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Ask the Wizard #296
relaxmax
The answer is 50/50. This will be true for any number of dice rolled, not just two.
A bit offtopic, but I've always thought an odd/even set of bets would be a good way to replace the dreaded big 6/8 bets in craps. To give the house an advantage, here are my proposed pay tables and analysis.
Odd Bet
Event  Pays  Combinations  Probability  Return 

3 or 11  1.5  4  0.111111  0.166667 
5 or 9  1  8  0.222222  0.222222 
7  0.5  6  0.166667  0.083333 
Even  1  18  0.500000  0.500000 
Total  36  1.000000  0.027778 
Even Bet
Event  Pays  Combinations  Probability  Return 

2 or 12  3  2  0.055556  0.166667 
4 or 10  1  6  0.166667  0.166667 
6 or 8  0.5  10  0.277778  0.138889 
Odd  1  18  0.500000  0.500000 
Total  36  1.000000  0.027778 
Please note that I claim all rights with this publication.
This question is raised and discussed in my forum at Wizard of Vegas.
"Anonymous" .
For those unfamiliar with the rules, the player is given four price tags and must place them on four items. When he is done he pulls a lever which gives the number of correct matches. If the player has less than four correct, then he may rearrange the tags and try again. The player may try as many times as he can within 45 seconds.
My advice is to always submit a selection that has a chance of winning, given the previous history of selections and scores. If the first score is 0, then don't reverse two sets of two tags, but instead move everything by one position in either direction.
If you're not able to do the logic on the spot, then I spell it out for you below. To use this strategy, assign the different tags the letters A, B, C, and D. Then place them in the order shown, from left to right on the stage. Always start with ABCD. Then look up the score history below and choose the sequence of tags indicated for that score sequence.
If 0, then BCDA
If 00, then CDAB
If 000, then DABC (must win)
If 01, then BDAC
If 010, then CADB (must win)
If 011, then CDBA
If 0110, then DCAB (must win)
If 02, then BADC
If 020, then DCBA (must win)
If 1, then ACDB
If 10, then BDCA
If 100, then CABD
If 1001, then CBAC (must win)
If 11, then BDCA
If 110, then CABD
If 1101, then CBAC (must win)
If 111, then BCAD (must win)
If 2, then ABDC
If 20, then BACD (must win)
If 21, then ACBD
If 210, then DBCA
If 211, then ADCB
If 2110, then CBAD (must win)
The following table shows the probability of each number of total turns. The bottom right cell shows an expected number of turns of 10/3.
Race Game
Turns  Combinations  Probability  Return 

1  1  0.041667  0.041667 
2  4  0.166667  0.333333 
3  8  0.333333  1.000000 
4  8  0.333333  1.333333 
5  3  0.125000  0.625000 
Total  24  1.000000  3.333333 
This question is discussed on my forum at Wizard of Vegas.
pokerbum
Let's first ask what is the probability that in a fourplayer game all four players have any aceking?
The answer to that question would be (4*4/combin(52,2)) * (3*3/combin(50,2)) * (2*2/combin(48,2)) * (1/combin(46,2)) = 1 in 3,292,354,406.
However, it is possible that some of these ace/king hands will be suited. To be exact, the probability that none of them are suited is 9/24. So lower the probability to 1 in 8,779,611,750.
However, it is a tenplayer game, and any of the combin(10,4)=210 sets of four players could be the four with nonsuited aceking. So, multiply that probability by 210 and the answer is 1 in 41,807,675.
This question is raised and discussed on my forum at Wizard of Vegas.