Question

A different game is also played with a 10-sided dice. The player initially bets $1 to go in their pot of money. If the player rolls a 5, their pot is multiplied by 5, and if they player doesn’t roll a 5, their pot is multiplied by (1/5). A player plays 12 games.

- Find the mean value of the pot at the end of 12 games
- Find the most probable value of the pot at the end of 12 games.
- How many games would the player have to play until the most probable value of the pot less than 1 cent—i.e. they lose all their money.

Answer #1

- If a player playes a game and the prob. to come 5 here are some cases ........

1st case - die rolled and 5 comes (1st times)

1/10(9\10*9\10*..................*9\10)

when 5 comes..... then the pot is multiplied by 5 and if 5 does't come .....then the pot multiplied by 1\5

5(1\5*1\5*..........*1\5)=5\5^11 =1\5^10

simillarly,(for 2 ,3,4,5,6,7,8,9,10,11,12 times)

1\5^9, 1\5^8 , 1\5^7 , 1\5^6 , 1\5^5 , 1\5^4 , 1\5^3 , 1\5^2 , 1\5 , 5 , 5^2

(a). mean of the pot at the end of 12 games

{1\5^10+1\5^9+1\5^8+1\5^7+1\5^6+1\5^5+1\5^4+1\5^3+1\5^2+1\5+5+25}\12 = $2.59

(b). most probable value of the pot at the end of 12 games

$2.59

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