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Half-right Mike Strikes Again!

Posted By: Taper_Mike
Date: Saturday, 17 August 2013, at 8:14 a.m.

In Response To: short bearoff/calculating (Taper_Mike)

Half-right Mike Strikes Again!

After I posted my original solution, I read through the other posts. I was feeling pretty good until I got to Dmitriy’s post. Reading it, I learned that 31 is a trouble number I overlooked in my calculations. Whoops!

Here is the do-over.

Trading 22 To Protect against 31

The two best plays are 6/off 2/1 and 6/off 3/2. It turns out there is almost no difference between the two.
• After moving 3/2, no matter what White rolls, Blue will win outright when he rolls 22 on his next turn.

After moving 2/1, Blue can still win when he rolls 22, but only when White rolls a pair of non-doublets.

• The downside to moving 3/2 comes on a roll of 31. After moving 3/2, that throw rips a checker off the 4pt, but leaves two more on the 2pt. If Blue rolls a non-doublet ace on the following turn, he will lose the game.

Playing 6/off 3/1 on this turn gives better protection against a roll of 31, but sacrifices the advantage described above when the roll is 22.

• All other sequences transpose. In particular, when Blue rolls 11 or 21 on his next turn, it does not matter whether he plays 3/2 or 2/1 now.

After Blue Plays 6/off, 3/2

Winning on the First Turn

Five doublets, 22, 33, 44, 55, and 66, give Blue an immediate victory. Blue gets these wins regardless of what White has rolled.

P( win on 1st turn ) = 5/36

Easy Wins on the Second Turn

By the time Blue gets his second turn—if he gets a second turn—White will have completed two turns himself. Only when White has failed to roll doublets will Blue get a second turn. By the same token, Blue won’t get a second turn if he wins by rolling doublets on his first turn.

Blue has a several trouble numbers on his first turn. They are 11, 21, and 31. So long as he does not roll one of them on his first turn, he will win whenever he gets a second turn.

P( win on 1st turn ) = 5/36
P( trouble on 1st turn ) = 5/36
P( no trouble & no win ) = 26/36

P( easy win on 2nd turn )
= (5/6)*(26/36)*(5/6)
= 650/1296

Winning after Trouble

When Blue rolls 11 on his first turn, he will be left with a single checker on the 4pt on his second turn. If he can avoid a roll of 21, he will win. That gives him 34 winning numbers.

When Blue rolls 21 on his first turn, he will leave checkers on his 1pt and 4pt. Subsequent rolls of 11, 21, 31, and 32 will cause him to lose. Otherwise, he wins when he rolls any of the other 29 numbers.

When Blue rolls 31 on his first turn, he will leave two checkers on his 2pt. Any non-doublet ace on the next turn will cost him the game. The other 26 rolls are winners.

As above, we require that White does not roll a doublet on either of his turns.

P( win after trouble 11 )
= (5/6)*(1/36)*(5/6)*(34/36)
= 850/46656

P( win after trouble 21 )
= (5/6)*(2/36)*(5/6)*(29/36)
= 1450/46656

P( win after trouble 31 )
= (5/6)*(2/36)*(5/6)*(26/36)
= 1300/46656

P( win after trouble )
= (850 + 1450 + 1300)/46656
= 3600/46656

Total Chance of Winning

The sum of the probabilities above gives the overall chance of winning.

P( win ) = P( win on 1st turn )
+ P( easy win on 2nd turn )
+ P( win after trouble )

= 5/36 + 650/1296 + 3600/46656
= 33480/46656
= 71.76%

After Blue Plays 6/off, 2/1

With two differences, calculations here are similar.

The first is that a roll of 22 no longer wins on the first turn. It leaves a checker on the 1pt. Blue will still win on the second turn provided White does not roll any doublets. So, the sequence White: non-doublet, Blue: 22, White: non-doublet is a winnner for Blue.

The other difference is a pick-up on 31. With checkers on the 1pt and 3pt, Blue rips two off when he rolls 31 on his first turn. Because his remaining checker will be on the 4pt, that still leaves trouble, but only when Blue rolls 21 on the following turn.

The other trouble numbers, 11 and 21, lead to identical positions—and probabilities—as they did above.

P( win on 1st turn ) = 4/36
P( trouble on 1st turn ) = 5/36
P( no trouble & no win ) = 27/36

P( easy win on 2nd turn )
= (5/6)*(27/36)*(5/6)
= 675/1296

P( win after trouble 11 ) = 850/46656

P( win after trouble 21 ) = 1450/46656

P( win after trouble 31 )
= (5/6)*(2/36)*(5/6)*(34/36)
= 1700/46656

P( win after trouble )
= (850 + 1450 + 1700)/46656
= 4000/46656

P( win ) = P( win on 1st turn )
+ P( easy win on 2nd turn )
+ P( win after trouble )

= 4/36 + 675/1296 + 4000/46656
= 33484/46656
= 71.77%

What’s the Difference?

In his original post, Bob asks what the difference is between plays. A simple subtraction gives the answer. It is only 4/46656, a little less than 1 part in 10,000.

P( difference ) = 33484/46656 – 33480/46656
= 4/46656
= 0.0086%

Mike

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