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## Wednesday, October 31, 2012

### dynamic dice game (Zhou Xinfeng book

P126 [[Zhou Xinfeng]] presents --
Game rule: you toss a fair dice repeatedly until you choose to stop or you lose everything due to a 6. If you get 1/2/3/4/5, then you earn an incremental \$1/\$2/\$3/\$4/\$5. This game has an admission price. How much is a fair price? In other words, how many dollars is the expected take-home earning by end of the game?

Let's denote the amount of money you take home as H. Your net profit/loss would be H minus admission price. If 555 reasonable/intelligent people play this game, then there would be 555 H values. What's the average? That would be the answer.

It's easy to see that if your cumulative earning (denoted h) is \$14 or less, then you should keep tossing.

Exp(H|h=14) is based on 6 equiprobable outcomes. Let's denote Exp(H|h=14) as E14
E14=1/6 \$0 + 1/6(h+1) + 1/6(h+2) + 1/6(h+3) + 1/6(h+4) + 1/6(h+5)=\$85/6= \$14.166

E15=1/6 \$0 + 1/6(h+1) + 1/6(h+2) + 1/6(h+3) + 1/6(h+4) + 1/6(h+5) where h=15, so E15=\$15 so when we have accumulated \$15, we can either stop or roll again.

It's trivial to prove that E16=\$16, E17=\$17 etc because we should definitely leave the game -- we have too much at stake.

How about E13? It's based on 6 equiprobable outcomes.
E13 = 1/6 \$0 +1/6(E14) + 1/6(E15) + 1/6(E16) + 1/6(E17) + 1/6(E18) = \$13.36111
E12 = 1/6 \$0 + 1/6(E13) +1/6(E14) + 1/6(E15) + 1/6(E16) + 1/6(E17) = \$12.58796296
...
E1 =  1/6 \$0 + 1/6(E2) +1/6(E3) + 1/6(E4) + 1/6(E5) + 1/6(E6)

Finally, at start of game, expected end-of-game earning is based on 6 equiprobable outcomes --
E0 =  1/6 \$0 + 1/6(E1) + 1/6(E2) +1/6(E3) + 1/6(E4) + 1/6(E5) = \$6.153737928