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Flipping Even-Odd (Posted on 2011-01-28) Difficulty: 3 of 5
A game is played starting with 6 fair coins laid out with heads face up.

Each round consists of flipping all of the coins showing heads.

If an even number of coins is flipped and an even number of heads comes up the player loses.

If an odd number of coins is flipped and an odd number of heads comes up the player loses.

Rounds continue until the player either loses or reaches zero heads.

The player wins by getting to zero heads without losing.

What is the probability of winning this game?

Examples:
6→3→1 would be a loss.
6→3→0 would be a win.
6→0 would be a loss (even→even rule is applied first.)

No Solution Yet Submitted by Jer    
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Solution solution detailed by computer Comment 3 of 3 |

In the table below, the probabilities are built up as Steve Herman started, recursing to previously calculated values:

 N                                                                                              Total    
 1       1/2 * 1  1/2 * 0                                                                        1/2
 2       1/4 * 0  1/2 * 1/2  1/4 * 0                                                             1/4
 3       1/8 * 1  3/8 * 0  3/8 * 1/4  1/8 * 0                                                    7/32
 4       1/16 * 0  1/4 * 1/2  3/8 * 0  1/4 * 7/32  1/16 * 0                                     23/128
 5       1/32 * 1  5/32 * 0  5/16 * 1/4  5/16 * 0  5/32 * 23/128  1/32 * 0                     563/4096
 6       1/64 * 0  3/32 * 1/2  15/64 * 0  5/16 * 7/32  15/64 * 0  3/32 * 563/4096  1/64 * 0  16793/131072
 


The first product on each row is the probability of getting zero heads times the probability of winning given that one has gotten zero heads. The second product is for getting 1 heads, etc. up to getting all heads.  Getting all heads is an immediate loss as the parity of N matches that of N, as it's the same number.

For example, in the non-parity matches for 6, as at the beginning, the probability of getting 1 heads is 3/32, and in that instance, the conditional probability of eventually winning is 1/2. For getting 3 heads, the probability is 5/16 and its conditional probability is 7/32. For getting 5 heads, the probability is 3/32 and the previously calculated probability is 563/4096. These products add up to 16793/131072, the final answer.

   5   point 10
   9   for Nbr=1 to 6
  10    print Nbr,1-fnPLose(Nbr)
  11   next Nbr
  20    end
  30
  40    fnPLose(N)
  50       local Pl,H,Ph,Totph
  60     Pl=0:if N=0 then goto 150
  70     for H=0 to N
  80       Ph=combi(N,H)//2^N
  85      if N=Nbr then print Ph;"*";
  90       if H @ 2<>N @ 2 then
 100        :Pl=Pl+Ph*fnPLose(H)
 101        :if N=Nbr then print 1-fnPLose(H);:endif
 110       :else
 120        :Pl=Pl+Ph
 121        :if N=Nbr then print 0;:endif
 130       :endif
 140     next H
 150    return(Pl)


  Posted by Charlie on 2011-01-29 15:45:03
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