Elo probability for a single game

[CRoberson]

This came up elsewhere but it seemed like a good topic for this forum.

The thought is that there is a difference between the probability
of winning a single game and the probability of winning a match.
The reason for this difference is the chance of draws.

In a match between A and B and A has an Elo of 202 pts greater
than B, then A has an Elo probability of 76%. This means that A should
pick up 76 points in a 100 game match. Thus, B should gain 24 points.

Due to the possibility of draws, the outcome could be A wins
52 games and draws 48 games. This means the probability of
B not losing to A in any given game is 48%. This makes the probability
of A winning any given game 52%.

That was the maximum outcome for B. The minimum outcome for
B is it loses 76 games and wins 24. Thus, A's maximum performance
is 76% probability of winning any given game.

If the two are averaged, we get a realistic probability for A winning
any given game (that is wins only not draws).
So, (Max(P(A win)) + Min(P(A win)))/2 = (76+52)/2 = 64% chance
of A winning any given game. This means P(B not losing (win or draw))
is 1-P(A win) = 36%. Thus, the probability of A winning a single game
is 64% not 76%.

So, for a match where A's Elo is 102 above B's the match probability
is 64% for A and the P(A winning a specific game) is (64+28)/2 = 46%.
Therefore, P(B not losing (win or draw)) = 1 - P(A wins the game) = 54%.
Thus, in a single game match a higher rating of 100 pts is insufficient
to expect a win and is only 46% instead of 64%.

[hgm]

The usual rating model describes win, draw and loss probabiliy separately as a function of the rating difference. See for instance the BayesElo website.

[bob]

This came up elsewhere but it seemed like a good topic for this forum.

The thought is that there is a difference between the probability
of winning a single game and the probability of winning a match.
The reason for this difference is the chance of draws.

In a match between A and B and A has an Elo of 202 pts greater
than B, then A has an Elo probability of 76%. This means that A should
pick up 76 points in a 100 game match. Thus, B should gain 24 points.

Due to the possibility of draws, the outcome could be A wins
52 games and draws 48 games. This means the probability of
B not losing to A in any given game is 48%. This makes the probability
of A winning any given game 52%.

That was the maximum outcome for B. The minimum outcome for
B is it loses 76 games and wins 24. Thus, A's maximum performance
is 76% probability of winning any given game.

If the two are averaged, we get a realistic probability for A winning
any given game (that is wins only not draws).
So, (Max(P(A win)) + Min(P(A win)))/2 = (76+52)/2 = 64% chance
of A winning any given game. This means P(B not losing (win or draw))
is 1-P(A win) = 36%. Thus, the probability of A winning a single game
is 64% not 76%.

So, for a match where A's Elo is 102 above B's the match probability
is 64% for A and the P(A winning a specific game) is (64+28)/2 = 46%.
Therefore, P(B not losing (win or draw)) = 1 - P(A wins the game) = 54%.
Thus, in a single game match a higher rating of 100 pts is insufficient
to expect a win and is only 46% instead of 64%.

There's another way to think of this. You have $10,000 and you want to walk into a casino and double your money. You are not an advantage player and don't count cards or shuffle track or hole card, so blackjack isn't a winning game. What is the best way to double your money:

(a) go in and play blackjack (-.26% house advantage for $10 a hand ) until you double or lose all your money;

(b) play black or red on the roulette wheel at $10 a bet (-5.26% house advantage) until you double or go broke;

(c) go in and play one hand of blackjack with a $10,000 bet.

(d) go in and play one roulette bet of $10,000.

The correct answer is (d).

If you try A, you are going to lose .26% of each $10 bet you make over time. You will go bankrupt with a high degree of certainty. How long it takes will vary, but you end up at zero most of the time.

If you try B, you lose even faster as in A. Except the house edge is now -5.26 rather than -.26. No good.

C is no good as you expect to win 42% of the hands and push/lose the rest. So you have a 42% chance of doubling.

D is the best option. You have a 18/38 chance of doubling, and 20/38 chance of losing it all. Note that there is a chance that depending on the odds given on craps, there can be a single bet that is a perfect 50-50, but not all casinos offer such high odds bets.

So one shot is your best chance, otherwise the statistics are going to catch up to you. If I am rated higher than my opponent, give me a match. If I am rated lower, give me one shot.

That statistics is on the side of the higher probability, the other side needs a single winner-take-all to keep from getting slowly ground down... In chess, your rating is what it is, so you can't shop around to find the best odds. But still, one winner-takes-all game gives you better odds than a match.

taken another way, if you are rated 200 points above me, you should win 3 of every 4 points played. What is the probability I can win such a 4 game match? Compared to the probability I just win the first game? On a single game, you can end up anywhere in the sample space. With a lot of samples, you are going to end up near the mean which is losing for me every time.