16 = 64 blackjack hand permutations, and 52*51 = 2652 total permutations of the 52 cards in the deck. 64/2652 = 2.41%, same as if you calculate with combinations.

[IsaacW]

I got it... the set of aces and the set of "tens" are mutually exclusive, so you don't divide by 2 to get the combinations... there are actually 128 permutations (4*16 + 16*4) of blackjack hands.

So right, what pzhon said.

[eobmtns]

Playing $5 per hand with real chips in a real casino, there is no way you could lose exactly $0.90. You could lose 0 or $2.50 (or, of course, you could win or lose hundreds).

[daryn]

just man up and play some damn blackjack. it's fun.

[TomCollins]

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Playing $5 per hand with real chips in a real casino, there is no way you could lose exactly $0.90. You could lose 0 or $2.50 (or, of course, you could win or lose hundreds).

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Someone needs a lesson in EV.