Card Sharks

[Nate tha' Great]

The following is a problem based on the game show Card Sharks. It isn't a poker problem per se, but is reasonably interesting to think about, and probably has some implications in terms of bankroll management.

The contestant is dealt a row of five cards from a standard 52 card deck. The cards are not replaced. The contestant starts with a bankroll of $1,000.

The cards are revealed one at a time. The contestant must determine two things as each card is flipped up:

a) A guess as to whether the next card will be higher or lower. Aces are high, and a push is considered a loss.
b) The contestant may bet any amount from $0 to his current bankroll on his guess.

For example, a typical game might go like this.

The first card is a 4. The contestant bets $800 that the next card will be higher.

The next card is a 5, bringing the contestant's bankroll to $1,800. He bets $700 that the next card will be higher.

The third card is a Q, bringing the contestant's bankroll to $2,500. He bets $2,000 that the next card will be lower.

The fourth card is a J, bringing the bankroll to $4,500. The contestant bets $3,000 that the next card will be lower.

The fifth and final card is a J. The push is conisidered a loss and the contestant finishes the game with $1,500.

Here is the problem: suppose that the first card is a T. Obviously, the contestant should guess that the next card will be lower. But what should his wager be?

Assumption: the contestant is trying to maximize his final total and is neither risk-averse nor risk-loving.

p.s. Nope, haven't been playing too many real cards lately. I'll be back in the grind soon enough.

[randomstumbl]

With a ten showing, there are 19 cards that lose and 32 cards that win. So, you bet everything. Obviously, it's a high variance strategy, but it's still the most profitable.

[Hawklet]

[ QUOTE ]
With a ten showing, there are 19 cards that lose and 32 cards that win. So, you bet everything. Obviously, it's a high variance strategy, but it's still the most profitable.

[/ QUOTE ]

If the game lasted only 1 round that would be true. But this game is +EV for 10 (or so) cards. Your strategy gives the highest EV for 1 card but if you bust out on that card you lose the EV for the other 9. If you know your going to get pocket Aces 5 times in a row, would you risk your entire bankroll all five times? No, because thats very bad managment, and its the same principle for this.

The correct ammount to wager is your edge odds times your bankroll. With a 10 its like 37/63 in your favor? So your edge is 26% ((.63-.5)+(.5-.37)). So bet 26% of your bankroll or 260 dollars.

[pzhon]

[ QUOTE ]
The correct ammount to wager is your edge odds times your bankroll. With a 10 its like 37/63 in your favor? So your edge is 26% ((.63-.5)+(.5-.37)). So bet 26% of your bankroll or 260 dollars.

[/ QUOTE ]
This is approximately what the Kelly criterion recommends, but that assumes you are risk averse, and the balance you have on the game show represents every asset you have in the world.

Under the original poster's assumption that you are not risk averse, you should bet everything and let it ride whenever you don't have an 8. randomstumbl and fnord_too are correct.

To make this clearer, suppose you get 5 chances to double-up as a 3:1 favorite. What fraction fi of your balance should you bet on the ith opportunity? Your average balance at the end is (1+f1/2)(1+f2/2)(1+f3/2)(1+f4/2)(1+f5/2). To maximize this, you want to maximize each term individually, which means fi=1. Conclusion: Bet everything and let it ride.

It is common for people to make the same mistake, applying something like the Kelly criterion to their stack in a tournament. They reason that they will get good opportunities in the future, and if they bust out, they will miss them. The result is that they often have too few chips to take full advantage of good opportunities later.

[bobman0330]

[ QUOTE ]
Bet everything and let it ride.

[/ QUOTE ]

I was going to disagree, but then I did the following math:

The EV of your fourth bet:
EV (4) = edge * bet + winning percentage * (EV (5) w/ winning bankroll) + losing percentage * (EV (5) w/ losing bankroll).
= (W-50%) * X + W((avg. edge) * (BR + X)) + (1 -W)((avg. edge) * (BR - X))
W = winning percentage; X = bet size; BR = bankroll

This equation is maximized by betting it all, even when it is a 50-50 proposition. (Most 8's are not even money because of prior cards).

I find this result surprising, as it this situation is exactly analogous to what Sklansky describes in TP4AP. It seems that the risk of not being able to take advantage of future +EV gambles is more than offset by the ability to take fuller advantage of those gambles by having a bigger roll. I guess in poker contexts, (assuming you're one of the few best players in the tournament), you'll usually have one of the biggest stacks, so having more money won't allow you to win more from future gambles. Not to hijack the thread, but does anyone else have comments on this situation v. TP4AP?

[pzhon]

[ QUOTE ]
Not to hijack the thread, but does anyone else have comments on this situation v. TP4AP?

[/ QUOTE ]
I don't recall any contradiction with what TPFAP says.

TPFAP: The very best players should be willing to give up small edges when they come with a lot of variance.

Standard misinterpretation: I can become a great player by passing up large advantages.

TPFAP: Be prepared for your opponents to value survival more than chips, whether or not this is correct for them.

[cardcounter0]

a bet of about 62% of your bankroll gives you the best long run results.

[Hawklet]

[ QUOTE ]
a bet of about 62% of your bankroll gives you the best long run results.

[/ QUOTE ]

No, with your logic (bet your chance to win), when your at 50/50, its the best to bet half your bankroll, which is wrong. And when your a 20/80 dog you should bet 20% of your bankroll, which is stupid.

Stop just making up numbers and think before you answer.

[cardcounter0]

[ QUOTE ]
With a 10 its like 37/63 in your favor?

[/ QUOTE ]

hahahahaha, shut up.

[PhantomeX]

You bet a percentage of your bankroll based on your odds. EG
first card 8, there are 27 cards that will lose if you bet lower, and 27 cards that will lose if you bet higher. This means that either way you are 24 to 27 against in other words bet 0 (higher or lower doesn't matter in this situation).
Second card is a Jack, there are now 15 cards that lose if you bet lower(35 that win), and 38 that lose if you bet higher (12 that win), you are now 70% favorite to win if you bet lower so you should bet an ammount based on 70 percent of your bankroll times your risk tolerance, if your hypothetical player is neither a risk taker nor a risk avoider this value coule be anywhere between .5 and 1.0, if you wanted you could add in a bell curve, for example

First card Ace, I would bet it all on lower myself the odds of another ace coming up being only 3 in 51 (about 6%) (94% of the time I make 1000 bucks and have the much larger stack to wager with for the rest of the game) (6% of the time I loose 1000 bucks and can't make any more money the rest of the game) If I could play this game 1,000,000 times (putting up my own money each new game) I would be make an average of $880.00 every time an ace came up as the first card. Whereas a standard play of 94% wager when Ace comes up as first card would average 827.20 of course with only being able to play once it doesn't matter too much what the possibilities are. This is probably why you would never see a setup like this in the Pit at any casino.