Statistical problem concerning session EV

Pensive Gerbil · #1 07-13-2003, 08:46 AM

I would appreciate help with a statistical problem of significance to some advantage gamblers (especially online bonus hustlers). Suppose Player A has a session bankroll of $100 and Player B has a session bankroll of $200. Both players will bet $5 per hand on the same randomly dealt videopoker game which has a house edge of X (a number between 0 and 100%) and a standard deviation of Y. Both players will bet on exactly 200 hands ($1000 in total wagers), unless they go bust before reaching 200 hands (in which case their gambling session will come to a premature end).

Since Player A's maximum potential loss is $100 less than the maximum potential loss of Player B, it should be clear that Player A will have a higher session EV (i.e., a smaller expected loss) than player B. One might also discern that the magnitide of the difference between the two player's EV's would be positively correlated with both variables "X" and "Y".

How would one calculate the difference between the session EV's of these two players?

-PG

Cyrus · #2 07-14-2003, 02:25 PM

"Suppose Player A has a session bankroll of $100 and Player B has a session bankroll of $200. Both players will bet $5 per hand on the same randomly dealt videopoker game which has a house edge of X (a number between 0 and 100%) and a standard deviation of Y. Both players will bet on exactly 200 hands ($1000 in total wagers), unless they go bust before reaching 200 hands (in which case their gambling session will come to a premature end).

Since Player A's maximum potential loss is $100 less than the maximum potential loss of Player B, it should be clear that Player A will have a higher session EV (i.e., a smaller expected loss) than player B."

No, the EV is the same for both players, in both percentage and monetary terms. If, for example the house edge is X=1.5%, both players' EV for their first 200 hands is -1.5% of their total action; in other words (-1.5%)*($1000)=-$15.

"One might also discern that the magnitude of the difference between the two player's EV's would be positively correlated with both variables "X" and "Y"."

The only difference between the two players' "expectation", is that player A should "expect" to go bust with a higher probbaility than player B, since A has a lower Bankroll than B, and since they are both betting the same flat amount.

Please also note that X (:edge) and Y (:SD) are the same for both players, who are, thus, playing a game with the same Reward-to-Risk ratio.

For more on this important issue, skip the economic journals' lingo and head for Don Schlesinger's "Blackjack Attack". Get the 2nd edition.

Copernicus · #3 07-14-2003, 02:56 PM

Since the mean (EV) of the difference is equal to the difference of the means, calculate the two players EVs and take the difference....with infinite bankrolls the difference is zero.

However the original point is that because of the difference in bankrolls, the minimum EV of player A is
-$100, while the minimum EV of player B is -$200...ie their situations are not symmetrical. If the house edge is so small that there is virtually no chance of ruin of player A, they are close enough to symmetrical and the EV difference is still 0. Once the house edge is large enough that there is a reasonable probability of ruin of A, then I agree, there should be some difference.

However, I don't see why this is an "important issue". Whats important to each is their own EVs and chances of ruin, and unless they have a side wager on the ruin of A, the fact that A has a smaller bankroll is of no concern to B. Its also of no concern to the house, since they don't care whether they bust A or B in particular, just that they earn their EV overall (unless a players bankroll and the bets are so big that even the casino's bankroll is at risk to a bad run...poor Bugsy Segal).

ChipWrecked · #4 07-14-2003, 06:09 PM

If the house edge is so small that there is virtually no chance of ruin of player A

$5 bet, video poker, let's say 9/6 Jacks or Better. A has $100, B has $200. I'm no mathematician, but I've played some vp. The variance is crushing. Both A and B are sure to go bust in short order. I know I'm in over my head here. This is just observation from "real world trials". [img]/forums/images/icons/frown.gif[/img]

Pensive Gerbil · #5 07-15-2003, 04:14 AM

The only difference between the two players' "expectation", is that player A should "expect" to go bust with a higher probbaility than player B

Since Player A's gambling session is more likely to end before he has wagered $1000, the expected sum of his wagers is less than the expected sum of Player B's wagers. Each player's expected loss (EV) for the gambling session can be calculated by multiplying the expected sum of their wagers times the house edge.

I am seeking help with how to calculate the expected sum of wagers.

-PG

Pensive Gerbil · #6 07-15-2003, 04:27 AM

You are correct Chip_rack. Even if the house edge is very small, there will be a some difference in the two player's session EV; the greater the variance, the greater the difference in EV. If the game has a player edge (over 100% return), however, then the player with a larger session bankroll would have the higher EV.

-PG