Reprinted from the September 96 issue of the JG Newsletter.
----------------------------- |= ' = ' =| | G " " - | | L I S ' ' - | |' U W ' - '| | T A - - | | " D I V E R C " " | |W A L F I D O ' | |A M I ' F E Z ' =| |R A N ' Y ' | |P I T " " " | | M Y - - | |' - ' - '| | - ' ' - | | - " " - | |= ' = ' =| -----------------------------
The following position and analysis is based on the Game of the Month in Scrabble Player News #125. The game is between Joel Wapnick and Jan Dixon from the 1995 Superstars of Scrabble tournament. The annotations are by Joe Edley. In the position on the left, Joel Wapnick made a fine play against Jan Dixon. With no possible bingos playable, he played the phony MYIASISE(S) 11B, 76. In his annotation to the game, Joe Edley writes, "A hard-to-challenge phoney. Dixon knows MYIASIS is good, but is the plural MYIASES or the play made? Unless she’s 90-100% sure, she shouldn’t challenge."
Firstly, I postulate that a player who knows MYIASIS probably knows MYIASES as well and the true question is whether there is an alternative plural. Secondly, I take exception to the 90-100% number given. If someone plays what may be a phony bingo against you, it certainly seems like the lack-of-faith level necessary to challenge is less than 90%. But rather than instinctively trot out a number of my own, I will try to demonstrate this rigorously.
The simplest case to analyze is if Jan does not challenge (Case NC). Her best play (with DEGNORR) after "MYIASISE(S)" is WRONG 4d for 11. WRONG’s simulated value is 23 points. The twelve points difference between score and value are accounted for by leave and board considerations. Here, the 12 points come from the DER leave and the right to first crack at this board is worth zero. The reason is that there are no great openings and any play chosen by the next player might even offer the opponent the better board the following play. Joel will be drawing to an empty rack in an about neutral bag, so I assign his residual as zero. Her Expected Value (EV) after a pair of turns is (-72) + (+23) = -49 . NC = -49
Now if Jan does challenge, there are two possibilities to analyze: the word is Acceptable (CA) or the word is Unacceptable (CU).
If the word is Acceptable (Case CA), her score for the play will be zero, but what will her value for the play be? We’ve assigned the board consideration as zero (nice how that’s simplified the analysis) so the only other factor is her leave. Since her best play before was only an eleven pointer (with value of 23, as opposed to typical one-turn progress being about 35 for best play -- though this board admittedly brings down that average), it’s not surprising that her overall rack leave is below average. The linear value of the letters is about -4 plus there is a slight unworkability of the rack due to consonant heaviness and the lack of thru-vowels on the board. Nack and Brian Sheppard have, I believe, stated that the ideal rack is +2 on consonants and subtract 2 points for every unit of deviation from this. There is some synergy, so assign a total value of -5 for the leave. As before, we have Joel drawing to an empty rack yielding a residual rack value of zero. Therefore, Jan’s EV for the challenge-Acceptable case is : (-72) + (-5) = -77. Note that the amount of one-turn progress forwent is only 28 points rather than the 35 I defined as the average above. This demonstrates the tight board dampening progress per turn. CA = -77
The trickiest case to analyze is the Challenge-Unacceptable case (Case CU). If we follow the structure of the analysis above, we compute Joel’s value as zero points + rack leave (once again, assuming zero residual board value). Linear rack leave would be computed: A=+1, E=+4, I=-1, 2nd I=-9, S=+7, 2nd S=0 and ?=+24 for a grand total of 26. However, some alterations are in order. Firstly, some of the S’s 7 points of value come from its ability to increase bingo chances which is made redundant by the presence of the blank. Off the cuff, I’ll discount it by 1 point. Also, there are just no hot or even lukewarm S spots on this board, so I’m going to discount it another 2 points. The above stated vowel-consonant ratio formula would have us discount the leave by 8 points ((+2C) - (-2C) * 2pts). However, the blank has the ability often to take a crummy vowel-heavy rack and turn it into a bingo (certainly part of the blank’s 24 value, but I give it a premium here for helping out in this way) so I’ll discount the -8 to -5. This gives AEIISS? a modified value of +18. Given that we have assessed Jan’s one-turn progress as +28 we could now compute the EV as -(0+18) + 28 = +10 . These numbers are not the final word, though. If Jan went strictly by this, challenged MYIASISE(S) and played WRONG, she would have allowed Joel to play (L)IAISES 3f for 65, a 17-point premium on average turn plus his residual! We have to analyze another pair of moves deep to come closer to the correct answer.
The highest value play for Jan that doesn’t allow Joel to bingo would instead be DENY 9f for 21 (which otherwise simulated 3 points behind WRONG). After DENY, Joel’s best play would be MYIASIS for 24 keeping E? for a total adjusted value of 52. Finally, how to compute Jan’s next turn? This is difficult to do with a computer so I’ll wing it (yes, I know that was what we are trying to demonstrate the foolishness of, but it’s a lot safer to do on the minutiae than the big picture). With MYIASIS on the board, average progress has probably perked up a little, so I’ll assign it 31. The GORR leave certainly is below average, -7 points. Also to be considered is that perhaps knowing Joel has E? will help her make better decisions next turn (as it did this turn -- not allowing (L)IAISES was worth 13 pts [65-52] ). It won’t be worth 13 points because she doesn’t have the total info, but giving her 5 points credit for this seems reasonable. That means Jan’s average value for next turn should be about 29. Now if we recompute the EV over 2 pair of moves for CU we get: -(0) + (21) -(+52) + (29) = -2, kind of a bummer that it’s still negative given that she even got a free turn, but that’s the way it goes. CU = -2
Now we can compute the value of challenging. The challenge should be made if the EV of challenging is greater than the EV of not challenging, that is C > NC, where C = the EV of challenging and NC = the EV of not challenging. C can be computed by C= (P)*(CA) + (1-P)*(CU), where P = your assessed probability of the word being Acceptable, CA = the EV computed for the word being Acceptable and CN = the EV of the word being Unacceptable. This can be solved for P. Remembering to flip the > sign because [CA-CU] is always negative, you should challenge if P < [NC - CU] / [CA - CU]. Although you may dispute some of the numbers and premises in the analysis above, THIS FORMULA IS TRUE and can be used over the board with your own numbers inside. Here we get P < 47/75 = 63%. So Jan should challenge here if she assesses the probability of the word being good at 63% chance or less, a far cry from the 0-10% suggested.
Last Modified 12/27/00
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