Calculating chance of 20% gain

Quote: virginiabreeze

If someone has access to a game simulator, I would greatly appreciate learning the results of the following.
Player vs House at a game where House has 1% House Edge for each game played (for comparison, I am using the term House Edge as in the statement that when betting the Pass Line in craps, the House Edge is 1.4%)
Player starts with $250 and plays $10 each game. A win earns $10 and a loss losses $10.
Player continues to play until one of the following occurs, he/she wins $50 (20% gain) or loses all $250 (100% loss) dollars. Either endpoint is considered a completed "set" of games.
If 1000 such sets are played:
How many sets does the player win 20% and what is the average number of games per winning set
How many sets does the player lose 100% and what is the average number of games per losing set

Then repeat simulator with a cutoff when player wins $70 (28%) as opposed to winning $50 (20%)

If there is a website or downloadable program where I can do such a calculation myself, please let me know and I will share the results.

Thank you.

Quote: virginiabreeze

If someone has access to a game simulator, I would greatly appreciate learning the results of the following.

Quote: virginiabreeze

Player vs House at a game where House has 1% House Edge for each game played (for comparison, I am using the term House Edge as in the statement that when betting the Pass Line in craps, the House Edge is 1.4%)

Quote: virginiabreeze

Player starts with $250 and plays $10 each game. A win earns $10 and a loss losses $10.

Quote: virginiabreeze

Player continues to play until one of the following occurs, he/she wins $50 (20% gain) or loses all $250 (100% loss) dollars. Either endpoint is considered a completed "set" of games.

Quote: virginiabreeze

If 1000 such sets are played:
How many sets does the player win 20% and what is the average number of games per winning set

Quote: virginiabreeze

How many sets does the player lose 100% and what is the average number of games per losing set

Quote: virginiabreeze

Then repeat simulator with a cutoff when player wins $70 (28%) as opposed to winning $50 (20%)

Quote: virginiabreeze

If there is a website or downloadable program where I can do such a calculation myself, please let me know and I will share the results.
Thank you.

=IF(p=q,b_x/bankAndgoal, ((q/p)^b_x - 1) / ((q/p)^bankAndgoal - 1))

=(L+W) / (q-p) * ((q/p)^(L+W)+ 1) / ((q/p)^(L+W) -1)  -  L/(q-p) * (((q/p))^L +1) / (((q/p))^L -1)

=(L+W) / (q-p) * ((q/p)^(L+W)+ 1) / ((q/p)^(L+W) -1)  -  W/(q-p) * (((q/p))^W +1) / (((q/p))^W -1)

Quote: ThatDonGuy

In your case, p = 0.49, q = 0.51, S = 5, and F = 25.

<snip>
Plugging in the numbers, the result is about 78.9%.

Quote: virginiabreeze

When I take your Excel sheet and change bankroll to $2000, the numbers come out 90.3% of the time achieving the goal of making $50 and 9.7% of the time losing $2000. Wow, just intuitively one would not expect that about 10% of the time you will lose 200 units before you find yourself a mere 5 units ahead of where you started.
Amazing, but not what I would have expected.

Quote: virginiabreeze

I assume that the term "Gambler's Ruin" implies that there is no reasonable bankroll size where the results favor the Gambler. For example, even if you started with a bankroll of $2000, the Gambler would still lose more than he wins?

Quote: mustangsally

think you used 0.495 in your formula

Quote: virginiabreeze

I assume that the term "Gambler's Ruin" implies that there is no reasonable bankroll size where the results favor the Gambler. For example, even if you started with a bankroll of $2000, the Gambler would still lose more than he wins?

Quote: virginiabreeze

When I take your Excel sheet and change bankroll to $2000, the numbers come out 90.3% of the time achieving the goal of making $50 and 9.7% of the time losing $2000.
Wow, just intuitively one would not expect that about 10% of the time you will lose 200 units before you find yourself a mere 5 units ahead of where you started.
Amazing, but not what I would have expected.

> # Gambler's Ruin Problem - mean time in transient states 
> # and probabilities to target goal and ruin
> p <- 0.495 ##Enter
> states <- 31 ##Enter bankroll + wingoal + 1
> 
> q <- 1-p
> target <- states - 1
> state2state <- states - 2
> 
> # enter transition probability matrix
> P <- matrix(rep(0,states^2), nrow=states, ncol=states)
> P[target,target] <- 1 # absorbing state target goal
> P[states,states] <- 1 # absorbing state ruin
> P[1,states] <- q # to ruin - not in diag
> 
> for (i in 1:state2state) {P[i,i-1] <- q
+ P[i,i+1] <- p}
> #P
> 
> # take sub-matrix for transient states to states
> P.s <- P[1:state2state, 1:state2state]
> #P.s
> # take sub-matrix for transient states to absorbing
> P.t <- P[1:state2state, target:states]
> #P.t
> 
> # compute mean number of revisits matrix S
> Q <- solve(diag(state2state)-P.s)
> #Q
> 
> # sum entries in Xth row of Q
> sum(Q[25,])
[1] 132.7509
> 
> Q %*% P.t
      [win goal]   [ruin]
 [1,] 0.02457203 0.97542797
 [2,] 0.04964046 0.95035954
 [3,] 0.07521532 0.92478468
 [4,] 0.10130685 0.89869315
 [5,] 0.12792548 0.87207452
 [6,] 0.15508186 0.84491814
 [7,] 0.18278685 0.81721315
 [8,] 0.21105154 0.78894846
 [9,] 0.23988723 0.76011277
[10,] 0.26930547 0.73069453
[11,] 0.29931801 0.70068199
[12,] 0.32993686 0.67006314
[13,] 0.36117428 0.63882572
[14,] 0.39304275 0.60695725
[15,] 0.42555504 0.57444496
[16,] 0.45872414 0.54127586
[17,] 0.49256332 0.50743668
[18,] 0.52708612 0.47291388
[19,] 0.56230635 0.43769365
[20,] 0.59823810 0.40176190
[21,] 0.63489574 0.36510426
[22,] 0.67229394 0.32770606
[23,] 0.71044767 0.28955233
[24,] 0.74937217 0.25062783
[25,] 0.78908303 0.21091697
[26,] 0.82959613 0.17040387
[27,] 0.87092767 0.12907233
[28,] 0.91309419 0.08690581
[29,] 0.95611257 0.04388743

> # Gambler's Ruin Problem - mean time in transient states 
> # and probabilities to target goal and ruin
> p <- 1 - 0.495 ##Enter
> states <- 31 ##Enter bankroll + wingoal + 1
> 
> q <- 1-p
> target <- states - 1
> state2state <- states - 2
> 
> # enter transition probability matrix
> P <- matrix(rep(0,states^2), nrow=states, ncol=states)
> P[target,target] <- 1 # absorbing state target goal
> P[states,states] <- 1 # absorbing state ruin
> P[1,states] <- q # to ruin - not in diag
> 
> for (i in 1:state2state) {P[i,i-1] <- q
+ P[i,i+1] <- p}
> #P
> 
> # take sub-matrix for transient states to states
> P.s <- P[1:state2state, 1:state2state]
> #P.s
> # take sub-matrix for transient states to absorbing
> P.t <- P[1:state2state, target:states]
> #P.t
> 
> # compute mean number of revisits matrix S
> Q <- solve(diag(state2state)-P.s)
> #Q
> 
> # sum entries in Xth row of Q
> sum(Q[25,])
[1] 116.2236
> 
> Q %*% P.t
      [win goal]   [ruin]
 [1,] 0.04388743 0.95611257
 [2,] 0.08690581 0.91309419
 [3,] 0.12907233 0.87092767
 [4,] 0.17040387 0.82959613
 [5,] 0.21091697 0.78908303
 [6,] 0.25062783 0.74937217
 [7,] 0.28955233 0.71044767
 [8,] 0.32770606 0.67229394
 [9,] 0.36510426 0.63489574
[10,] 0.40176190 0.59823810
[11,] 0.43769365 0.56230635
[12,] 0.47291388 0.52708612
[13,] 0.50743668 0.49256332
[14,] 0.54127586 0.45872414
[15,] 0.57444496 0.42555504
[16,] 0.60695725 0.39304275
[17,] 0.63882572 0.36117428
[18,] 0.67006314 0.32993686
[19,] 0.70068199 0.29931801
[20,] 0.73069453 0.26930547
[21,] 0.76011277 0.23988723
[22,] 0.78894846 0.21105154
[23,] 0.81721315 0.18278685
[24,] 0.84491814 0.15508186
[25,] 0.87207452 0.12792548
[26,] 0.89869315 0.10130685
[27,] 0.92478468 0.07521532
[28,] 0.95035954 0.04964046
[29,] 0.97542797 0.02457203
> 

Quote: ThatDonGuy

Still trying to find that "expected length of winning and losing runs" formula.