You get 15 cards. I get the 16th one. Odds that i beat you?

saw this in the british tv drama 'Lucky Man.'

standard 52card deck.
you get the 1st 15 cards off the top after the shuffle/cut.
he gets the 16th.

he was giving 10:1 odds (I think). (his $5k for $50k?)

what are the true odds that he beats you?

When you say "he beats you" do you mean that the his card (the 16th card) is higher than any of your 15 cards?

Need more information.

The way you described it doesn't sound right, unless it was some kind of con or scam and the 16th card is always guaranteed to win...

Does he lose ties?

So if any of the first 15 cards is an Ace, I can’t lose ?

So you get 15 and he gets 1 and he is giving the odds? My non methetmarical brain says that is too good to be true.

I'll assume the guy was able to cut the deck and can pull an Ace.

This is like some John Scarne fanfiction....

If a tie is a redo the odds are 15 to 1.

Quote: 100xOdds

saw this in the british tv drama 'Lucky Man.'

standard 52card deck.
you get the 1st 15 cards off the top after the shuffle/cut.
he gets the 16th.

he was giving 10:1 odds (I think). (his $5k for $50k?)

what are the true odds that he beats you?

Assuming he neither wins nor loses on ties, my answer is that to make the game fair, he should be paid about 27.3 to 1 on his wins.

I find your edge is 54% if the payout is only 10:1.

Quote: ChesterDog

Assuming he neither wins nor loses on ties, my answer is that to make the game fair, he should be paid about 27.3 to 1 on his wins.

I find your edge is 54% if the payout is only 10:1.

Ah, I just realized how I was misthinking this. It is not analogous to a game with 16 people each getting one card, because if you tie it’s a push, but if two other people tie, you lose.

Quote: ChesterDog

Assuming he neither wins nor loses on ties, my answer is that to make the game fair, he should be paid about 27.3 to 1 on his wins.

I find your edge is 54% if the payout is only 10:1.

how did you come up with 27.3 to 1?

I get it that if a tie is a win then 10 for 1 (note "for" not "to") is reasonable as the chances are 9.940% - this seems likely as 15 cards is a key number to get this. (Interestingly it's almost 7 to 1 with 10 cards, 4 to 1 with 5 cards.)

If it is normal that you have to beat the dealer to win then
(i) Chances of winning = 3.304%
(ii) Chances of drawing = 6.636%.
As has been said fair odds are about 29/1 (or slightly lower if ties are a push).

Quote: charliepatrick

I get it that if a tie is a win then 10 for 1 (note "for" not "to") is reasonable as the chances are 9.940% - this seems likely as 15 cards is a key number to get this. (Interestingly it's almost 7 to 1 with 10 cards, 4 to 1 with 5 cards.)

If it is normal that you have to beat the dealer to win then
(i) Chances of winning = 3.304%
(ii) Chances of drawing = 6.636%.
As has been said fair odds are about 29/1 (or slightly lower if ties are a push).

im lost.

first you said 10:1 (9:1?) is reasonable if a tie = win.
then you fair odds is 29:1???

Quote: 100xOdds

how did you come up with 27.3 to 1?

Assuming neither player loses on ties, I get 0.03304 as his probability of winning, 0.06636 as his probability of tying, and 0.90060 as his probability of losing.


To get the above, I considered each card, ace down to two, and found the number of ways for him to win and ways for him to tie. For example, if his card is a ten, there are 3 tens remaining, 16 cards that will beat him, and 32 cards that he will beat. The number of ways he can win would then be combin(32,15).


In the above example, you can tie him by having either 1, 2, or 3 tens with lower cards. So, the number of ways to tie is combin(3,1)*combin(32,14) + combin(3,2)*combin(32,13) + combin(3,3)*combin(32,12).

The number of ways for him to lose is combin(51,15) - number of ways to win - number of ways to tie.


Divide the number of ways by combin(51,15) to convert to probabilities. Do this for his 13 possible cards, add, and divide by 13 to find his overall probabilities of winning, tying, and losing.


His EV is then P(win) * odds - P(lose). For odds of 27.258 to 1, his EV would be near 0.


If he loses on ties, then the odds would have to be 29.27 to 1 for his EV to be near 0, which agrees with Charlie Patrick's answer.

Quote: 100xOdds

im lost.

first you said 10:1 (9:1?) is reasonable if a tie = win.
then you fair odds is 29:1???

If you have to beat the dealer to be paid out and all other outcomes (equal or lower) lose, then fair odds are about 29/1 (you have about 3.304% chance of being paid out).

If you can win or tie the dealer to be paid out (i.e. higher or equal) and only lose if your card is lower, then fair odds are about 9/1; this equates to 10 for 1 (you have about 9.940% of being paid out).

It is because the OP mentioned "10" and that 15 cards was the actual number of cards to get such a close result, I wondered whether this was the rules of the game especially as the House edge is only 0.6%!

It’s the pushes that kill you here. If suits count in ranking like they do in determining the button at the beginning of a hold em poker game (Spades > Hearts > Diamonds > Clubs), then the odds of winning jump to 15:1.