Please help me with
Player A and Player B roll a dice. They roll 5 times each. Who dices "5" first will win the game? If player A plays first, what is prob of Player A winning the game. Is it unfair?
If A plays first and the one who rolls a 5 first wins: If A wins, B gets fewer rolls than A. If B wins, A gets the same number of rolls as B. So the game is unfair (in favor of A) and [prob A winning]>[prob B winning]. There is also the possibility that there is no winner within these 5 rounds of rolls.
- roll 1: win A= 1/6 win B= (5/6)*(1/6) no win= 1-(11/36)
- roll 2: win A= (1/6)*(25/36) win B= (5/6)*(1/6)*(25/36) no win= 1-(275/1296)
- roll 3: win A= (1/6)*(1021/1296) win B= (5/6)*(1/6)*(1021/1296) no win= 1-(11231/46656)
- roll 4: win A= (1/6)*(35425/46656) win B= (5/6)*(1/6)*(35425/46656) no win= 1-(389675/1679616)
- roll 5: win A= (1/6)*(1289941/1679616) win B= (5/6)*(1/6)*(1289941/1679616) no win= 1-(14189351/60466176)
Totals:
- win A= (40406838/60466176)/5= 0.133651044841995630747345424986028552558045013463 3948
- win B= (33672365/60466176)/5= 0.111375870701663025622787854155023793798370844552 8290
- no win= (168361825/60466176)/5= 0.754973084456341343629866720858947653643584141983 7761
Reply #2 many well be correct. But I cannot read it. That is clearly evidence that serious helpers should know LaTeX.
Let $\displaystyle X$ be the number of rolls on which the game is over.
Note that $\displaystyle X=1,2,\cdots,10$.
A wins if $\displaystyle X=1,3,5,7,9$ and B wins if $\displaystyle X=2,4,6,8,10$
So $\displaystyle \mathcal{P}(X=k)=\left( {\frac{1}{6}} \right)\left( {\frac{5}{6}} \right)^{ k-1} $.
Indeed you left out the calculations.
I am referring to the fact that we could have skipped this meaningless exchange of replies if you had just pointed out the policy violation from the beginning instead of arguing LaTeX.
In reply #3 you made a point about text readability, not solution-provision policy. And thank you for deciding on whether I know LaTeX or not from my first ever reply on this forum.
This is very similar to this Probability Puzzles: A Coin Tossing Game: Optimal Strategy puzzle. You can follow the same approach. Yes the game is biased towards the person playing first