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.

1. Find the mean value of the pot at the end of 12 games
2. Find the most probable value of the pot at the end of 12 games.
3. 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

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