Ask The Wizard #171
Let’s look at an example. Suppose it is your turn with the following scenario.
The following table breaks down each row into its binary components.
Player’s Turn 1
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
9 | 1 | 0 | 0 | 1 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
25 | 1 | 0 | 0 | 1 | 1 |
Total | 3 | 1 | 3 | 2 | 1 |
You can see that there is an odd number of ones, twos, fours, and sixteens. Clearly we need to get the row of 25 under 16 to eliminate the 16 unit. To keep the total of the binary components even we need to remove the 1, add a 2, add a 4, keep the 8, and remove the 16. That means the best play is 2+4+8=14 in the last row. Leaving 14 in the bottom row we have the following.
Computer’s Turn 1
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
9 | 1 | 0 | 0 | 1 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
14 | 0 | 1 | 1 | 1 | 0 |
Total | 2 | 2 | 4 | 2 | 0 |
The computer takes its turn, leaving us with this.
Here is the binary breakdown of that.
Player’s Turn 2
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
9 | 1 | 0 | 0 | 1 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
14 | 0 | 1 | 1 | 1 | 0 |
Total | 2 | 3 | 3 | 2 | 0 |
Here we need to remove a 2 and a 4, to get those totals even. There is only one row, the 14, which has both components. So remove 6 from that, leaving 8.
Computer’s Turn 2
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
9 | 1 | 0 | 0 | 1 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
8 | 0 | 0 | 0 | 1 | 0 |
Total | 2 | 2 | 2 | 2 | 0 |
The computer takes its turn, leaving us with this.
Now we need to change the 1, 4, and 8 columns.
Player’s Turn 3
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
8 | 0 | 0 | 0 | 1 | 0 |
Total | 1 | 2 | 3 | 1 | 0 |
That can be done by changing the row of 8 to 5 as follows.
Computer’s Turn 3
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
Total | 2 | 2 | 4 | 0 | 0 |
The computer takes its turn, leaving us with this.
Now we need to change the 2 and 4 totals.
Player’s Turn 4
Row | 1 | 2 | 4 | 8 | 16 |
6 | 0 | 1 | 1 | 0 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
3 | 1 | 1 | 0 | 0 | 0 |
Total | 2 | 3 | 3 | 0 | 0 |
This can be done by changing the 6 to a 0.
Computer’s Turn 4
Row | 1 | 2 | 4 | 8 | 16 |
0 | 0 | 0 | 0 | 0 | 0 |
4 | 0 | 0 | 1 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
3 | 1 | 1 | 0 | 0 | 0 |
Total | 2 | 2 | 2 | 0 | 0 |
The computer takes its turn, leaving us with this.
Now we need to change the 2s and 4s.
Player’s Turn 5
Row | 1 | 2 | 4 | 8 | 16 |
0 | 0 | 0 | 0 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
5 | 1 | 0 | 1 | 0 | 0 |
3 | 1 | 1 | 0 | 0 | 0 |
Total | 2 | 3 | 1 | 0 | 0 |
This can be accomplished by changing the row of 5 to 3. If you can ever get your opponent to an x,x,y,y situation you can’t help but win, if you can maintain the same situation until the end.
Computer’s Turn 5
Row | 1 | 2 | 4 | 8 | 16 |
0 | 0 | 0 | 0 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
2 | 0 | 1 | 0 | 0 | 0 |
3 | 1 | 1 | 0 | 0 | 0 |
3 | 1 | 1 | 0 | 0 | 0 |
Total | 2 | 4 | 0 | 0 | 0 |
The next few moves I keep the computer on x,x,y,y patterns. Here the computer leaves me with 2,2,3,2; so I leave it with 2,2,2,2.
The computer then gives me 2,2,1,2. I leave it with 2,2,1,1.
The computer then leaves me with 2,2,1. I leave it with 2,2. If you can ever get your opponent to two equal rows you can’t help but win, just keep the rows equal.
The computer then leaves me with a single pile of 2, and I remove 1.
Here is the end of the game.
I recently witnessed a situation at the local tribal casino involving protocol at the craps table that puzzled me, and I'm wondering if you clarify it for me. On his come out roll, Player ’A’ threw a number and established a point. For the next roll, the player next to ’A’ (Player ’B’) picked up the dice and shot. It’s not clear why he did this, or if he even knew ’A’. One of the other players objected, pointing out that ’B’ was not the original shooter. After much discussion and head scratching by the dealers and the boxman, the dice were passed to the next player (Player ’C’) who finished the hand (he eventually sevened out).
Was this the correct protocol for this situation, and if so, what is the logic behind it? If Player ’A’ for some reason simply didn't want to shoot any more, why shouldn’t he be allowed to relinquish the dice? If Player ’B’ picked up the dice because he didn't understand the game, or if the stickman mistakenly put them in front of him, shouldn't the dice go back to ’A’ to finish the hand?
I asked the Bone Man at nextshooter.com this one. Here is what he said.
If a player other than the correct shooter picks up the dice and rolls them, it should be a no-call, no-roll and the dice should be returned to the correct shooter. Though this is indeed the proper ruling, the boxman in some instances may allow for the roll if the result is to the favor of all or most of the players. In some instances, the result of the roll may not effect any of the player(s) wagers. Also... Any player can request that the dice be sent to the next shooter to finish a hand. In such cases the same dice may be sent out or the new shooter can request new dice. Upon the completion of the roll, the SAME shooter can then shoot HIS/HER OWN HAND, thereby having more than one hand.
log(b) = 5.520696
b = 105.520696
b = $331,662.50.
So this hypothetical player should be indifferent at a bank offer of $331,662.50. The lesser your wealth going into the game the more conservatively you should play. Usually in the late stages of the game the bank offers are close to expected value, sometimes a little more bit more. The only rational case where a player could win the million is if he had a lot of wealth going into the game and/or the bank offers were unusually stingy. The producers seem to like hard-working middle class people, so we’re unlikely to see somebody who can afford to be cavalier when large amounts are involved. I have also never seen the bank make offers under 90% of expected value late in the game. The time when we will see somebody win the million is when a degenerate gambler gets on the show who can’t stop. When that happens I will be rooting for the banker.
I agree that a "2/4" game means the bets are in $2 units before the turn and $4 after the turn. However, the small blind is likely be $1. It sounds like when Arizona Charlie's is calling a game "2/4" they are referring to the blinds, which would mean a small blind of $2, and a big blind of $4, which would imply $8 bets after the turn. I've never heard this usage before so I don’t blame you for being surprised.
p.s. I later received the following from Anthony, a poker room supervisor.
I am a poker room supervisor. I was just writing to let you know about the correct wording for the poker limits in Texas Hold 'em. If as game is referred to as $2/4. It is referring to the bets. (A game is only referred to by the blinds if it is No-Limit.) $2 pre flop and post flop,$4 on the turn and the river.A game that is listed as $2/$4/$8 (which is exactly what I suspect the previous letter writer was actually referring to, is $2 pre-flop,$4 post-flop,$8 turn and river .It is also possible to have a game listed as $2/$4/$6/$8. Just passing the information on, good poker supervisors want people to have as much information as possible.